STT 455-6: Actuarial ModelsSTT 455-6: Actuarial Models Albert Cohen Actuarial Sciences Program Department of Mathematics Department of Statistics and Probability C336 Wells Hall Michigan

STT 455-6: Actuarial Models

Albert Cohen

Actuarial Sciences ProgramDepartment of Mathematics

Department of Statistics and ProbabilityC336 Wells Hall

Michigan State UniversityEast Lansing MI

[email protected]

[email protected]

Albert Cohen (MSU) STT 455-6: Actuarial Models MSU 2020 1 / 348

Copyright Acknowledgment

Many examples and theorem proofs in these slides, and on in class exampreparation slides, are taken from our textbook ”Actuarial Mathematics forLife Contingent Risks” by Dickson, Hardy, and Waters.

Please note that Cambridge owns the copyright for that material.No portion of the Cambridge textbook material may be reproducedin any part or by any means without the permission of thepublisher. We are very thankful to the publisher for allowing postingof these notes on our class website.

Also, we will from time-to-time look at problems from releasedprevious Exams MLC by the SOA. All such questions belong incopyright to the Society of Actuaries, and we make no claim onthem. It is of course an honor to be able to present analysis of suchexamples here. Copyright (2007-present) The Society of Actuaries,Schaumburg, Illinois. Reproduced with permission.


Survival Models

An insurance policy is a contract where the policyholder pays apremium to the insurer in return for a benefit or payment later.

The contract specifies what event the payment is contingent on. Thisevent may be random in nature

Assume that interest rates are deterministic, for now

Consider the case where an insurance company provides a benefitupon death of the policyholder. This time is unknown, and so theissuer requires, at least, a model of of human mortality


Survival Models

Define (x) as a human at age x . Also, define that person’s future lifetimeas the continuous random variable Tx . This means that x + Tx representsthat person’s age at death.

Define the lifetime distribution

Fx(t) = P[Tx ≤ t] (1)

the probability that (x) does not survive beyond age x + t years, and it’scomplement, the survival function Sx(t) = 1− Fx(t).


Conditional Equivalence

We have an important conditional relationship

P[Tx ≤ t] = P[T0 ≤ x + t | T0 > x ]

=P[x < T0 ≤ x + t]

P[T0 > x ]

(2)

and so

Fx(t) =F0(x + t)− F0(x)

1− F0(x)

Sx(t) =S0(x + t)

S0(x)

(3)

In general we can extend this to

Sx(t + u) = Sx(t)Sx+t(u) (4)




P[Tx ≤ t] = P[T0 ≤ x + t | T0 > x ]

=P[x < T0 ≤ x + t]

P[T0 > x ]

(2)

and so

Fx(t) =F0(x + t)− F0(x)

1− F0(x)

Sx(t) =S0(x + t)

S0(x)

(3)


Sx(t + u) = Sx(t)Sx+t(u) (4)




P[Tx ≤ t] = P[T0 ≤ x + t | T0 > x ]

=P[x < T0 ≤ x + t]

P[T0 > x ]

(2)

and so

Fx(t) =F0(x + t)− F0(x)

1− F0(x)

Sx(t) =S0(x + t)

S0(x)

(3)


Sx(t + u) = Sx(t)Sx+t(u) (4)


Conditions and Assumptions

Conditions on Sx(t)

Sx(0) = 1

limt→∞ Sx(t) = 0 for all x ≥ 0

Sx(t1) ≥ Sx(t2) for all t1 ≤ t2 and x ≥ 0

Assumptions on Sx(t)ddt Sx(t) exists ∀t ∈ R+

limt→∞ t · Sx(t) = 0 for all x ≥ 0

limt→∞ t2 · Sx(t) = 0 for all x ≥ 0

The last two conditions ensure that E[Tx ] and E[T 2x ] exist, respectively.


Example 2.1

Assume that F0(t) = 1−(1− t

120

) 16 for 0 ≤ t ≤ 120. Calculate the

probability that

(0) survives beyond age 30

(30) dies before age 50


P[(0) survives beyond age 30] = S0(30) = 1− F0(30)

=

(1− 30

120

) 16

= 0.9532

P[(30) dies before age 50] = F30(20)

=F0(50)− F0(30)

1− F0(30)= 0.0410

P[(40) survives beyond age 65] = S40(25) =S0(65)

S0(40)= 0.9395

(5)


Example 2.1


120

) 16 for 0 ≤ t ≤ 120. Calculate the

probability that


(30) dies before age 50


P[(0) survives beyond age 30] = S0(30) = 1− F0(30)

=

(1− 30

120

) 16

= 0.9532

P[(30) dies before age 50] = F30(20)

=F0(50)− F0(30)

1− F0(30)= 0.0410

P[(40) survives beyond age 65] = S40(25) =S0(65)

S0(40)= 0.9395

(5)


The Force of Mortality

Recall from basic probability that the density of Fx(t) is defined asfx(t) := d

dt Fx(t).

It follows that

f0(x) :=d

dxF0(x) = lim

dx→0+

F0(x + dx)− F0(x)

dx

= limdx→0+

P[x < T0 ≤ x + dx ]

dx

(6)



Recall from basic probability that the density of Fx(t) is defined asfx(t) := d

dt Fx(t).

It follows that

f0(x) :=d

dxF0(x) = lim

dx→0+

F0(x + dx)− F0(x)

dx

= limdx→0+

P[x < T0 ≤ x + dx ]

dx

(6)



However, we can find the conditional density, also known as the Force ofMortality via

µx = limdx→0+

P[x < T0 ≤ x + dx | T0 > x ]

dx

= limdx→0+

P[Tx ≤ dx ]

dx= lim

dx→0+

1− Sx(dx)

dx

= limdx→0+

1− Sx(dx)

dx

= limdx→0+

1− S0(x+dx)S0(x)

dx=

1

S0(x)lim

dx→0+

S0(x)− S0(x + dx)

dx

= − 1

S0(x)

d

dxS0(x) =

f0(x)

S0(x)

(7)



In general, we can show

µx+t = − 1

Sx(t)

d

dtSx(t) =

fx(t)

Sx(t)(8)

and integration of this relation leads to

Sx(t) =S0(x + t)

S0(x)

=e−

∫ x+t0 µsds

e−∫ x

0 µsds

= e−∫ x+tx µsds

= e−∫ t

0 µx+sds

(9)


Example 2.2


120

) 16 for 0 ≤ t ≤ 120. Calculate µx

d

dxS0(x) =

1

6·(

1− x

120

)− 56 ·(− 1

120

)∴ µx = − 1(

1− x120

) 16

·(

1

6·(

1− x

120

)− 56 ·(− 1

120

))=

1

720− 6x

(10)


Example 2.2


120

) 16 for 0 ≤ t ≤ 120. Calculate µx

d

dxS0(x) =

1

6·(

1− x

120

)− 56 ·(− 1

120

)∴ µx = − 1(

1− x120

) 16

·(

1

6·(

1− x

120

)− 56 ·(− 1

120

))=

1

720− 6x

(10)


Gompertz’ Law / Makehams’s Law

One model of human mortality, postulated by Gompertz, is µx = Bcx ,where (B, c) ∈ (0, 1)× (1,∞). This is based on the assumption thatmortality is age dependent, and that the growth rate for mortality isproportional to it’s own value. Makeham proposed that there should alsobe an age independent component, and so Makeham’s Law is

µx = A + Bcx (11)



Of course, when A = 0, this reduces back to Gompertz’ Law.

By definition,

Sx(t) = e−∫ x+tx µsds = e−

∫ x+tx (A+Bcs)ds

= e−At−B

ln ccx (ct−1)

(12)

Keep in mind that this is a multivariable function of (x , t) ∈ R2+

Some online resources:

CDC National Vital Statistics report, Dec. 2002 .


http://www.cdc.gov/nchs/data/nvsr/nvsr51/nvsr51_03.pdf


Of course, when A = 0, this reduces back to Gompertz’ Law.By definition,

Sx(t) = e−∫ x+tx µsds = e−

∫ x+tx (A+Bcs)ds

= e−At−B

ln ccx (ct−1)

(12)

Keep in mind that this is a multivariable function of (x , t) ∈ R2+

Some online resources:

CDC National Vital Statistics report, Dec. 2002 .


http://www.cdc.gov/nchs/data/nvsr/nvsr51/nvsr51_03.pdf

Comparison with US Gov’t data


Actuarial Notation

Actuaries make the notational conventions

tpx = P[Tx > t] = Sx(t)

tqx = P[Tx ≤ t] = Fx(t)

u|tqx = P[u < Tx ≤ u + t] = Sx(u)− Sx(u + t)

(13)

u|tqx , also known as the deferred mortality probability, is the probabilitythat (x) survives u years, and then dies in the subsequent t years.

Another convention is that px := 1px and qx := 1qx .


Actuarial Notation-Relationships

Consequently,

tpx + tqx = 1

u|tqx = upx − u+tpx

t+upx = tpx · upx+t

µx =−1

xp0

d

dx(xp0)

(14)

Similarly,

µx+t =−1

tpx

d

dttpx ⇒

d

dttpx = µx+t · tpx

µx+t =fx(t)

Sx(t)⇒ fx(t) = µx+t · tpx

tpx = e−∫ t

0 µx+sds

(15)



Also, since Fx(t) =∫ t

0 fx(s)ds, we have as a linear approximation

tqx =

∫ t

0spx · µx+sds

qx =

∫ 1

0spx · µx+sds

=

∫ 1

0e−

∫ s0 µx+vdv · µx+sds

≈∫ 1

0µx+sds

≈ µx+ 12

(16)



Also, since Fx(t) =∫ t

0 fx(s)ds, we have as a linear approximation

tqx =

∫ t

0spx · µx+sds

qx =

∫ 1

0spx · µx+sds

=

∫ 1

0e−

∫ s0 µx+vdv · µx+sds

≈∫ 1

0µx+sds

≈ µx+ 12

(16)


Mean and Standard Deviation of Tx

Actuaries make the notational definition ex := E[Tx ], also known as thecomplete expectation of life. Recall fx(t) = tpx · µx+t = − d

dt tpx , and

ex =

∫ ∞0

t · fx(t)dt

=

∫ ∞0

t · tpx · µx+tdt

=

∫ ∞0

t · − d

dttpxdt

=

∫ ∞0

tpxdt

E[T 2x ] =

∫ ∞0

t2 · fx(t)dt =

∫ ∞0

2t · tpxdt

V [Tx ] := E[T 2x ]− (ex)2

(17)




dt tpx , and

ex =

∫ ∞0

t · fx(t)dt

=

∫ ∞0

t · tpx · µx+tdt

=

∫ ∞0

t · − d

dttpxdt

=

∫ ∞0

tpxdt

E[T 2x ] =

∫ ∞0

t2 · fx(t)dt =

∫ ∞0

2t · tpxdt

V [Tx ] := E[T 2x ]− (ex)2

(17)




dt tpx , and

ex =

∫ ∞0

t · fx(t)dt

=

∫ ∞0

t · tpx · µx+tdt

=

∫ ∞0

t · − d

dttpxdt

=

∫ ∞0

tpxdt

E[T 2x ] =

∫ ∞0

t2 · fx(t)dt =

∫ ∞0

2t · tpxdt

V [Tx ] := E[T 2x ]− (ex)2

(17)




dt tpx , and

ex =

∫ ∞0

t · fx(t)dt

=

∫ ∞0

t · tpx · µx+tdt

=

∫ ∞0

t · − d

dttpxdt

=

∫ ∞0

tpxdt

E[T 2x ] =

∫ ∞0

t2 · fx(t)dt =

∫ ∞0

2t · tpxdt

V [Tx ] := E[T 2x ]− (ex)2

(17)


Example 2.6

Assume that F0(x) = 1−(1− x

120

) 16 for 0 ≤ x ≤ 120. Calculate ex ,V [Tx ]

for a.)x = 30 and b.)x = 80.

Since S0(x) =(1− x

120

) 16 , it follows that in keeping with the model where

survival is constrained to be less than 120,

tpx =S0(x + t)

S0(x)=

(

1− t120−x

) 16

: x + t ≤ 120

0 : x + t > 120


Example 2.6

Assume that F0(x) = 1−(1− x

120

) 16 for 0 ≤ x ≤ 120. Calculate ex ,V [Tx ]

for a.)x = 30 and b.)x = 80.

Since S0(x) =(1− x

120

) 16 , it follows that in keeping with the model where

survival is constrained to be less than 120,

tpx =S0(x + t)

S0(x)=

(

1− t120−x

) 16

: x + t ≤ 120

0 : x + t > 120


Example 2.6

So,

ex =

∫ 120−x

0

(1− t

120− x

) 16

dt =6

7· (120− x)

E[T 2x ] =

∫ 120−x

02t ·

(1− t

120− x

) 16

dt

=

(6

7− 6

13

)· 2(120− x)2

(18)

and

(e30, e80) = (77.143, 34.286)

(V [T30],V [T80]) =((21.396)2, (9.509)2

) (19)


Exam MLC Spring 2007: Q8

Kevin and Kira excel at the newest video game at the local arcade,“Reversion”. The arcade has only one station for it. Kevin is playing. Kirais next in line. You are given:

(i) Kevin will play until his parents call him to come home.

(ii) Kira will leave when her parents call her. She will start playing assoon as Kevin leaves if he is called first.

(iii) Each child is subject to a constant force of being called: 0.7 perhour for Kevin; 0.6 per hour for Kira.

(iv) Calls are independent.

(v) If Kira gets to play, she will score points at a rate of 100,000 perhour.



Calculate the expected number of points Kira will score before she leaves.

(A) 77,000

(B) 80,000

(C) 84,000

(D) 87,000

(E) 90,000



Define

tpx = P [Kevin still there]

tpy = P [Kira still there](20)

and so

E [Kira’s playing time] =

∫ ∞0

(1− tpx) · tpydt

=

∫ ∞0

(1− e−0.7t

)· e−0.6tdt

=

∫ ∞0

(e−0.6t − e−1.3t

)dt

=1

0.6− 1

1.3= 0.89744 hrs.

(21)



It follows that

E [Kira’s winnings] = 100000$

hr· E [Kira’s playing time]

= $89744.(22)

Hence, we choose (E ).


Numerical Considerations for Tx

In general, computations for the mean and SD for Tx will requirenumerical integration. For example,

Table: 2.1: Gompertz Model Statistics: (B, c) = (0.0003, 1.07)

x ex SD[Tx ] x + ex0 71.938 18.074 71.93810 62.223 17.579 72.22320 52.703 16.857 72.70330 43.492 15.841 73.49240 34.252 14.477 74.75250 26.691 12.746 76.69160 19.550 10.693 79.55070 13.555 8.449 83.55580 8.848 6.224 88.84890 5.433 4.246 95.433100 3.152 2.682 103.152


Curtate Future Lifetime

Define

Kx := bTxc (23)

and so

P [Kx = k] = P [k ≤ Tx < k + 1]

= k|qx

= kpx − k+1px

= kpx − kpx · px+k

= kpx · qx+k

(24)



E [Kx ] := ex =∞∑k=0

k · P [Kx = k]

=∞∑k=0

k · (kpx − k+1px)

=∞∑k=1

kpx by telescoping series..

E[K 2x

]=∞∑k=0

k2 · P [Kx = k]

= 2 ·∞∑k=1

k · kpx −∞∑k=1

kpx

= 2 ·∞∑k=1

k · kpx − ex

(25)



E [Kx ] := ex =∞∑k=0

k · P [Kx = k]

=∞∑k=0

k · (kpx − k+1px)

=∞∑k=1

kpx by telescoping series..

E[K 2x

]=∞∑k=0

k2 · P [Kx = k]

= 2 ·∞∑k=1

k · kpx −∞∑k=1

kpx

= 2 ·∞∑k=1

k · kpx − ex

(25)


Relationship between ex and ex

Recall

ex =

∫ ∞0

tpxdt =∞∑j=0

∫ j+1

jtpxdt (26)

By trapezoid rule for numerical integration, we obtain∫ j+1j tpxdt ≈ 1

2 (jpx + j+1px), and so

ex ≈∞∑j=0

1

2(jpx + j+1px)

=1

2+∞∑j=1

jpx =1

2+ ex

(27)

As with all numerical schemes, this approximation can be refined whennecessary.


Comparison of ex and ex

Approximation matches well for small values of x

Table: 2.2: Gompertz Model Statistics: (B, c) = (0.0003, 1.07)

x ex ex0 71.438 71.93810 61.723 62.22320 52.203 52.70330 42.992 43.49240 34.252 34.75250 26.192 26.69160 19.052 19.55070 13.058 13.5580 8.354 8.84890 4.944 5.433100 2.673 3.152


Notes

An extension of Gompertz - Makeham Laws is the GM(r , s) formulaµx = h1

r (x) + eh2s (x), where h1

r (x), h2s (x) are polynomials of degree r

and s, respectively.

Hazard rate in survival analysis and failure rate in reliability theory isthe same as what actuaries call force of mortality.


Homework Questions

HW: 2.1, 2.2, 2.5, 2.6, 2.10, 2.13, 2.14, 2.15


Life Tables

Define for a model with maximum age ω and initial age x0 the radix lx0 ,where

lx0+t = lx0 · tpx0 (28)


Life Tables

It follows that

lx+t = lx0 · x+t−x0px0

= lx0 · x−x0px0 · tpx

= lx · tpx

tpx =lx+t

lx

(29)

We assume a binomial model where Lt is the number of survivors to agex + t.


Life Tables

It follows that

lx+t = lx0 · x+t−x0px0

= lx0 · x−x0px0 · tpx

= lx · tpx

tpx =lx+t

lx

(29)

We assume a binomial model where Lt is the number of survivors to agex + t.


Life Tables

So, if there are lx independent individuals aged x with probability tpx ofsurvival to age x + t, then we interpret lx+t as the expected number ofsurvivors to age x + t out of lx independent individuals aged x .Symbolically,

E[Lt | L0 = lx ] = lx+t = lx · tpx (30)

Also, define the expected number of deaths from year x to year x + 1 as

dx := lx − lx+1 = lx ·(

1− lx+1

lx

)= lx · (1− px) = lxqx (31)


Life Tables

So, if there are lx independent individuals aged x with probability tpx ofsurvival to age x + t, then we interpret lx+t as the expected number ofsurvivors to age x + t out of lx independent individuals aged x .Symbolically,

E[Lt | L0 = lx ] = lx+t = lx · tpx (30)

Also, define the expected number of deaths from year x to year x + 1 as

dx := lx − lx+1 = lx ·(

1− lx+1

lx

)= lx · (1− px) = lxqx (31)


Example 3.1

Table: 3.1: Extract from a life table

x lx dx

30 10000.00 34.7831 9965.22 38.1032 9927.12 41.7633 9885.35 45.8134 9839.55 50.2635 9789.29 55.1736 9734.12 60.5637 9673.56 66.4938 9607.07 72.9939 9534.08 80.11


Example 3.1

Calculate:

a.) l40

b.) 10p30

c.) q35

d.) 5q30

e.) P [(30) dies between age 35 and 36]


Example 3.1

Calculate:

a.) l40 = l39 − d39 = 9453.97

b.) 10p30 = l40l30

= 0.94540

c.) q35 = d35l35

= 0.00564

d.) 5q30 = l30−l35l30

= 0.02107

e.) P [(30) dies between age 35 and 36] = l35−l36l30

= 0.00552


Example 3.1

Calculate:

a.) l40 = l39 − d39 = 9453.97

b.) 10p30 = l40l30

= 0.94540

c.) q35 = d35l35

= 0.00564

d.) 5q30 = l30−l35l30

= 0.02107


= 0.00552


Example 3.1

Calculate:

a.) l40 = l39 − d39 = 9453.97

b.) 10p30 = l40l30

= 0.94540

c.) q35 = d35l35

= 0.00564

d.) 5q30 = l30−l35l30

= 0.02107


= 0.00552


Example 3.1

Calculate:

a.) l40 = l39 − d39 = 9453.97

b.) 10p30 = l40l30

= 0.94540

c.) q35 = d35l35

= 0.00564

d.) 5q30 = l30−l35l30

= 0.02107


= 0.00552


Example 3.1

Calculate:

a.) l40 = l39 − d39 = 9453.97

b.) 10p30 = l40l30

= 0.94540

c.) q35 = d35l35

= 0.00564

d.) 5q30 = l30−l35l30

= 0.02107


= 0.00552


Fractional Age Assumptions

So far, the life table approach has mirrored the survival distributionmethod we encountered in the previous lecture. However, in detailing thelife table, no information is presented on the cohort in between wholeyears. To account for this, we must make some fractional ageassumptions. The following are equivalent:

UDD1 For all (x , s) ∈ N× [0, 1), we assume that sqx = s · qx

UDD2 For all x ∈ N, we assume

Rx := Tx − Kx ∼ U(0, 1)Rx is independent of Kx .


Fractional Age Assumptions

So far, the life table approach has mirrored the survival distributionmethod we encountered in the previous lecture. However, in detailing thelife table, no information is presented on the cohort in between wholeyears. To account for this, we must make some fractional ageassumptions. The following are equivalent:

UDD1 For all (x , s) ∈ N× [0, 1), we assume that sqx = s · qx

UDD2 For all x ∈ N, we assume

Rx := Tx − Kx ∼ U(0, 1)Rx is independent of Kx .


Proof of Equivalence

Proof:

UDD1 ⇒ UDD2: Assume for all (x , s) ∈ N× [0, 1), we assume that

sqx = s · qx . Then

P [Rx ≤ s] =∞∑k=0

P [Rx ≤ s,Kx = k]

=∞∑k=0

P [k ≤ Tx ≤ k + s]

=∞∑k=0

kpx · sqx+k =∞∑k=0

kpx · s · qx+k

= s ·∞∑k=0

kpx · qx+k = s ·∞∑k=0

P [Kx = k] = s

(32)

and so Rx ∼ U(0, 1).


Proof of Equivalence

To show independence of Rx and Kx ,

P [Rx ≤ s,Kx = k] = P [k ≤ Tx ≤ k + s]

= kpx · sqx+k

= s · kpx · qx+k

= P[Rx ≤ s] · P[Kx = k]

(33)

UDD2 ⇒ UDD1: Assuming UDD2 is true, then for (x , s) ∈ N× [0, 1)we have

sqx = P [Tx ≤ s]

= P [Kx = 0,Rx ≤ s]

= P[Rx ≤ s] · P[Kx = 0]

= s · qx

(34)

QED


Corollary

Recall that sqx = lx−lx+s

lx. It follows now that

sqx = sqx = sdx

lx=

lx − lx+s

lx

lx+s = lx − s · dx

which is a linear decreasing function of s ∈ [0, 1)

qx =d

ds[sqx ] = fx(s) = spx · µx+s

(35)

But, since qx is constant in s, we have fx(s) is constant for s ∈ [0, 1).

Read over Examples 3.2− 3.5


Corollary



sqx = sqx = sdx

lx=

lx − lx+s

lxlx+s = lx − s · dx


qx =d


(35)




Corollary



sqx = sqx = sdx

lx=

lx − lx+s

lxlx+s = lx − s · dx


qx =d


(35)




Constant Force of Mortality for Fractional Age

For all (x , s) ∈ N× [0, 1), we assume that µx+s does not depend on s, andwe denote µx+s := µ∗x . It follows that

px = e−∫ 1

0 µx+sds = e−µ∗x

spx = e−∫ s

0 µ∗x du = e−µ

∗x s = (px)s

spx+t = e−∫ s

0 µ∗x du = (px)s when t + s < 1

qx = 1− e−µ∗x =

∞∑k=1

(−1)k+1 (µ∗x)k

k!≈ µ∗x

sqx = 1− e−µ∗x ·t ≈ µ∗x · t,

(36)

where the last two lines assume µ∗x 1.

Read Examples 3.6, 3.7 and Sections 3.4, 3.5, 3.6.


Homework Questions

HW: 3.1, 3.2, 3.4, 3.7, 3.8, 3.9, 3.10


Contingent Events

We have spent the previous two lectures on modeling human mortality.The need for such models in insurance pricing arises when designingcontracts that are event-contingent. Such events include reachingretirement before the end of the underlying life (x) .

However, one can also write contracts that are dependent on a life (x)being admitted to college (planning for school), and also on (x)

′s externalportfolios maintaining a minimal value over a time-interval (insuringexternal investments.)


Contingent Events: General Case

Consider a probability space (Ω,F ,P) and an event A ∈ F .If we are working with a force of interest δs(ω) and the time of event Was τW , then we have under the stated probability measure P the ExpectedPresent Value of a payoff K (ω) contingent upon W

EPV = E[K (ω)e−∫ τW

0 δs(ω)ds ] (37)

Actuarial Encounters of the Third Kind !!


Some Initial Simplifying Assumptions

K (ω) = 1 for all ω ∈ Ω (a.s.)

δs(ω) = δ for all ω ∈ Ω (a.s.)

W := event that (x) dies ⇒ τW := Tx

P is obtained via historical observation and is thus a physicalmeasure. Specifically, we use tpx obtained from life tables or viamodels of human mortality

We do not assume now that a unique risk-neutral pricing measure Pexists.

Standard Ultimate Survival Model with assumes Makeham’s lawwith (A,B, c) = (0.00022, 2.7× 10−6, 1.124)


Recall...

The equivalent interest rate i := eδ − 1 per year

The discount factor v := 11+i = e−δ per year

The nominal interest rate i (p) = p ·(

(1 + i)1p − 1

)compounded p

times per year

The effective rate of discount d := 1− v = i · v = 1− e−δ per year

The nominal rate of discount d (p) := p ·(

1− v1p

)compounded p

times per year


Whole Life Insurance: Continuous Case

Consider now the random variable

Z = vTx = e−δTx (38)

which represents the present value of a dollar upon death of (x). We areinterested in statistical measures of this quantity:

E[Z ] = Ax := E[e−δTx ] =

∫ ∞0

e−δt tpxµx+tdt

E[Z 2] = 2Ax := E[e−2δTx ]

=

∫ ∞0

e−2δttpxµx+tdt

Var(Z ) = 2Ax −(Ax

)2

P[Z ≤ z ] = P[

Tx ≥− ln (z)

δ

](39)


Whole Life Insurance: Yearly Case

Assuming payments are made at the end of the death year, our randomvariable is now Z = vKx+1 = e−δKx−δ and so

E[Z ] = Ax := E[vKx+1] =∞∑k=0

vk+1P[Kx = k]

=∞∑k=0

vk+1k|qx

E[Z 2] =∞∑k=0

v 2k+2k|qx

Var(Z ) = 2Ax − (Ax)2

P[Z ≤ z ] = P[

Kx ≥−δ − ln (z)

δ

](40)


Whole Life Insurance: 1m

thlyCase

Instead of only paying at the end of the last whole year lived, an insurancecontract might specify payment upon the end of the last period lived. Inthis case, if we split a year into m periods, and define

K(m)x =

1

mbmTxc (41)

For example, if Kx = 19.78, then

K(m)x =

19 m = 119 1

2 = 19.5 m = 219 3

4 = 19.75 m = 419 9

12 = 19.75 m = 12



thlyCase

It follows that we need ∀r ∈

0, 1m ,

2m , ...,

m−1m , 1, m+1

m , ...

P[K

(m)x = r

]= P

[r ≤ Tx < r +

1

m

]= r | 1

mqx (42)

to compute statistics for our random variable

Z = vK(m)x + 1

m (43)



thlyCase

E[Z ] = A(m)x := E[vK

(m)x + 1

m ] =∞∑k=0

vk+1m k

m| 1m

qx

E[Z 2] = 2A(m)x =

∞∑k=0

v2k+2m k

m| 1m

qx

Var(Z ) = 2A(m)x −

(A

(m)x

)2

P[Z ≤ z ] = P[

K(m)x ≥ − ln (z)

δ− 1

m

](44)


Recursion Method

One of the computational tools we share directly with quantitative financeis the method of backwards-pricing. In option pricing, we assume thecontract has a finite term. Here, we assume a finite lifetime maximum ofω <∞. It follows that

Aω−1 = E[vKω−1+1

]= E

[v 1]

= v (45)

At age ω − 2, we have P[Kω−2 = 0] = qω−2 and so

Aω−2 = E[vKω−2+1

]= qω−2 · v + pω−2 · E

[v (1+Kω−1)+1

]= qω−2 · v + pω−2 · v · E

[vKω−1+1

]= qω−2 · v + pω−2 · v 2

(46)


Recursion Method

One of the computational tools we share directly with quantitative financeis the method of backwards-pricing. In option pricing, we assume thecontract has a finite term. Here, we assume a finite lifetime maximum ofω <∞. It follows that

Aω−1 = E[vKω−1+1

]= E

[v 1]

= v (45)

At age ω − 2, we have P[Kω−2 = 0] = qω−2 and so

Aω−2 = E[vKω−2+1

]= qω−2 · v + pω−2 · E

[v (1+Kω−1)+1

]= qω−2 · v + pω−2 · v · E

[vKω−1+1

]= qω−2 · v + pω−2 · v 2

(46)


Recursion Method

In general, we have the recursion equation for a life (x) that satisfies

Ax = vqx + vpxAx+1

Aω−1 = v(47)

in the whole life case, and

A(m)x = v

1m 1

mqx + v

1m 1

mpxA

(m)

x+ 1m

A(m)

ω− 1m

= v1m

(48)

in the 1m

thlycase.


Recursion Method

Recall that for Makeham’s law we have, respectively

1m

px = e− A

m− Bcx

ln (c)

(c

1m−1

)

1px = e−A− Bcx

ln (c)(c−1)

(49)

and for the power law of survival, S0(x) =(1− x

ω

)afor some a, ω > 0

1m

px =

(ω − x − 1

m

ω − x

)a

1px =

(ω − x − 1

ω − x

)a(50)


Recursion Method

For any positive integer m, it follows that for Makeham’s law:

A(m)x = v

1m

(1− e

− Am− Bcx

ln (c)

(c

1m−1

))+ v

1m e− A

m− Bcx

ln (c)

(c

1m−1

)A

(m)

x+ 1m

A(m)

ω− 1m

= v1m

(51)


Recursion Method

For any positive integer m, it follows that for Power law:

A(m)x = v

1m

(1−

(ω − x − 1

m

ω − x

)a)

+ v1m

(ω − x − 1

m

ω − x

)a

A(m)

x+ 1m

A(m)

ω− 1m

= v1m

(52)

HW Project: For Power law with a = 35 and ω = 101

Generate a spreadsheet like Table 4.1 in the text, including values for2A

(m)x

Repeat Example 4.3 with the Power law model


Term Insurance: Continuous Case

Consider now the case where payment is made in the continuous case, anddeath benefit is payable to the policyholder only if Tx ≤ n. Then, we areinterested in the random variable

Z = e−δTx 1Tx≤n (53)

and so

A1x :n = E[Z ] =

∫ n

0e−δt tpxµx+tdt

2A1x :n = E

[Z 2]

=

∫ n

0e−2δt

tpxµx+tdt

(54)


Term Insurance: 1m

thlyCase

Consider again the case where the death benefit is payable at the end of

the 1m

thlyperiod in the death year to the policyholder only if

K(m)x + 1

m ≤ n. Then, we are interested in the random variable

Z = e−δ(K(m)x + 1

m)1

K(m)x + 1

m≤n (55)

and so

A(m)1x :n = E[Z ] =

mn−1∑k=0

vk+1m k

m| 1m

qx

2A(m)1x :n = E

[Z 2]

=mn−1∑k=0

v2k+2m k

m| 1m

qx

(56)


Pure Endowment

Pure endowment benefits depend on the survival policyholder (x) until atleast age x + n. In such a contract, a fixed benefit of 1 is paid at time n.This is expressed via

Z = e−δn1Tx≥n

A 1x :n = E[Z ] = vn

npx

= e−δnnpx

(57)


Endowment Insurance

Endowment insurance is a combination of term insurance and pureendowment. In such a policy, the amount is paid upon death if it occurswith a fixed term n. However, if (x) survives beyond n years, the suminsured is payable at the end of the nth year. The corresponding presentvalue random variable is

Z = e−δminTx ,n

E[Z ] = Ax :n

=

∫ n

0e−δt tpxµx+tdt + e−δnnpx

= A1x :n + A 1

x :n

Z = e−δminKx+1,n

⇒ Ax :n = A1x :n + A 1

x :n .

(58)

This can also be extended to the 1m

thlycase and for E[Z 2]


Endowment Insurance


Z = e−δminTx ,n

E[Z ] = Ax :n

=

∫ n


= A1x :n + A 1

x :n

Z = e−δminKx+1,n

⇒ Ax :n = A1x :n + A 1

x :n .

(58)




Endowment Insurance


Z = e−δminTx ,n

E[Z ] = Ax :n

=

∫ n


= A1x :n + A 1

x :n

Z = e−δminKx+1,n

⇒ Ax :n = A1x :n + A 1

x :n .

(58)




Deferred Insurance Benefits

Suppose policyholder on a life (x) receives benefit 1 if u ≤ Tx < u + n.Then

Z = e−δTx 1u≤Tx<u+n

E[Z ] = u|A1x :n

=

∫ u+n

ue−δt tpxµx+tdt

=

∫ n

0e−δ(s+u)

s+upxµx+s+uds

= e−δu∫ n

0e−δsupx · spx+uµx+s+uds

= e−δuupx A 1x+u:n

= A1x :u+n − A1

x :u

(59)


Relationships

By definition, we have

Ax = A1x :n + n|Ax

= A1x :n + vn

npxAx+n

(60)

What about relationship between Ax and Ax?


Relationships

By definition, we have

Ax = A1x :n + n|Ax

= A1x :n + vn

npxAx+n

(60)

What about relationship between Ax and Ax?


Employing UDD: Ax vs Ax

If expected values are computed via information derived from life tables,then certainly Ax must be approximated using techniques from previouslecture.

Recall that by the definition of spx and the UDD, we have

spxµx+s = fx(s)

=d

dsP[Tx ≤ s]

=d

ds(sqx) =

d

ds(s · qx)

= qx

(61)



If expected values are computed via information derived from life tables,then certainly Ax must be approximated using techniques from previouslecture.

Recall that by the definition of spx and the UDD, we have

spxµx+s = fx(s)

=d

dsP[Tx ≤ s]

=d

ds(sqx) =

d

ds(s · qx)

= qx

(61)



It follows that using the UDD approximation leads to

Ax =

∫ ∞0

e−δt tpxµx+tdt =∞∑k=0

∫ k+1

ke−δt tpxµx+tdt

=∞∑k=0

kpxvk+1 ·∫ 1

0eδe−δs spx+kµx+k+sds

≈∞∑k=0

kpxvk+1qx+k ·∫ 1

0eδe−δsds

=∞∑k=0

vk+1P[Kx = k] ·∫ 1

0eδe−δsds = Ax ·

∫ 1

0eδe−δsds

= Ax ·i

δ

A(m)x ≈ i

i (m)Ax =

i

m ·(

(1 + i)1m − 1

) · Ax

(62)



It follows that using the UDD approximation leads to

Ax =

∫ ∞0

e−δt tpxµx+tdt =∞∑k=0

∫ k+1

ke−δt tpxµx+tdt

=∞∑k=0

kpxvk+1 ·∫ 1

0eδe−δs spx+kµx+k+sds

≈∞∑k=0

kpxvk+1qx+k ·∫ 1

0eδe−δsds

=∞∑k=0

vk+1P[Kx = k] ·∫ 1

0eδe−δsds = Ax ·

∫ 1

0eδe−δsds

= Ax ·i

δ

A(m)x ≈ i

i (m)Ax =

i

m ·(

(1 + i)1m − 1

) · Ax

(62)


Claims Acceleration Approach : A(m)x vs Ax

Consider now a policy that pays the holder at the end of the 1m

thlyperiod

of death. In this case, the benefit is paid at one of the times r where

r ∈

Kx +1

m,Kx +

2

m, ...,Kx +

m

m

(63)

and so under the UDD,

E [Tpayment | Kx = k] =m∑j=1

1

m·(

k +j

m

)= k +

m + 1

2m(64)


Claims Acceleration Approach : A(m)x vs Ax

Once again, it follows using the UDD approximation

A(m)x = E[vK

(m)x + 1

m ] =∞∑k=0

vk+1m k

m| 1m

qx

≈∞∑k=0

vE[Tpayment |Kx=k]P [Kx = k]

=∞∑k=0

vk+m+12m k|qx = (1 + i)

m−12m ·

∞∑k=0

vk+1k|qx

= (1 + i)m−12m · Ax → (1 + i)

12 · Ax

(65)

as m→∞. Note that using the UDD approximation, both AxAx

and A(m)xAx

are independent of x .


Variable Insurance Benefits

Upon death, we have considered policies that pay the holder a fixedamount. What varied was the method and time of payment. If, however,the actual payoff amount depended on the time Tx of death for (x), thenwe term such a contract a Variable Insurance Contract.

Specifically, if the payoff amount dependent on Tx is h(Tx), then

Z = h(Tx)e−δTx

E[Z ] =

∫ ∞0

h(t)e−δt tpxµx+tdt(I A)x

:=

∫ ∞0

te−δt tpxµx+tdt(I A)

1x :n :=

∫ n

0te−δt tpxµx+tdt

(66)


Example 4.8

Consider an n−year term insurance issued to (x) under which the deathbenefit is paid at the end of the year of death. The death benefit if deathoccurs between ages x + k and x + k + 1 is valued at (1 + j)k . Hence,using the definition i∗ := 1+i

1+j − 1,

Z = vKx+1(1 + j)Kx

E[Z ] =n−1∑k=0

vk+1(1 + j)kk|qx

=1

1 + j·n−1∑k=0

vk+1(1 + j)k+1k|qx

=1

1 + j·n−1∑k=0

k|qx(1+i1+j

)k+1=

1

1 + j· A1

x :n

(67)


Homework Questions

HW: 4.1, 4.2, 4.3, 4.7, 4.9, 4.11, 4.12, 4.14, 4.15, 4.16, 4.17, 4.18


Life Annuities

A Life Annuity refers to a series of payments to or from an individual aslong as that person is still alive. For a fixed rate i and term n , we recallthe deterministic pricing theory:

an i = 1 + v + ...+ vn−1 =1− vn

d

an i = v + ...+ vn = an i − 1 + vn =1− vn

i

an i =

∫ n

0v tdt =

1− vn

δ

a(m)n i =

1

m·(

1 + v1m + ...+ vn− 1

m

)=

1− vn

d (m)

a(m)n i =

1

m·(

v1m + ...+ vn− 1

m + vn)

=1− vn

i (m)

(68)


Whole Life Annuity Due

Consider the case where 1 is paid out at the beginning of every perioduntil death. Our present random variable is now

Y := aKx+1 =1− vKx+1

d(69)

and so

ax = E[Y ] = E[

1− vKx+1

d

]=

1− Ax

d(70)

V [Y ] = V

[1− vKx+1

d

]=

1

d2V [1] +

1

d2V [vKx+1]

= 0 +2Ax − A2

x

d2

(71)


Whole Life Annuity Due

The present value random variable can also be represented as

Y =∞∑k=0

vk1Tx>k (72)

As P[Tx > k] = tpx , we have the alternate expression for ax

ax = E[Y ] = E

[ ∞∑k=0

vk1Tx>k

]

=∞∑k=0

E[vk1Tx>k]

=∞∑k=0

vkkpx =

∞∑k=0

k|qx ak+1

(73)


Term Annuity Due

Define the present value random variable

Y =

aKx+1 : Kx ∈ 0, 1, 2, ..., n − 1an : Kx ∈ n, n + 1, n + 2, ...

Another expression is

Y = aminKx+1,n =1− v minKx+1,n

d(74)

and so

ax :n = E[Y ] =1− E

[v minKx+1,n]

d

=1− Ax :n

d

=n−1∑t=0

v ttpx =

n−1∑k=0

k|qx ak+1 + npx an

(75)


Whole Life and Term Immediate Annuity

Define Y ∗ =∑∞

k=1 vk1Tx>k. Then we have an annuity immediate thatbegins payment one unit of time from now. It follows that

ax = ax − 1

V [Y ∗] = V [Y ](76)

Also, if we define Y = aminKx ,n, then

ax :n =n∑

t=1

v ttpx = ax :n − 1 + vn

npx (77)


Whole Life Continuous Annuity

Define

Y = aTx=

1− vTx

δ=

∫ ∞0

e−δt1Tx>tdt

ax = E[Y ] =1− Ax

δ=

∫ ∞0

e−δt tpxdt

(78)

Note that if δ = 0, then ax = ex


Term Continuous Annuity

Define Y = aminTx ,n .

Then

Y =1− v minTx ,n

δ

ax :n = E [Y ] =1− Ax :n

δ

=

∫ n

0e−δt tpxdt

(79)


Deferred Annuity

Consider now the case of an annuity for (x) that will pay 1 at the end ofeach year, beginning at age x + u and will continue until death agex + Tx . We define u|ax to be the Expected Present Value of this policy. Itshould be apparent that

u|ax = ax − ax :u

=∞∑t=u

v ttpx

= vuupx ·

∞∑t=0

v ttpx+u

= vuupx ax+u

(80)

holds in the discrete case, and similarly in the continuous case,

u|ax = ax − ax :u (81)


Term Deferred Annuity

For the cases of an annuity for (x) that will pay 1 at the end of each year,beginning at age x + u and will continue until death age x + Tx up to a

term of length n, or annuity-due payable 1m

thly. Then

u|ax :n = vuupxax+u:n

u|a(m)x = vu

upx a(m)x+u

(82)

respectively.These combine with the previous slide to reveal the useful formulae:

ax :n = ax − vnnpx ax+n

a(m)x :n = a

(m)x − vn

npx a(m)x+n

(83)


Linearly Increasing Annuities

Define an annuity where the payments increase linearly at timest = 0, 1, 2, .. provided that (x) is alive at time t

(I a)x =∞∑t=0

(t + 1) · v ttpx

(I a)x :n =n−1∑t=0

(t + 1) · v ttpx

(84)


Linearly Increasing Annuities

If then the annuity is payable continuously, with payments increasing by 1at each year end and the rate of payment in the tth year constant andequal to t for t ∈ 1, 2, ..m, .., n, then h(t) = (m + 1)1m≤t<m+1, andthe EPV is

(I a)x :n =n−1∑m=0

(m + 1)m|ax :1 (85)

If h(t) = t, then

(I a)x :n =

∫ n

0te−δt tpxdt. (86)


Evaluating Annuities Using Recursion

By recursion, we observe

ax = 1 + vpx + v 22px + v 3

3px + ....

= 1 + vpx

(1 + vpx+1 + v 2

2px+1 + v 33px+1 + ....

)= 1 + vpx ax+1

a(m)x =

1

m+ v

1m 1

mpx a

(m)

x+ 1m

(87)

Consider the case where there is a maximum age in the model, and so

aω−1 = 1

a(m)

ω− 1m

=1

m

(88)




ax = 1 + vpx + v 22px + v 3

3px + ....

= 1 + vpx

(1 + vpx+1 + v 2

2px+1 + v 33px+1 + ....

)= 1 + vpx ax+1

a(m)x =

1

m+ v

1m 1

mpx a

(m)

x+ 1m

(87)


aω−1 = 1

a(m)

ω− 1m

=1

m

(88)




ax = 1 + vpx + v 22px + v 3

3px + ....

= 1 + vpx

(1 + vpx+1 + v 2

2px+1 + v 33px+1 + ....

)= 1 + vpx ax+1

a(m)x =

1

m+ v

1m 1

mpx a

(m)

x+ 1m

(87)


aω−1 = 1

a(m)

ω− 1m

=1

m

(88)


Evaluating Annuities Using UDD

Recall that under the UDD assumption,

A(m)x =

i

i (m)Ax

Ax =i

δAx

(89)

and by definition,

ax =1− Ax

d

a(m)x =

1− A(m)x

d (m)

ax =1− Ax

δ

(90)



It follows that

a(m)x =

1− A(m)x

d (m)

=1− i

i (m) Ax

d (m)

=i (m) − iAx

i (m)d (m)

=i (m) − i(1− dax)

i (m)d (m)

=id

i (m)d (m)ax −

i − i (m)

i (m)d (m)

:= α(m)ax − β(m)

ax =id

δ2ax −

i − δδ2

(91)



For term annuities, we have

a(m)x :n = a

(m)x − vn

npx a(m)x+n

= α(m)ax − β(m)− vnnpx · (α(m)ax+n − β(m))

= α(m) ·(

ax − vnnpx a

(m)x+n

)− β(m) · (1− vn

npx)

= α(m) · ax :n − β(m) · (1− vnnpx)

≈ ax :n −m − 1

2m· (1− vn

npx)

(92)


Guaranteed Annuities

There are instances where an age (x) wishes to buy a policy wherepayments are guaranteed to continue upon death to a beneficiary. In thiscase, define the present random variable as Y = an + Y1, where

Y1 =

0 : Kx ∈ 0, 1, 2, ..., n − 1aKx+1 − an : Kx ∈ n, n + 1, n + 2, ...

and so

E[Y1] = E[(

aKx+1 − an)

1Kx≥n

]= n|ax = vn

npx ax+n

E[Y ] := ax :n = an + vnnpx ax+n

and E[Y (m)] := a(m)

x :n= a

(m)n + vn

npx a(m)x+n

(93)


Guaranteed Annuities

There are instances where an age (x) wishes to buy a policy wherepayments are guaranteed to continue upon death to a beneficiary. In thiscase, define the present random variable as Y = an + Y1, where

Y1 =

0 : Kx ∈ 0, 1, 2, ..., n − 1aKx+1 − an : Kx ∈ n, n + 1, n + 2, ...

and so

E[Y1] = E[(

aKx+1 − an)

1Kx≥n

]= n|ax = vn

npx ax+n

E[Y ] := ax :n = an + vnnpx ax+n

and E[Y (m)] := a(m)

x :n= a

(m)n + vn

npx a(m)x+n

(93)


Example 5.4

A pension plan member is entitled to a benefit of 1000 per month, inadvance, for life from age 65, with no guarantee. She can opt to take alower benefit with a 10−year guarantee. The revised benefit is calculatedto have equal EPV at age 65 to the original benefit. Calculate the revisedbenefit using the Standard Ultimate Survival Model, with interest at 5%per year.


Example 5.4

Let B denote the revised monthly benefit. Then the two options are

12000 per year, paid per month with Present Value Y1

12B per year, paid per month with Present Value Y2

Hence E[Y1 − Y2] = 0 implies

12000a(12)65 = 12Ba

(12)

65:10

= 12B ·(

a(12)

10+ v 10

10p65a(12)75

)

∴ B = 1000 ·a

(12)65

a(12)

10+ v 10

10p65a(12)75

= 1000 · 13.0870

13.3791= 978.17

V [Y1 − Y2] = 0?

(94)


Example 5.4





12000a(12)65 = 12Ba

(12)

65:10

= 12B ·(

a(12)

10+ v 10

10p65a(12)75

)∴ B = 1000 ·

a(12)65

a(12)

10+ v 10

10p65a(12)75

= 1000 · 13.0870

13.3791= 978.17

V [Y1 − Y2] = 0?

(94)


Example 5.4





12000a(12)65 = 12Ba

(12)

65:10

= 12B ·(

a(12)

10+ v 10

10p65a(12)75

)∴ B = 1000 ·

a(12)65

a(12)

10+ v 10

10p65a(12)75

= 1000 · 13.0870

13.3791= 978.17

V [Y1 − Y2] = 0?

(94)


Example 5.4 - Extended Cut!

A pension plan member is entitled to a benefit of 12000 per year, inadvance, for life from age 65, with no guarantee.

She can opt to take a lower benefit with a 10−year guarantee, butthen returns to 12000 per year for life from age 75 with no guarantee.

The revised benefit is calculated to be equal to the EPV at age 65 ofthe original benefit.

Calculate the revised benefit using the MSU Model:A 1

65:10= v 10

10p65 = 0.55

a(12)65 = 12.45

a(12)75 = 11.35

a(12)

10= 8.11

Note that a(12)

10= 8.11 doesn’t depend on the mortality model used,

only on i .


Example 5.4 - Extended Cut!

Let B denote the revised monthly benefit. Then

12000a(12)65 = 12Ba

(12)

10+ v 10

10p65 · 12000a(12)75 . (95)

It follows that

B = 1000 ·a

(12)65 − v 10

10p65 · a(12)75

a(12)

10

= 1000 · 12.45− 0.55 · 11.35

8.11= 765.41.

(96)

HW: Now calculate this revised benefit if the insurance company is allowedto keep a 10% commission of the EPV at age 65 of the original benefit.


HW: Example 5.4 - Now with Commissions!

Let B denote the revised monthly benefit. Then

0.90×12000a(12)65 = 12Ba

(12)

10+ v 10

10p65 · 12000a(12)75 . (97)

It follows that

B =10800a

(12)65 − 12000v 10

10p65 · a(12)75

12a(12)

10

=(10800)(12.45)− (0.55)(12000)(11.35)

(12)(8.11)

= 1000 · 0.90 · 12.45− 0.55 · 11.35

8.11= 611.90.

(98)


Woolhouse’s Formula

Consider a function g : R+ → R such that limt→∞ g(t) = 0, then

∫ ∞0

g(t)dt = h ·∞∑k=0

g(kh)− h

2g(0) +

h2

12g ′(0)− h4

720g ′′(0) + ... (99)



Define

g(t) = v ttpx

∴ g ′(t) = −tpxδe−δt − v ttpxµx+t

∴ g ′(0) = −δ − µx

(100)

and so for h = 1,

ax ≈∞∑k=0

g(k)− 1

2+

1

12g ′(0)

=∞∑k=0

vkkpx −

1

2− 1

12(δ + µx)

= ax −1

2− 1

12(δ + µx)

(101)



Correspondingly, for h = 1m ,

ax ≈1

m

∞∑k=0

g

(k

m

)− 1

2m+

1

12m2g ′(0)

=∞∑k=0

vkm k

mpx −

1

2m− 1

12m2(δ + µx)

= a(m)x − 1

2m− 1

12m2(δ + µx)

(102)



Equating the previous two approximations for ax , we obtain

a(m)x ≈ ax −

m − 1

2m− m2 − 1

12m2(δ + µx) (103)



For term annuities, we obtain the approximation

a(m)x :n ≈ ax :n −

m − 1

2m(1−vn

npx)−m2 − 1

12m2(δ+µx−vn

npx(δ+µx+n)) (104)



Letting m→∞, we get

ax ≈ ax −1

2− 1

12(δ + µx)

ax :n ≈ ax :n −1

2(1− vn

npx)− 1

12(δ + µx − vn

npx(δ + µx+n))

(105)

For ax with δ = 0, the approximation above reduces further to

ex ≈ (ex + 1)− 1

2− 1

12µx (106)

NB: For life tables, we can compute these quantities using theapproximation µx ≈ −1

2 [ln (px) + ln (px+1)]


Select and Ultimate Survival Models

Notation:

Aggregate Survival Models: Models for a large population, where

tpx depends only on the current age x .

Select (and Ultimate) Survival Models: Models for a select groupof individuals that depend on the current age x and

Future survival probabilities for an individual in the group depend onthe individual’s current age and on the age at which the individualjoined the group∃d > 0 such that if an individual joined the group more than d yearsago, future survival probabilities depend only on current age. So, afterd years, the person is considered to be back in the aggregatepopulation.

Ultimately, a select survival model includes another event upon whichprobabilities are conditional on.


Select and Ultimate Survival Models

Notation:

d is the select period

The mortality applicable to lives after the select period is over isknown as the ultimate mortality.

A select group should have a different mortality rate, as they have beenoffered (selected for) life insurance. A question, of course, is the effect onmortality by maintaining proper health insurance.


Example 3.8

Consider men who need to undergo surgery because they are sufferingfrom a particular disease. The surgery is complicated and

P[survive one year after surgery] = 0.5

(d , l60, l61, l70) = (1, 89777, 89015, 77946)(107)

Calculate P[A],P[B],P[C ], where

A = (60),about to have surgery, will be alive at age 70B = (60),had surgery at age 59, will be alive at age 70C = (60),had surgery at age 58, will be alive at age 70


Example 3.8

P[A] = P[(60),about to have surgery, alive at age 61] · l70

l61

= 0.5 · 77946

89015= 0.4378

P[B] =l70

l60= 0.8682

P[C ] =l70

l60= 0.8682

(108)


Select Survival Models

S[x]+s(t) = P[(x + s) selected at (x),survives to(x + s + t)]

tq[x]+s = P[(x + s) selected at (x),dies before(x + s + t)]

µ[x]+s = force of mortality at (x + s) for select at (x)

= limh→0+

(1− S[x]+s(h)

h

)tp[x]+s = 1− tq[x]+s = S[x]+s(t)

= e−∫ t

0 µ[x]+s+udu

(109)

For t < d , we refer to to the above as part of the select model. Fort ≥ d , they are part of the ultimate model. Please read through sectionon Select Life Tables.


Select Life Tables

Sometimes, we wish to compute values from life tables. Consider again amodel where x ≥ x0, where x0 is the initial age, and 0 ≤ t ≤ d . Then

lx+d = d−tp[x]+t · l[x]+t (110)


Example 3.9

Theorem

Consider y ≥ x + d > x + s > x + t ≥ x ≥ x0. Then

y−x−tp[x]+t =ly

l[x]+t

s−tp[x]+t =l[x]+s

l[x]+t

(111)


Example 3.9

Proof.

y−x−tp[x]+t = y−x−dp[x]+d · d−tp[x]+t

= y−x−dpx+d · d−tp[x]+t

=ly

lx+d

lx+d

l[x]+t

=ly

l[x]+t

s−tp[x]+t =d−tp[x]+t

d−sp[x]+s

=lx+d

l[x]+t

l[x]+s

lx+d

=l[x]+s

l[x]+t

(112)


Example 3.11

A select survival model has a select period of three years. Its ultimatemortality is equivalent to the US Life Tables, 2002 Females of which anextract is shown below. Information given is that for all x ≥ 65,(

p[x], p[x−1]+1, p[x−2]+2

)= (0.999, 0.998, 0.997). (113)

Table: 3.5: Extract from US LIfe Tables, 2002 Females

x lx70 8055671 7902672 7741073 7566674 7380275 71800


Example 3.11

Calculate the probability that a woman currently aged 70 will survive toage 75 given that

1 she was select at age 67:

2 she was select at age 68




Example 3.11

5p[70−3]+3 = 5p70 =l75

l70= 0.8913

5p[70−2]+2 =l[68]+2+5

l[68]+2=

l75

l[68]+2

=l75

l[68]+3

1p[68]+2

=l75

l71· 1p[68]+2

= 4p71 · 1p[68]+2

=71800

79026· 0.997 = 0.9058

(114)


Example 3.11

5p[70−1]+1 =l[69]+1+5

l[69]+1=

l75

l[69]+1

=l75

l[69]+3

(1p[69]+1)·(1p[69]+2)

=l75

l72· (1p[69]+1) · (1p[69]+2)

=71800

77410· 0.997 · 0.998 = 0.9229

5p[70] =l75

l73· (1p[70]) · (1p[70]+1) · (1p[70]+2)

=71800

75666· 0.997 · 0.998 · 0.999 = 0.9432

(115)


Example 3.12

Given a table of values for q[x], q[x−1]+1, qx and the knowledge that themodel incorporates a 2−year select period, compute

4p[70]

3q[60]+1

4p[70] = p[70]p[70]+1p[70]+2p[70]+3 = p[70]p[70]+1p72p73

=(1− q[70]

)·(1− q[70]+1

)· (1− q72) · (1− q73)

3q[60]+1 = q[60]+1 + p[60]+1q62 + p[60]+1p62q63

= q[60]+1 +(1− q[60]+1

)· q62

+(1− q[60]+1

)· (1− q62) · q63

(116)


Example 3.12


4p[70]

3q[60]+1

4p[70] = p[70]p[70]+1p[70]+2p[70]+3 = p[70]p[70]+1p72p73

=(1− q[70]

)·(1− q[70]+1

)· (1− q72) · (1− q73)

3q[60]+1 = q[60]+1 + p[60]+1q62 + p[60]+1p62q63

= q[60]+1 +(1− q[60]+1

)· q62

+(1− q[60]+1

)· (1− q62) · q63

(116)


Example 3.12


4p[70]

3q[60]+1

4p[70] = p[70]p[70]+1p[70]+2p[70]+3 = p[70]p[70]+1p72p73

=(1− q[70]

)·(1− q[70]+1

)· (1− q72) · (1− q73)

3q[60]+1 = q[60]+1 + p[60]+1q62 + p[60]+1p62q63

= q[60]+1 +(1− q[60]+1

)· q62

+(1− q[60]+1

)· (1− q62) · q63

(116)


Example 3.13

A select survival model has a two-year select period and is specified asfollows. The ultimate part of the model follows Makeham’s law, where(A,B, c) = (0.00022, 2.7× 10−6, 1.124):

µx = A + B · cx (117)

The select part of the model is such that for 0 ≤ s ≤ 2,

µ[x]+s = 0.92−sµx+s (118)

and so for 0 ≤ t ≤ 2,

tp[x] = e−∫ t

0 µ[x]+sds = exp

[−∫ t

0

(0.92−s (A + B · cx+s

))ds

](119)


Example 3.13

It follows that given an initial cohort at age x0, that is given lx0 , we cancompute the entries of a select life table via

lx = px−1lx−1

l[x]+1 =lx+2

p[x]+1

l[x] =lx+2

2p[x]

(120)


Example: Linear µx Transformation

A select group has µ[x]+s = 12 · µx+s , for all s ≥ 0. It follows that

(a) tp[x] = 0.5 · tpx .

(b) tp[x] = (tpx)2.

(c) tp[x] =√

tpx .

(d) tp[x] = 2 · tpx .

(e) None of the above.

Answer: (c)

tp[x] = e−∫ t

0 µ[x]+sds = e−12

∫ t0 µx+sds

= (e−∫ t

0 µx+sds)12 =√

tpx .(121)




(a) tp[x] = 0.5 · tpx .

(b) tp[x] = (tpx)2.

(c) tp[x] =√

tpx .

(d) tp[x] = 2 · tpx .


Answer: (c)

tp[x] = e−∫ t

0 µ[x]+sds = e−12

∫ t0 µx+sds

= (e−∫ t

0 µx+sds)12 =√

tpx .(121)




(a) tp[x] = 0.5 · tpx .

(b) tp[x] = (tpx)2.

(c) tp[x] =√

tpx .

(d) tp[x] = 2 · tpx .


Answer: (c)

tp[x] = e−∫ t

0 µ[x]+sds = e−12

∫ t0 µx+sds

= (e−∫ t

0 µx+sds)12 =√

tpx .(121)


Example: Linear px Transformation

For a 2-year select mortality model, you are given:

p[x]+1 = 0.10 + 0.90× px+1

l76 = 98000

l77 = 96000

(122)

Solve for l[75]+1.



Answer:First note that

q[x]+1 = 1− p[x]+1 = 1− (0.10 + 0.90× px+1)

= 0.90− 0.90× px+1 = 0.90qx+1.(123)

In general, if q[x]+1 = α× qx+1, then

1− lx+2

l[x]+1= α

(1− lx+2

lx+1

)⇒ l[x]+1 =

lx+2

(1− α) + α lx+2

lx+1

.(124)

So, for our case, with α = 0.9, it follows that

l[75]+1 =96000

0.1 + 0.9 · 9600098000

= 97796.2577963 ≈ 97796.3. (125)




q[x]+1 = 1− p[x]+1 = 1− (0.10 + 0.90× px+1)

= 0.90− 0.90× px+1 = 0.90qx+1.(123)


1− lx+2

l[x]+1= α

(1− lx+2

lx+1

)⇒ l[x]+1 =

lx+2

(1− α) + α lx+2

lx+1

.(124)


l[75]+1 =96000

0.1 + 0.9 · 9600098000

= 97796.2577963 ≈ 97796.3. (125)




q[x]+1 = 1− p[x]+1 = 1− (0.10 + 0.90× px+1)

= 0.90− 0.90× px+1 = 0.90qx+1.(123)


1− lx+2

l[x]+1= α

(1− lx+2

lx+1

)⇒ l[x]+1 =

lx+2

(1− α) + α lx+2

lx+1

.(124)


l[75]+1 =96000

0.1 + 0.9 · 9600098000

= 97796.2577963 ≈ 97796.3. (125)




q[x]+1 = 1− p[x]+1 = 1− (0.10 + 0.90× px+1)

= 0.90− 0.90× px+1 = 0.90qx+1.(123)


1− lx+2

l[x]+1= α

(1− lx+2

lx+1

)⇒ l[x]+1 =

lx+2

(1− α) + α lx+2

lx+1

.(124)


l[75]+1 =96000

0.1 + 0.9 · 9600098000

= 97796.2577963 ≈ 97796.3. (125)


Example: Martian Annuities!

You are given the extract from a 3−year select mortality table:

Table: Extract from Martian Life Tables

[x ] l[x] l[x]+1 l[x]+2 lx+3 x+3

50 10000 9985 9975 9965 5351 9990 9975 9965 9955 5452 9980 9965 9955 9945 5553 9970 9955 9945 9935 5654 9960 9945 9935 9925 5755 9950 9935 9925 9915 58

Calculate, as polynomials in v :

a[50]:4 .

(I a)[50]:4 .

v · ddv

(a[50]:4

).


Example: Martian Insurance!

Answer:

a[50]:4 = vl[50]+1

l[50]+ v 2 l[50]+2

l[50]+ v 3 l53

l[50]+ v 4 l54

l[50]

= v9985

10000+ v 2 9975

10000+ v 3 9965

10000+ v 4 9955

10000= 0.9985v + 0.9975v 2 + 0.9965v 3 + 0.9955v 4.

(I a)[50]:4 = 1 + 2vl[50]+1

l[50]+ 3v 2 l[50]+2

l[50]+ 4v 3 l53

l[50]

= 1 + 1.997v + 2.9925v 2 + 3.986v 3.

v · d

dva[50]:4 = v · d

dv

(v

l[50]+1

l[50]+ v 2 l[50]+2

l[50]+ v 3 l53

l[50]+ v 4 l54

l[50]

)= v

l[50]+1

l[50]+ 2v 2 l[50]+2

l[50]+ 3v 3 l53

l[50]+ 4v 4 l54

l[50]=: (Ia)[50]:4

(126)Albert Cohen (MSU) STT 455-6: Actuarial Models MSU 2020 117 / 348

Homework Questions

HW: 3.1, 3.2, 3.4, 3.7, 3.8, 3.9, 3.10, 5.1, 5.3, 5.5, 5.6, 5.11, 5.14.


What is a Premium?

When entering into a contract, the financial obligations of all parties mustbe specified. In an insurance contract, the insurance company agrees topay the policyholder benefits in return for premium payments. Thepremiums secure the benefits as well as pay the company for expensesattached to the administration of the policy


Premium Types

A Net Premium does not explicitly allow for company’s expenses, while aOffice or Gross Premium does. There may be a Single Premium or or aseries of payments that could even match with the policyholder’s salaryfrequency.

It is important to note that premiums are paid as soon as the contract issigned, otherwise the policyholder would attain coverage before paying forit with the first premium. This could be seen as an arbitrage opportunity -non-zero probability of gain with no money up front.


Premium Types

Premiums cease upon death of the policyholder. The premium payingterm is the maximum length of time that premiums are required.Certainly, premium term can be fixed so that upon retirement, say, nomore payments are required.

Also, the benefits can be secured in the future (deferred) by a singlepremium payment up front. For example, pay now to secure annuitypayments upon retirement until death.


Assumptions

Recall the life model used in Example 3.13 : The select survival model hasa two-year select period and is specified as follows. The ultimate part ofthe model follows Makeham’s law, where(A,B, c) = (0.00022, 2.7× 10−6, 1.124):

µx = 0.00022 + (2.7× 10−6) · (1.124)x (127)

The select part of the model is such that for 0 ≤ s ≤ 2,

µ[x]+s = 0.92−sµx+s (128)

and so for 0 ≤ t ≤ 2,

tp[x] = e−∫ t

0 µ[x]+sds = exp

[0.92−t

(1− 0.9t

ln (0.9)+

ct − 0.9t

ln(

0.9c

) )] (129)


Assumptions

Furthermore, we can extend the recursion principle when using a select lifemodel to obtain

ax = 1 + vpx ax+1

a[x]+1 = 1 + vp[x]+1ax+2

a[x] = 1 + vp[x]a[x]+1

(130)


Basic Model

In general, an insurance company can expect to have a total benefit paidout, along with expense loading and other related costs. We represent thistotal benefit as Z . Similarly, to fund Z , the company can expect thepolicyholder to make a single payment, or stream of payments, that haspresent value P · Y . Here, P represents the level premium P and Yrepresents the present value associated to a unit payment or paymentstream.


Future Loss Random Variable

For life contingent contracts, there is an outflow and inflow of moneyduring the term of the agreement. The premium income is certain, butsince the benefits are life contingent, the term and total income may notbe certain up front. To account for this, we define the Net Future LossLn

0 (which includes expenses) and the Gross Future Loss Lg0 (which does

not includes expenses) as

Ln0 = PV [benefit outgo]− PV [net premium income]

Lg0 = PV [benefit outgo] + PV [expenses]

− PV [gross premium income]

(131)


Example 6.2

An insurer issues a whole life insurance to [60], with sum insured Spayable immediately upon death. Premiums are payable annually inadvance, ceasing at 80 or on earlier death. The net annual premium is P.What is the net future loss random variable Ln

0 for this contract in terms oflifetime random variables for [60]?

Ln0 = SvT[60] − Pa

minK[60]+1,20 (132)


Example 6.2

An insurer issues a whole life insurance to [60], with sum insured Spayable immediately upon death. Premiums are payable annually inadvance, ceasing at 80 or on earlier death. The net annual premium is P.What is the net future loss random variable Ln

0 for this contract in terms oflifetime random variables for [60]?

Ln0 = SvT[60] − Pa

minK[60]+1,20 (132)


Equivalence Principle

Absent a risk-neutral type pricing measure, insurers price theseevent-contingent contracts by setting the average value of the loss to bezero. Symbolically, this is simply (for net premiums) find P such that

E [Ln0] = 0 (133)

Note that this value P does not necessarily set Var [Ln0] = 0

Returning to our general set-up, we see that the equivalence pricingprinciple can be summarized as

P =E[Z ]

E[Y ](134)



As an introductory example, consider λ > 0 and a contract where (underno selection)

Z = vTx

Y = aTx

tpx = e−λt(135)

Hence, we have a unit whole-life insurance payable immediately upon deathof (x), where mortality is modeled to be exponential with parameter λ.



As an introductory example, consider λ > 0 and a contract where (underno selection)

Z = vTx

Y = aTx

tpx = e−λt(135)

Hence, we have a unit whole-life insurance payable immediately upon deathof (x), where mortality is modeled to be exponential with parameter λ.



We obtain

Px =E[vTx]

E[aTx

] =Ax

ax= δ

Ax

1− Ax

= δ

∫∞0 e−δtλe−λtdt

1−∫∞

0 e−δtλe−λtdt

= δλλ+δ

1− λλ+δ

= λ

(136)

HW: repeat the above calculation if S0(x) = ω−xω for a finite lifetime

model with maximal age ω.



We obtain

Px =E[vTx]

E[aTx

] =Ax

ax= δ

Ax

1− Ax

= δ

∫∞0 e−δtλe−λtdt

1−∫∞

0 e−δtλe−λtdt

= δλλ+δ

1− λλ+δ

= λ

(136)

HW: repeat the above calculation if S0(x) = ω−xω for a finite lifetime

model with maximal age ω.



If we repeat the previous example, but now for the case of of a unitwhole-life insurance contract with level annual premium payment andbenefit paid at the end of the death year, then

Z = vKx+1

Y = aKx+1

tpx = e−λt(137)



It follows that

Px = dAx

1− Ax= d

∑∞k=0 e−δ(k+1) · (kpx − k+1px)

1−∑∞

k=0 e−δ(k+1) · (kpx − k+1px)

= d(1− e−λ) ·

∑∞k=0 e−δ(k+1)e−λk

1− (1− e−λ) ·∑∞

k=0 e−δte−λk

= d(1− e−λ) · e−δ · 1

1−e−(δ+λ)

1− (1− e−λ) · e−δ · 11−e−(δ+λ)

= (1− e−λ) · e−δ

(138)


”Deterministic” Insurance

Consider an endowment insurance with sum insured 100000 issued to anage (x) where 20 premiums are paid in return for the benefit 100000 paidat the end of year 20. Assume v = 1

1.05 . Then

Z = 100000v 20

Y = a20 =1− v 20

1− v

⇒ Pd =100000v 20(

1−v20

1−v

)= (1− v)× 100000× v 20

1− v 20

≈ 2880.25.

(139)


”Deterministic” Insurance

Consider an endowment insurance with sum insured 100000 issued to anage (x) where 20 premiums are paid in return for the benefit 100000 paidat the end of year 20. Assume v = 1

1.05 . Then

Z = 100000v 20

Y = a20 =1− v 20

1− v

⇒ Pd =100000v 20(

1−v20

1−v

)= (1− v)× 100000× v 20

1− v 20

≈ 2880.25.

(139)


Example 6.5

Now, consider an endowment insurance with sum insured 100000 issued toa select life aged [45] with term 20 years under which the death benefit ispayable at the end of of the year of death. Using the Standard SelectSurvival Model with interest at 5% per year, calculate the total amount ofnet premium payable in a year if premiums are payable annually.


Example 6.5

By the EPP, the fact that d = 1− 11.05 , and tables 6.1 and 3.7, we have

100000 · A[45]:20 = P · a[45]:20

⇒ P = 100000 ·A[45]:20

a[45]:20

=100000 ·

(1− da[45]:20

)a[45]:20

= 100000 ·

(1− d

(a[45] − l65

l[45]v 20a65

))a[45] − l65

l[45]v 20a65

= 100000 · 0.383766

12.94092= 2965.52

(140)

Is it reasonable that P = 2965.52 > 2880.25 = Pd?


New Business Strain

Starting up an insurance company requires start-up capital like most othercompanies. Agents are charged with drumming up new business in theform of finding and issuing new life insurance contracts. This helps todiversify risk in the case of a large loss on one contract (more on thislater.)

However, new contracts can incur larger losses up front in the first fewyears even without a benefit payout. This is due to initial commisionpayments to agents as well as contract preparation costs. Periodicmaintenance costs can also factor into the premium calculation.


An Example..

Consider offering an n−year endowment policy to an age (x) in theaggregate population where the benefit B is paid at the end of the year ofdeath or on maturity. There are periodic renewal expenses of r per policy.

Then the premium P is calculated via the EPP as

Pax :n = B · Ax :n + r ax :n

⇒ P = B · Ax :n

ax :n+ r

(141)

and we see that periodic expenses are simply passed on to the consumer!


An Example..

Consider offering an n−year endowment policy to an age (x) in theaggregate population where the benefit B is paid at the end of the year ofdeath or on maturity. There are periodic renewal expenses of r per policy.

Then the premium P is calculated via the EPP as

Pax :n = B · Ax :n + r ax :n

⇒ P = B · Ax :n

ax :n+ r

(141)

and we see that periodic expenses are simply passed on to the consumer!


Another Example..

Consider offering an n−year endowment policy to an age (x) in theaggregate population where the benefit B is paid at the end of the year ofdeath or on maturity. There are periodic renewal expenses of r per policyand an initial preparation expense of z per contract.

Then the premium P is calculated via

P = B · Ax :n

ax :n+ r +

z

ax :n(142)

and so the initial preparation expense is amortized over the lifetime of thecontract.


Another Example..

Consider offering an n−year endowment policy to an age (x) in theaggregate population where the benefit B is paid at the end of the year ofdeath or on maturity. There are periodic renewal expenses of r per policyand an initial preparation expense of z per contract.

Then the premium P is calculated via

P = B · Ax :n

ax :n+ r +

z

ax :n(142)

and so the initial preparation expense is amortized over the lifetime of thecontract.


Example 6.6

An insurer issues a 25−year annual premium endowment insurance withsum insured 100000 to a select life aged [30]. The insurer incurs initialexpenses of 2000 plus 50% of the first premium and renewable expenses of2.5% of each subsequent premium. The death benefit is payableimmediately upon death. What is the annual premium P?


Example 6.6

We can see that

Lg0 = 100000v minT[30],25 + 2000 + 0.475P

+ 0.025PaminK[30]+1,25 − Pa

minK[30]+1,25

⇒ P =100000 · E

[v minT[30],25

]+ 2000

0.975 · E[

aminK[30]+1,25

]− 0.475

=100000 · A[30]:25 + 2000

0.975 · a[30]:25 − 0.475

=100000 · (0.298732) + 2000

0.975 · (14.73113)− 0.475

= 2295.04.

(143)


Some more worked examples

Consider the following net loss random variables:

Ln0 = vTx − PaminTx ,t

Ln0 = v minTx ,n − PaminTx ,t

Ln0 = vKx+1 − PaminKx+1,t

(144)

What are the fair premiums under the EPP?

P =Ax

ax :t

P =Ax :n

ax :t

P =Ax

ax :t

(145)







(144)


P =Ax

ax :t

P =Ax :n

ax :t

P =Ax

ax :t

(145)







(144)


P =Ax

ax :t

P =Ax :n

ax :t

P =Ax

ax :t

(145)







(144)


P =Ax

ax :t

P =Ax :n

ax :t

P =Ax

ax :t

(145)







(144)


P =Ax

ax :t

P =Ax :n

ax :t

P =Ax

ax :t

(145)







(144)


P =Ax

ax :t

P =Ax :n

ax :t

P =Ax

ax :t

(145)


Refunded Deferred Annual Whole-Life Annuity Due

Consider the case where a n−year deferred annual whole-life annuity dueof 1 on a life (x) where if the death occurs during the deferral period, thesingle benefit premium is refunded without interest at the end of theyear of death. What is this single benefit premium P?



The net loss random variable is

Ln0 = PvKx+1 · 1Kx+1≤n + vn1Kx+1>n · aKx+1−n − P (146)

and by the EPP we have

0 = PA1x :n + n|ax − P (147)

This implies that

P =n|ax

1− A1x :n

=Ax :n − A1

x :n

1− A1x :n

· ax+n

(148)



The net loss random variable is

Ln0 = PvKx+1 · 1Kx+1≤n + vn1Kx+1>n · aKx+1−n − P (146)

and by the EPP we have

0 = PA1x :n + n|ax − P (147)

This implies that

P =n|ax

1− A1x :n

=Ax :n − A1

x :n

1− A1x :n

· ax+n

(148)


Profit

Consider a 1−year term insurance contract issued to a select life [x ], withsum insured S = 1000, interest rate i = 0.05, and mortalityq[x] = P[T[x] ≤ 1] = 0.01

It follows that L0, the future loss random variable calculated at the time ofissuance, is

L0 = 1000v 11T[x]≤1 − P

⇒ P = E[1000v 11T[x]≤1] = 1000v · P[T[x] ≤ 1]

=(1000)(0.01)

1.05= 9.52

(149)


Profit

Consider a 1−year term insurance contract issued to a select life [x ], withsum insured S = 1000, interest rate i = 0.05, and mortalityq[x] = P[T[x] ≤ 1] = 0.01

It follows that L0, the future loss random variable calculated at the time ofissuance, is

L0 = 1000v 11T[x]≤1 − P

⇒ P = E[1000v 11T[x]≤1] = 1000v · P[T[x] ≤ 1]

=(1000)(0.01)

1.05= 9.52

(149)


Profit

Consider that the company has issued a lot of these contracts, say N 1,to independent select lives [x ]. Let D[x] be the random variablerepresenting the number of deaths in a year of this population, and assume

D[x] ∼ Bin(N, q[x]) (150)


Profit

In general, we have the event that the insurer turns a profit on this groupof policies is Profit =

D[x] ≤ N · q[x]

, and so as N →∞,

P[Profit] = P[D[x] ≤ N · q[x]]

= P[D[x] ≤ E[D[x]]]

= P

D[x] − E[D[x]]√Var

[D[x]

] ≤ 0

→ Φ(0) =

1

2by the CLT.

(151)

HW: Compute

E[Profit | D[x] ≤ N · q[x]

]N · P

(152)


Profit

Consider now a whole-life contract issued to [x ] with sum insured S andannual premium P. Then

P[Profit] = P[L0 < 0] = P[SvK[x]+1 − PaK[x]+1 < 0]

= P[SvK[x]+1 < P · 1− vK[x]+1

d]

= P[

vK[x]+1 <P

P + d · S

]= P

[K[x] + 1 >

1

δln

(P + d · S

P

)]= P

[K[x] ≥ b

1

δln

(P + d · S

P

)c]

= b 1δ

ln ( PP+d·S )cp[x]

(153)


Conditional Sums

Define L0(k) = PV [Loss | K[x] = k]. For a contract with a term n, itfollows that

E[L0] = E[PV [Loss]] = E

[n−1∑k=0

PV [Loss | K[x] = k] · 1K[x]=k

]+ E

[PV [Loss | K[x] ≥ n] · 1K[x]≥n

]=

n−1∑k=0

(PV [Loss | K[x] = k] · P[K[x] = k]

)+ L0(n) · P[K[x] ≥ n]

=n−1∑k=0

L0(k) · k|q[x] + L0(n)np[x]

(154)Q: Can we use this for the case n =∞ ?


Example 6.9

A life insurer is about to issue a 25−year endowment insurance with abasic sum insured S = 250000 to a select life aged exactly [30]. Premiumsare payable annually throughout the term of the policy. Initial expenses are1200 plus 40% of the first premium and renewal expenses are 1% of thesecond and subsequent premiums. The insurer allows for a compoundreversionary bonus of 2.5% of the basic sum insured, vesting on eachpolicy anniversary (including the last.) The death benefit is payable at theend of the year of death. Assume the Standard Select Survival Model withinterest rate 5% per year.


Example 6.9

In this case, we have the reversionary bonus exponentially grow the suminsured:

L0 = B0(K[x]) + E0 − P0(K[x])

B0(k) = 250000 · 1.025kvk+1 for k ∈ 0, 1, 2, ..., 24B0(25) = 250000 · 1.02525v 25

E0 = 1200 + 0.39P

P0(k) = 0.99Pamin k+1,25

(155)


Example 6.9

For 1 + j = 1+i1.025 , we have j = 0.02439 and so

E[L0] = E[B0(K[x])] + E0 − E[P0(K[x])]

=24∑k=0

B0(k)k|q[30] + B0(25)25p[30] + E0 − 0.99Pa[30]:25

=24∑k=0

250000 · (1.025)k

(1.05)k+1 k|q[30] + B0(25)25p[30]

+ E0 − 0.99Pa[30]:25

=250000

1.025A1

[30]25 j+ B0(25) · 25p[30] + 1200 + 0.39P − 14.5838P

(156)

Under the EPP, we have P = 9764.44.

HW: Replicate Table 6.3 in the text by using a spreadsheet program.Compare this example with Example 6.10.


Example 6.9

For 1 + j = 1+i1.025 , we have j = 0.02439 and so

E[L0] = E[B0(K[x])] + E0 − E[P0(K[x])]

=24∑k=0

B0(k)k|q[30] + B0(25)25p[30] + E0 − 0.99Pa[30]:25

=24∑k=0

250000 · (1.025)k

(1.05)k+1 k|q[30] + B0(25)25p[30]

+ E0 − 0.99Pa[30]:25

=250000

1.025A1

[30]25 j+ B0(25) · 25p[30] + 1200 + 0.39P − 14.5838P

(156)

Under the EPP, we have P = 9764.44.

HW: Replicate Table 6.3 in the text by using a spreadsheet program.Compare this example with Example 6.10.


Portfolio Percentile Premium Principle

Assume once again that a company is about to issue insurance to Nindependent lives [x ], each with loss L0,i for i ∈ 1, 2, ..,N . In this case

L0 =N∑i=1

L0,i

E[L0] = E

[N∑i=1

L0,i

]=

N∑i=1

E [L0,i ] = N · E [L0,1]

Var [L0] = N · Var [L0,1]

(157)


Portfolio Percentile Premium Principle

If we require P such that P[L0 < 0] = α, we can use CLT once again toshow

α = P[L0 < 0]

= P

[L0 − E[L0]√

Var [L0]< − E[L0]√

Var [L0]

]

→ Φ

(− E[L0]√

Var [L0]

)as N →∞

(158)

For an individual present value of loss, stated wlog as L0,1, we recover theEPP as N →∞

E[L0,1] ≈ −Φ−1(α)

√Var [L0,1]

√N

→ 0 (159)


Example 6.11

An insurer issues whole life insurance policies to select lives aged [30]. Thesum insured S = 100000 is paid at the end of the month of death andlevel monthly premiums are payable throughout the term of the policy.Initial expenses, incurred at the issue of the policy, are 15% of the total ofthe first year’s premiums. Renewal expenses are 4% of every premium,including those in the first year. Assume the SSSM with interest at 5% peryear.

Calculate the monthly premium P using the EPP and

Calculate the monthly premium P using the PPPP such thatα = 0.95 and N = 10000.


Example 6.11

For the EPP calculation, we have

E[PV (Premiums)] = 12Pa(12)[30] = 227.065P

E[PV (Benefits)] = 100000A(12)[30] = 7866.18

E[PV (Expenses)] = (0.15)(12P) + (0.04)(12Pa(12)[30] )

= 10.8826P

E[PV (Premiums)] = E[PV (Benefits)] + E[PV (Expenses)]

⇒ P = 36.39

(160)


Example 6.11

For the EPP calculation, we have

E[PV (Premiums)] = 12Pa(12)[30] = 227.065P

E[PV (Benefits)] = 100000A(12)[30] = 7866.18

E[PV (Expenses)] = (0.15)(12P) + (0.04)(12Pa(12)[30] )

= 10.8826P

E[PV (Premiums)] = E[PV (Benefits)] + E[PV (Expenses)]

⇒ P = 36.39

(160)


Example 6.11

For all i ∈ 1, 2, ..,N, we have the i.i.d. PV(Loss) random variables

L0,i = 100000vK

(12)[30]

+ 112 + (0.15)(12P)

− (0.96)

(12Pa

(12)

K(12)[30]

+ 112

)E[L0,i ] = 100000A

(12)[30] + (0.15)(12P)− (0.96)(12Pa

(12)[30] )

= 7866.18− 216.18P

(161)


Example 6.11

To find the variance, we rewrite

L0,i =

(100000 +

(0.96)(12P)

d (12)

)vK

(12)[30]

+ 112

+ (0.15)(12P)− (0.96)(12P)

d (12)

Var [L0,i ] =

(100000 +

(0.96)(12P)

d (12)

)2

·(

2A(12)[30] −

(A

(12)[30]

)2)

= (100000 + 236.59P)2 · (0.0053515)

(162)


Example 6.11

Collecting our results, we now have

0.95 = α = P[L0 < 0]

≈ Φ

(− E[L0]√

Var [L0]

)

= Φ

(−√

N · E[L0,1]√Var [L0,1]

)

= Φ

(√10000 · 216.18P − 7866.18

(100000 + 236.59P) ·√

0.0053515

)(163)


Example 6.11

It follows that

1.645 = Φ−1(0.95)

≈√

10000 · 216.18P − 7866.18

(100000 + 236.59P) ·√

0.0053515

⇒ P = 36.99

(164)

For general N, we have

216.18P − 7866.18

(100000 + 236.59P) ·√

0.0053515=

1.645√N

(165)

and as N →∞, we have P → 36.39, recovering the EPP premium asexpected.


Independent Exponential RV’s

Imagine that a fully continuous whole life insurance is offered to N

individuals aged [x ] with T(i)[x] ∼ exp(λ) for all i ∈ 1, ..,N. For each

insured, the i.i.d. loss random variables are

L0,i = SvT

(i)[x] − Pa

T(i)[x]

=δS + P

δe−δT (i)

[x] − P

δ(166)

and so for L0 =∑N

i=1 L0,i , the PPPP seeks to determine P such that

P

[N∑i=1

(δS + P

δe−δT (i)

[x] − P

δ

)< 0

]= α (167)



In this case, we can rewrite this as

P

[1

N

N∑i=1

e−δT (i)

[x] <P

P + δS

]= α (168)

In this case, for each i if we define Yi := e−δT (i)

[x] , then we know that

P [Yi > y ] = P[

e−δT (i)

[x] > y

]= 1− y

λδ .

HW Using convolution techniques, find the above probability

P

[1

N

N∑i=1

Yi <P

P + δS

]= α. (169)

Are there any ergodic theory results that we can use?



In this case, we can rewrite this as

P

[1

N

N∑i=1

e−δT (i)

[x] <P

P + δS

]= α (168)

In this case, for each i if we define Yi := e−δT (i)

[x] , then we know that

P [Yi > y ] = P[

e−δT (i)

[x] > y

]= 1− y

λδ .

HW Using convolution techniques, find the above probability

P

[1

N

N∑i=1

Yi <P

P + δS

]= α. (169)

Are there any ergodic theory results that we can use?


Capped Maximal Loss

Another principle is to find P such that for any α ∈ (0, 1), the probability

that any contract suffers a loss of β < SαδλN < S is set to α for a set of

i.i.d. exponentially distributed times T(i)[x] :

P[

maxi=1..N

δS + P

δe−δT (i)

[x] − P

δ

< β

]= α (170)

We rewrite this as

α = P[

mini=1..N

T

(i)[x]

> −1

δln

(P + δβ

P + δ

)]=(

e−λ[− 1δ

ln (P+δβP+δ )]

)N=

(P + δβ

P + δS

)λNδ

P = δ · SαδλN − β

1− αδλN

→∞ as N →∞

(171)

For this risk measure, is there a number of policy holders N that is toohigh?


Capped Maximal Loss




P[

maxi=1..N

δS + P

δe−δT (i)

[x] − P

δ

< β

]= α (170)

We rewrite this as

α = P[

mini=1..N

T

(i)[x]

> −1

δln

(P + δβ

P + δ

)]=(

e−λ[− 1δ

ln (P+δβP+δ )]

)N=

(P + δβ

P + δS

)λNδ


1− αδλN

→∞ as N →∞

(171)



Capped Maximal Loss




P[

maxi=1..N

δS + P

δe−δT (i)

[x] − P

δ

< β

]= α (170)

We rewrite this as

α = P[

mini=1..N

T

(i)[x]

> −1

δln

(P + δβ

P + δ

)]=(

e−λ[− 1δ

ln (P+δβP+δ )]

)N=

(P + δβ

P + δS

)λNδ


1− αδλN

→∞ as N →∞

(171)



Comments on PPPP

Notice that the PPPP only guarantees that the probability of a loss is1− α.

It says nothing about the size of what that loss could be if itarises.

This is a big problem if the loss is extremely large and bankrupts theinsurer. It may seem very unlikely, but recent economic events haveshown otherwise.

Further improvements to this model can be seen in the ERM forStrategic Management (Status Report) by Gary Venter, posted onthe SOA.org website

Also, there is a close link, perhaps to be explored in a project, withVAR in the financial world. Click here for an informative article in theNY Times TM for an article on VAR and the recent financial crisis.


http://www.soa.org/library/monographs/other-monographs/2008/april/mono-2008-m-as08-1-venter.pdf



http://www.nytimes.com/2009/01/04/magazine/04risk-t.html?pagewanted=all

http://www.nytimes.com/2009/01/04/magazine/04risk-t.html?pagewanted=all

Homework Questions

HW: 6.1, 6.2, 6.5, 6.7, 6.8, 6.12, 6.14, 6.15


Policy Value Basis

When entering into a contract, the financial obligations of all partiesshould be specified at the time the agreement is signed. This includesdisclosure of health status, age, and premium payments expected to fundbenefits and expenses associated with the contract.The Policy Value tV is the expected value of the future loss randomvariable Lt at time t:

tV = E[Lt ] = E[ Loss | T[x] > t] (172)


Policy Value Basis

Definition

The gross premium policy value for a policy in force at durationt ≥ 0 years after it was purchased is the expected value at that timeof the gross future loss random variable on a specified basis. Thepremiums used in the calculation are the actual premiums payableunder the contract.

The net premium policy value for a policy in force at durationt ≥ 0 years after it was purchased is the expected value at that timeof the net future loss random variable on a specified basis (whichmakes no allowance for expenses.) The premiums used in thecalculation are the net premiums calculated on the policy value basisusing the equivalence principle, not the actual premiums payable

It is important to note that usual practice dictates that when calculating

tV , premiums and premium-related expenses due at t are regarded asfuture payments and any insurance benefits and related expenses as pastpayments.


Recursion

Define

Pt as the premium payable at time t

et as the premium-related expense payable at time t

St+1 as the sum insured payable at time t + 1

Et+1 as the expense of paying the sum insured at time t + 1

t+1V as the gross premium policy value for a policy in force at timet + 1

Lt as the gross future loss random variable at time t

it as the interest rate from time t to time t + 1.


Recursion

Then, using recursion, we obtain

tV = et − Pt + q[x]+t ·St+1 + Et+1

1 + it+ p[x]+t ·

t+1V

1 + it. (173)

Notice that if there is a fixed term to the contract, such as an endowmentor term insurance, then we have the boundary condition

nV = 0 (174)

Also, if the premium is calculated using the EPP and the policy basis isthe same as the premium basis, then

0V = E[L0] = 0 (175)


BC : Endowment

For an endowment insurance contract with sum insured S , however, wehave the pair of boundary conditions

n−V = limε→0+

n−εV = S

nV = 0(176)

In calculating n−1V , we actually use n−V instead of nV . See next example!


Example 7.7

Consider a zero-expense, 20 year endowment policy purchased by a lifeaged 50. Level premiums of 23500 per year are payable annuallythroughout the term of the policy. A sum insured of 700000 is payable atthe end of the term if the life survives to age 70. On death before age 70,a sum insured is payable at the end of the year of death equal to the policyvalue at the start of the year in which the policyholder dies. Assuming theSSSM with interest at 3.5% per year, calculate 15V , the policy value inforce at the start of the 16th year.


Example 7.7

It follows that

St+1 = tV

et = 0 = Et

S = 700000

Pt = 23500

tV = −23500 + q[50]+ttV

1.035+ p[50]+t

t+1V

1.035

(177)


Example 7.7

Combining with our boundary value, we obtain the difference equation

tV =p[50]+t · t+1V − 24322.50

p[50]+t + 0.035

20−V = 700000.

(178)

Our initial iteration actually uses 20−V to obtain

19V =p69 · (20−V )− 24322.50

p69 + 0.035= 652401 (179)

Use tables or spreadsheet to calculate SSSM values and obtain

15V = 478063.


Example 7.7


tV =p[50]+t · t+1V − 24322.50

p[50]+t + 0.035

20−V = 700000.

(178)


19V =p69 · (20−V )− 24322.50

p69 + 0.035= 652401 (179)


15V = 478063.


Example 7.7


tV =p[50]+t · t+1V − 24322.50

p[50]+t + 0.035

20−V = 700000.

(178)


19V =p69 · (20−V )− 24322.50

p69 + 0.035= 652401 (179)


15V = 478063.


Example 7.1

Consider a 20−year endowment policy purchased by a life aged 50. Levelpremiums are payable annually throughout the term of the policy and thesum insured, S = 500000, is payable at the end of the year of death or atthe end of the term, whichever is sooner. The basis used by the insurancecompany for all calculations is under the SSSM with 5% per year interestand no allowance for expenses. Calculate P under the EPPP and thecorresponding policy values


Example 7.1

Substituting the information contained in the problem formation, we obtain

tV = −P + q[50]+t ·500000

1.05+ p[50]+t ·

t+1V

1.05

= p[50]+tt+1V − 500000

1.05+

500000

1.05− P

0V = 0 = E[L0] = 500000A[50]:20 − Pa[50]:20

20−V = 500000

(180)

Solving for P, we obtain P = 15114.33. Iteration of the resultingdifference equation delivers the remaining policy values.


Example 7.4

A man aged 50 purchases a deferred annuity policy. The annuity will bepaid annually for life, with the first payment on his 60th birthday. Eachannuity payment will be 10000. Level premiums of 11900 are payableannually for at most 10 years. On death before age 60, all premiums paidwill be returned, without interest, at the end of the year of death. Theinsurer uses the following basis for calculation of policy values:

SSSM with 5% interest per year

Expenses of 10% of the first premium, 5% of subsequent premiums,25 each time an annuity payment is paid, and 100 when a death claimis paid.

Calculate tV for t ∈ 0, 1, ..., 9


Example 7.4

Our initial policy value is

0V = P · (IA) 1[50]:10

+ 100A 1[50]:10

+ 10025v 1010p[50]a60

−(

0.95a[50]:10 − 0.05)

P

= 485 > 0

(181)

This of course can now be used to forward iterate to find tV 9t=1. Since

0V = 485 > 0, the premiums charged correspond to a valuation basis thatis more conservative than the premium basis.

In general, we have for t ∈ 1, 2, ..., 9,

tV = P · (IA) 1[50]+t:10−t + (tP + 100)A 1

[50]+t:10−t

+ 10025v 10−t10−tp[50]+t a60 − 0.95Pa[50]+t:10−t

(182)


Example 7.4

Our initial policy value is

0V = P · (IA) 1[50]:10

+ 100A 1[50]:10

+ 10025v 1010p[50]a60

−(

0.95a[50]:10 − 0.05)

P

= 485 > 0

(181)

This of course can now be used to forward iterate to find tV 9t=1. Since

0V = 485 > 0, the premiums charged correspond to a valuation basis thatis more conservative than the premium basis.

In general, we have for t ∈ 1, 2, ..., 9,

tV = P · (IA) 1[50]+t:10−t + (tP + 100)A 1

[50]+t:10−t

+ 10025v 10−t10−tp[50]+t a60 − 0.95Pa[50]+t:10−t

(182)


Example 7.4

For 1 ≤ t ≤ 9, we can see our recursion equation is

tV = −0.95P + q[50]+t ·(t + 1) · P + 100

1.05+ p[50]+t ·

t+1V

1.05(183)

For t ≥ 10, we have

t−V = 10025a[50]+t

t+V = 10025a[50]+t = t−V − 10025(184)

Which do we use to find 9V , 10−V or 10+V ? Recall that from (182), wehave

9V = P · (IA) 159:1

+ (9P + 100)A 159:1

+ 10025p59v a60 − 0.95Pa59:1

= Pvq59 + (9P + 100)vq59 + 10025p59v a60 − 0.95P

= −0.95P + q5910P + 100

1.05+ p59

10025a60

1.05.


Notes

The previous example shows that sometimes we need to calculate theinitial value, given the information contained in the problemstatement, to iterate forward, especially if there is no term n andcorresponding boundary condition nV . Also, no annuity paymentshave occurred yet and this reflects in the expenses.

It is likely that DSAR := St+1 + Et+1 − t+1V 6= 0. The Death StrainAt Risk, or DSAR, is the extra amount needed to increase the policyvalue to the death benefit at time t + 1. This is a capital based riskmeasure, as it is a direct measure of what the insurer may be at riskof needing to close out a contract if a benefit must be paid. If theDSAR is large enough, management may want to purchasereinsurance in case a large DSAR (even with low probability) occurs.


Example 7.3

A woman aged 60 purchases a 20 year endowment insurance with a suminsured S = 100000 payable at the end of the year of death or on survivalto age 80, whichever occurs first. An annual premium of 5200 ispayable for at most 10 years. The insurer uses the following basis forcalculation of policy values:

SSSM with 5% interest per year

Expenses of 10% of the first premium, 5% of subsequent premiums,and 200 on payment of the sum insured.

Let’s set up an algorithm to calculate the policy values 0V , 1V , ..., 9V .


Example 7.3

Our initial recursion equation is

0V = −0.9P + q[60] ·100200

1.05+ p[60] ·

1V

1.05. (186)

For 1 ≤ t ≤ 9, we have

tV = −0.95P + q[60]+t ·100200

1.05+ p[60]+t ·

t+1V

1.05. (187)

For t = 10, we have

10V = E [L10] = E[100200v minK70+1,10

]= 100200A70:10 = 63703.

(188)

HW Compute 9V and 12V explicitly.


Backwards or Forwards (Q16 from SOA Sample Set)

A whole life insurance contact (wl) is issued to an age 40, with deathbenefit 1000, has P40 defined as the premium for a whole lifeinsurance issued to age (40) with benefit 1, under the equivalencepricing principle.

Now, assume that in fact a modified whole life (mwl) contractissued to (40), where now the death benefit is 1000 for the first 20years, 5000 for the next 5 years, and 1000 thereafter. The annual netpremium is 1000P40 for the first 20 years, 5000P40 for the next 5years, and a value π thereafter.

Assume

(A40,A60,A1

60:5) = (0.16132, 0.36913, 0.06674)

(q60,v20

20p40, v5

5p60) = (0.01376, 0.27414, 0.68756)

(a40, a40:20 , a60, a60:5 , a65) = (14.8166, 11.7612, 11.1454, 4.3407, 9.8969)(189)

Given the information above, how can we calculate 20V mwl and 21V mwl?Albert Cohen (MSU) STT 455-6: Actuarial Models MSU 2020 180 / 348


Going forward, i.e. valuing the cash flows prospectively, and substitutingthe data found in (189), leads to

0 = 0V wl = 1000× A40 − 1000P40a40

⇒ 1000P40 = 1000× A40

a40= 1000× 0.16132

14.8166= 10.89.

(190)

Similarily,

0 = 0V mwl =(

1000A40 + 4000v 2020p40A 1

60:5

)−(1000P40a40:20 + 5000P40v 20

20p40a60:5 + πv 2020p40v 5

5p60a65

)⇒ π = 22.32.

(191)



Continuing our prospective analysis, we calculate

20V mwl = 4000A 160:5

+ 1000A60 − 5000P40a60:5 − πv 55p60a65

= 247.86 ≈ 248

⇒ 21V mwl =(20V + 5000P40) (1.06)− 5000q60

p60

=(248 + (5× 10.89)) (1.06)− (5000× 0.01376)

1− 0.01376

= 255.31 ≈ 255.

(192)



However, notice that 0 = 0V mwl and that, like a regular whole lifecontract, for t ∈ 0, 1, ..., 19

tVmwl = −1000P40 + q40+t

1000

1 + rt+ p40+t

t+1V mwl

1 + rt. (193)

Iterating forward, we see that

20V mwl = 20V wl = 1000A60 − 1000P40a60

= 1000×(

A60 −A40

a40a60

)= 1000×

(1− da60 −

1− da40

a40a60

)= 1000×

(a40 − a60

a40

)= 1000×

(14.8166− 11.1454

14.8166

)≈ 248.

(194)


Annual Profit - Example 7.8

An insurer issues a large number of policies identical to the policy inExample 7.3 to women aged 60. Five years after they were issued, a totalof 100 of these policies were still in force. In the following year, one persondied (d5 = 1) and

expenses of 6% of each premium paid were incurred - i.e.eactual5 = 0.06P5

interest was earned at 6.5% on all assets - i.e. iactual5 = 0.065

expenses of E actual6 = 250 were incurred on the payment of the sum

insured for the policyholder who died.

Calculate a.) the profit or loss on the group of policies for this year and b.)determine how much of this profit or loss is attributable to profit or lossfrom mortality, from interest, and from expenses.



To solve, we compute the difference between the growth of assets fromt = 5 to t = 6 and subtract the total asset value at t = 6:

Profit = N · (5V + P5 − 0.06P5) · (1 + i5)1

− (d5 · (S + E6) + (N − d5) · 6V )

= 100 · (5V + (0.94)(5200)) · (1.065)1

− (1 · (S + E6) + 99 · 6V )

= 106.5 · (5V + 4888)− (100250 + 99 · 6V )

(195)



Furthermore,

5V = E[L5] = 100200A65:15 − 0.95 · 5200a65:5

= 29068

6V = E[L6] = 100200A66:14 − 0.95 · 5200a66:4

= 35324

∴ Profit = 106.5 · (29068 + 4888)− (100250 + (99)(35324))

= 18919

(196)

HW: Read the solution for part b.) in the textbook.



Furthermore,

5V = E[L5] = 100200A65:15 − 0.95 · 5200a65:5

= 29068

6V = E[L6] = 100200A66:14 − 0.95 · 5200a66:4

= 35324

∴ Profit = 106.5 · (29068 + 4888)− (100250 + (99)(35324))

= 18919

(196)

HW: Read the solution for part b.) in the textbook.



Define ASt as the share of the insurer’s assets attributable to each policyin force at time t. Consider now a policy identical to the policy studied inExample 7.4 and suppose that this policy has now been in force for fiveyears. Suppose that over the past five years, the insurer’s experience inrespect of similar policies has realized annual interest on investments as(i1, i2, i3, i4, i5) = (0.048, 0.056, 0.052, 0.049, 0.047).



Furthermore,

Expenses at the start of the year in which a policy was issued were15% of the premium

Expenses at the start of the year after the year in which a policy wasissued were 6% of the premium

The expense of paying a death claim was, on average, 120

The mortality rate q[50]+t ≈ 0.0015 for t ∈ 0, 1, 2, 3, 4Calculate ASt using the convention that ASt does not include thepremium and related expense due at time t. (This of course means thatAS0 = 0, as no premiums paid in and no benefits paid out yet.)



We calculate AS1 here and refer to Table 7.1 for the complete set ofcalculations

At time 0, insurer receives premiums minus expenses of0.85 · 11900N = 10115N.

At time 1, this accumulates to 10115N · (1 + i1) = 10601N.

A total of 0.0015N policy holders die in the first year and deathclaims are 0.0015N · (11900 + 120) = 18N.

Therefore, the value of the fund at the end of the first year is10601N − 18N = 10582N.



It follows that

AS1 =Fund Value at time 1

Number of Policies in Force at time 1

=10582N

0.9985N= 10598

(197)

Now, read Section 7.4 on computing Valuation between premium dates.


Policy Values - Cts Cash Flows

Recall that

tV = et − Pt + q[x]+t ·St+1 + Et+1

1 + it+ p[x]+t ·

t+1V

1 + it. (198)

Now, consider that t is real-valued and define

Pt as the annual rate of premium payable at time t

et as the annual rate of premium-related expense payable at time t

St as the sum insured payable at time t if the policy holder diesexactly at t

Et as the expense of paying the sum insured at time t

µ[x]+t as the force of mortality at age [x ] + t

δt as the force of interest assumed at time t

tV as the policy value at time t.



Now, as the force of interest varies, we have

v(t) = e−∫ t

0 δudu

v(t)

v(s)= e−

∫ ts δudu

(199)

and so

tV =

∫ ∞0

v(t + s)

v(t)·([St+s + Et+s ] · sp[x]+tµ[x]+t+s

)ds

−∫ ∞

0

v(t + s)

v(t)·([Pt+s − et+s ] · sp[x]+t

)ds

(200)

Q: What happens if there is a finite term to contract?



By the product rule and the identities

r−tp[x]+t =rp[x]

tp[x]

d

dt

(tp[x]

)= −tp[x]µ[x]+t

v ′(t) = −δtv(t)

(201)

we obtain the ODE

d

dt(tV ) = δt · tV + Pt − et − (St + Et − tV )µ[x]+t (202)



Boundary Conditions:

For S sum insured, we have

limt→n−

tV = S for an endowment policy with term n years.

limt→n−

tV = 0 for a term policy with term n years.

limt→ω−

tV = S for a whole life policy with upper limit ω years.

(203)

Forward Euler:

t+hV = tV + h ·(δt · tV + Pt − et − (St + Et − tV )µ[x]+t

)(204)



As an example, consider the case where for S sum insured, we have

St+s = S

et+s = 0 = Et+s

δt = δ

µ[x]+t+s = λ

Pt+s = Pe−γ(t+s)

(205)

Then it follows that

tV =

∫ ∞0

Se−δs · λe−λsds −∫ ∞

0e−δs · Pe−γ(t+s)e−λsds

=Sλ

λ+ δ− Pe−γt

λ+ δ + γ

(206)


Policy Alterations

In many cases, policyholders may wish to change the terms of theircontract if it is still in effect. For example:

They may wish to stop making premiums, or to change the terms oftheir benefit payout.

They may wish to cash out their position, or simply wish to shortenthe time remaining until payout.

One may argue that the insurer is under no obligation to make suchchanges if they are not written expressly into the initial contract. Forexample:

The policyholder (but not insurer) may know something about theirhealth status that would make it better for them to cash out now.

By having to liquidate assets that cover the policy, the insurer mayhave to take a loss to be able to settle the alteration, and this couldaffect other policyholders adversely.


Policy Alterations

In many cases, policyholders may wish to change the terms of theircontract if it is still in effect. For example:

They may wish to stop making premiums, or to change the terms oftheir benefit payout.

They may wish to cash out their position, or simply wish to shortenthe time remaining until payout.

One may argue that the insurer is under no obligation to make suchchanges if they are not written expressly into the initial contract. Forexample:

The policyholder (but not insurer) may know something about theirhealth status that would make it better for them to cash out now.

By having to liquidate assets that cover the policy, the insurer mayhave to take a loss to be able to settle the alteration, and this couldaffect other policyholders adversely.


Policy Alterations

Because of these concerns, the lender may agree to alter the terms of thecontract, but only paying a Surrender (Cash) Value Ct of a fraction of

tV or ASt .

Ct = E [PVt(future benefits + expenses, altered contract)]

− E [PVt(future premiums, altered contract)](207)

In allowing the policy to lapse, the policy holder is cashing out a policyand using the proceeds to enter into a new contract. If the period betweenlapsing and entering into a new contract is too short, then the insurer maysuffer from not earning enough income to cover the new business strain ofwriting the first contract. Hence, some countries including the US havenon-forfeiture laws that allow for zero cash values for early surrenders.


Policy Alterations

Because of these concerns, the lender may agree to alter the terms of thecontract, but only paying a Surrender (Cash) Value Ct of a fraction of

tV or ASt .

Ct = E [PVt(future benefits + expenses, altered contract)]

− E [PVt(future premiums, altered contract)](207)

In allowing the policy to lapse, the policy holder is cashing out a policyand using the proceeds to enter into a new contract. If the period betweenlapsing and entering into a new contract is too short, then the insurer maysuffer from not earning enough income to cover the new business strain ofwriting the first contract. Hence, some countries including the US havenon-forfeiture laws that allow for zero cash values for early surrenders.


Policy Alterations: Example 7.13

Consider the policy discussed in Examples 7.4 and 7.9. Given theexperience of the insurer detailed in Example 7.9, at the start of the 6th

year but before paying the premium due, the policyholder requests that thepolicy be altered in one of the following ways:



1 The policy is surrendered immediately

2 No more premiums are paid, and a reduced annuity is payable fromage 60. In this case, all premiums paid are refunded at the end of theyear of death if the policyholder dies before age 60.

3 Premiums continue to be paid, but the benefit is altered from anannuity to a lump sum (pure endowment) payable on reaching age 60.Expenses and benefits on death before age 60 follow the originalpolicy terms. There is an expense of 100 associated with paying thesum insured at the new maturity date.



Calculate the surrender value, the reduced annuity, and sum insuredassuming the insurer uses

90% of the asset share less a charge of 200 or

90% of the policy value less a charge of 200

together with the assumptions in the policy value basis when calculatingrevised benefits and premiums. Use the associated values

5V = 65470

AS5 = 63509(208)



1

C assetshare5 = 0.9 · AS5 − 200 = 56958

Cpolicyvalue5 = 0.9 · 5V − 200 = 58723

(209)

2

C5 = 5 · 11900A 155:5

+ 100A 155:5

+ (X + 25) · v 55p55 · a60

X assetshare = 4859

X policyvalue = 5012

(210)

3

C5 + 0.95 · 11900a55:5 = 11900(

(IA) 155:5

+ 5A 155:5

)+ 100A 1

55:5+ v 5

5p55 (S + 100)

Sassetshare = 138314

Spolicyvalue = 140594

(211)



1

C assetshare5 = 0.9 · AS5 − 200 = 56958

Cpolicyvalue5 = 0.9 · 5V − 200 = 58723

(209)

2

C5 = 5 · 11900A 155:5

+ 100A 155:5

+ (X + 25) · v 55p55 · a60

X assetshare = 4859


(210)

3

C5 + 0.95 · 11900a55:5 = 11900(

(IA) 155:5

+ 5A 155:5

)+ 100A 1

55:5+ v 5

5p55 (S + 100)



(211)



1

C assetshare5 = 0.9 · AS5 − 200 = 56958

Cpolicyvalue5 = 0.9 · 5V − 200 = 58723

(209)

2

C5 = 5 · 11900A 155:5

+ 100A 155:5

+ (X + 25) · v 55p55 · a60

X assetshare = 4859


(210)

3

C5 + 0.95 · 11900a55:5 = 11900(

(IA) 155:5

+ 5A 155:5

)+ 100A 1

55:5+ v 5

5p55 (S + 100)



(211)


Related Project Topics

Over- or underestimated interest rates are only one risk factor foractuarial reserving. Another very real factor is known as longevityrisk, which is due to the possibility that a pensioner may live longerthan expected. Hedging against such a possibility is extremelyimportant, Please consult the paper by Tsai, Tzeng, and Wang onHedging Longevity Risk When Interest Rates Are Uncertain


https://www.soa.org/Library/Journals/NAAJ/2011/june/naaj-2011-vol15-no2-tsai.aspx

https://www.soa.org/Library/Journals/NAAJ/2011/june/naaj-2011-vol15-no2-tsai.aspx

Related Project Topics

For those of you interested in more sophisticated, cutting edge codingmethods for reserving, code.google.com has a site dedicated toChainLadder (google code name chainladder) that contains an Rpackage providing methods which are typically used in insuranceclaims reserving. Links to slides explaining the method are also on thesite

An Introduction to R: Examples for Actuaries by Nigel de Silva is avery nice primer on using R.


http://code.google.com/p/chainladder/





http://toolkit.pbworks.com/f/R Examples for Actuaries v0.1-1.pdf

Homework Questions

HW: 7.1, 7.2, 7.4, 7.5, 7.8, 7.12, 7.14, 7.15


Two State Model: Alive or Dead

Recall that for the survival time Tx of an individual (x), we have

Sx(t) = 1− Fx(t) = 1− P[Tx ≤ t] (212)

We now extend the model to include multiple states, but first we definethe random variable Y (t) ∈ 0, 1 as the state of the individual (x). If (x)is alive at time x + t, then Y (t) = 0. Otherwise, Y (t) = 1.Hence, we can define

Tx = max t | Y (t) = 0 (213)

and the model flow 0→ 1.


Accidental Death Model

We can also define

Y (t) =

0 if (x) is alive at time x + t1 if (x) is dead at time x + t of accidental cause2 if (x) is dead at time x + t of other cause

0

>>>>>>>>

1 2

Figure: ADM Flow Chart

There is a sum insured upon leaving state 0, but that sum is dependent onentering state 1 or 2.


Permanent Disability Model

However, we can go even further and define

Y (t) =

0 if (x) is alive at time x + t1 if (x) is disabled at time x + t2 if (x) is dead at time x + t

0 //

>>>>>>>>1

2

Figure: PDM Flow Chart

There is a lump sum paid upon entering state 1, an annuity paid while instate 1, and a lump sum paid upon entering state 2.


Disability Income Insurance Model

However, we can go even further and define

Y (t) =

0 if (x) is alive and healthy at time x + t1 if (x) is alive and sick at time x + t2 if (x) is dead at time x + t

0 //

>>>>>>>>1oo

2

Figure: DIIM Flow Chart

Premium is paid while in state 0, is a lump sum paid upon entering state1, an annuity paid while in state 1, and a lump sum paid upon enteringstate 2.


Joint Life - Last Survivor

Define

Joint Life Annuity - annuity that pays until the first death among agroup of lives

Last Survivor Annuity - annuity that pays until the last deathamong a group of lives

A common feature is payment rate decreases upon each death

Reversionary Annuity - life annuity that starts payment upon deathof a specified life, as long as another member of group is alive

Joint Life Insurance - life annuity that starts payment upon firstdeath of a member of group

Usually, group consists of two members, a husband and a wife


Joint Model

For example, consider a policy issued to a group (H,W ) of age (x , y).Then,

Y (t) =

0 if H is alive at x + t and W is alive at y + t1 if H is alive at x + t and W is dead at y + t2 if H is dead at x + t and W is alive at y + t3 if H is dead at x + t and W is dead at y + t

0 //

1

2 // 3

Figure: Joint Model Flow Chart


Notation

Assuming that the group (which can consist of 1,2, or more individuals)can be found in an of the the n + 1 states 0, 1, 2, ..., n − 1, n, we definethe event

Y (t) = i (214)

to mean the group is in state i at time t.

It follows that Y (t)t≥0 is a discrete valued stochastic process.


Assumptions

We make the following assumptions and definitions about transitionsbetween states and their associated probabilities:

P[Y (x + t) = j | Y (x) = i ] := tpijx (Markovity)

P[Y (x + s) = i for all s ∈ [0, t] | Y (x) = i ] := tpiix

limh→0

P[2 or more transitions in interval of length h]

h= 0

limh→0+

hpijx

h:= µijx

d

dt

(tp

ijx

)exists for all t ≥ 0

(215)


Assumptions




limh→0


h= 0

limh→0+

hpijx

h:= µijx

d

dt

(tp

ijx


(215)


Assumptions




limh→0


h= 0

limh→0+

hpijx

h:= µijx

d

dt

(tp

ijx


(215)


Assumptions




limh→0


h= 0

limh→0+

hpijx

h:= µijx

d

dt

(tp

ijx


(215)


Assumptions




limh→0


h= 0

limh→0+

hpijx

h:= µijx

d

dt

(tp

ijx


(215)


Note:

tp00x = tpx

tp01x = tqx

tp10x = 0

0pijx = 1i=j

µ01x = µx

hpijx = h · µijx + o(h)

tpiix ≤ tp

iix

(216)

As an example, we can show that for the permanent disability model

up01x =

∫ u

0tp

00x · µ01

x+t · u−tp11x+tdt. (217)


Note:

tp00x = tpx

tp01x = tqx

tp10x = 0

0pijx = 1i=j

µ01x = µx


tpiix ≤ tp

iix

(216)


up01x =

∫ u

0tp

00x · µ01

x+t · u−tp11x+tdt. (217)


Note:

tp00x = tpx

tp01x = tqx

tp10x = 0

0pijx = 1i=j

µ01x = µx


tpiix ≤ tp

iix

(216)


up01x =

∫ u

0tp

00x · µ01

x+t · u−tp11x+tdt. (217)


P[Remaining in State i from age x to x + t]

Theorem

hpiix = 1− h ·

n∑j=0,j 6=i

µijx + o(h)

tpiix = exp

−∫ t

0

n∑j=0,j 6=i

µijx+sds

(218)


P[Remaining in State i from age x to x + t]

Proof.

P[(x , i)→ (x + h, i)] = 1− P[(x , i) 9 (x + h, i)]

= 1− h ·n∑

j=0,j 6=i

µijx + o(h).

∴ t+hpiix = tp

iix · hpii

x+t = tpiix ·(

1− h ·n∑

j=0,j 6=i

µijx+t + o(h))

⇒ d

dt

(tp

iix

)= lim

h→0

t+hpiix − tp

iix

h= −tp

iix ·

n∑j=0,j 6=i

µijx+t

⇒ tpiix = 0pii

x · e−∫ t

0

∑i 6=j µ

ijx+sds = e−

∫ t0

∑i 6=j µ

ijx+sds .

(219)


Kolmogorov Forward Equations

Using the vocabulary of probabilists, we define the Kolmogorov forwardequations for the evolution of the densities of the birth death Markovprocess Y as

d

dttp

ijx =

n∑k=0,k 6=j

tpikx µ

kjx+t − tp

ijx

n∑k=0,k 6=j

µjkx+t (220)



In matrix notation, for a fixed x , define the matrices P(t),Q(t) such that

[P(t)]i ,j = tpijx

Q(t) =

−∑n

k=1 µ0kx+t µ01

x+t · · · µ0nx+t

µ10x+t −

∑nk=0,k 6=1 µ

1kx+t · · · µ1n

x+t...

.... . .

...

µn0x+t µn1

x+t · · · −∑n−1

k=0 µnkx+t

(221)

and the corresponding ODE system is

P ′(t) = P(t)Q(t)

P(0) = I = Identity Matrix.(222)



In matrix notation, for a fixed x , define the matrices P(t),Q(t) such that

[P(t)]i ,j = tpijx

Q(t) =

−∑n

k=1 µ0kx+t µ01

x+t · · · µ0nx+t

µ10x+t −

∑nk=0,k 6=1 µ

1kx+t · · · µ1n

x+t...

.... . .

...

µn0x+t µn1

x+t · · · −∑n−1

k=0 µnkx+t

(221)

and the corresponding ODE system is

P ′(t) = P(t)Q(t)

P(0) = I = Identity Matrix.(222)



Here, Q is referred to as the transition intensity matrix. We can workwith off diagonal entries as the diagonal entries are dependent on them.Also, a whole row of the matrix is filled by zeros if there is no transitionout of the state corresponding to the row.



Consider the case where Q is time-independent. Also, consider thediagonalization of Q via

Q = UDU−1

D =

λ1 0 · · · 00 λ2 · · · 0...

.... . .

...0 0 · · · λn

(223)

where λ1, λ2, · · · , λn are the eigenvalues of Q and U is the matrixcomposed of the corresponding eigenvectors.



Then P(t) = etQP(0), where etQ =∑∞

k=0tk

k! Qk .

If Q is diagonalizable, then

etQ = UetDU−1

etD =

etλ1 0 · · · 0

0 etλ2 · · · 0...

.... . .

...0 0 · · · etλn

(224)

Question: What if Q is in fact time dependent?


0→ 1 transition matrix

For the regular alive-dead model with constant force of mortality µ, therate matrix is

Q =

(−µ µ0 0

)=

(1 10 1

)(−µ 00 0

)(1 −10 1

)(225)

It follows that we retain

P(t) = etQ =

(e−µt 1− e−µt

0 1

)(226)

as in the previous sections.


Zombies!!

Consider a fun toy problem that remains in the 2× 2 matrix setting:

A zombie population has the rate matrix

Q =

(−1 12 −2

)=

(−1 12 1

)(−3 00 0

)(−1

313

23

13

)(227)

It follows that

P(t) = etQ =

(−1 12 1

)(e−3t 0

0 1

)(−1

313

23

13

)(228)

Question As t →∞, can you say anything about the percentage ofhumans in the total population?


Zombies!!

Consider a fun toy problem that remains in the 2× 2 matrix setting:

A zombie population has the rate matrix

Q =

(−1 12 −2

)=

(−1 12 1

)(−3 00 0

)(−1

313

23

13

)(227)

It follows that

P(t) = etQ =

(−1 12 1

)(e−3t 0

0 1

)(−1

313

23

13

)(228)

Question As t →∞, can you say anything about the percentage ofhumans in the total population?


Zombies - the general case..

For a general zombie population constant rate matrix

Q =

(−a ab −b

)=

(1 − a

b1 1

)(0 00 −(a + b)

)( ba+b

ab

ba+b

− ba+b

ba+b

)(229)

It follows that

P(t) =

(1 − a

b1 1

)(1 0

0 e−(a+b)t

)( ba+b

ab

ba+b

− ba+b

ba+b

)=

(b+ae−(a+b)t

a+ba−ae−(a+b)t

a+bb−be−(a+b)t

a+ba+be−(a+b)t

a+b

)

→( b

a+ba

a+bb

a+ba

a+b

) (230)




Q =

(−a ab −b

)=

(1 − a

b1 1

)(0 00 −(a + b)

)( ba+b

ab

ba+b

− ba+b

ba+b

)(229)

It follows that

P(t) =

(1 − a

b1 1

)(1 0

0 e−(a+b)t

)( ba+b

ab

ba+b

− ba+b

ba+b

)

=

(b+ae−(a+b)t

a+ba−ae−(a+b)t

a+bb−be−(a+b)t

a+ba+be−(a+b)t

a+b

)

→( b

a+ba

a+bb

a+ba

a+b

) (230)




Q =

(−a ab −b

)=

(1 − a

b1 1

)(0 00 −(a + b)

)( ba+b

ab

ba+b

− ba+b

ba+b

)(229)

It follows that

P(t) =

(1 − a

b1 1

)(1 0

0 e−(a+b)t

)( ba+b

ab

ba+b

− ba+b

ba+b

)=

(b+ae−(a+b)t

a+ba−ae−(a+b)t

a+bb−be−(a+b)t

a+ba+be−(a+b)t

a+b

)

→( b

a+ba

a+bb

a+ba

a+b

) (230)




Q =

(−a ab −b

)=

(1 − a

b1 1

)(0 00 −(a + b)

)( ba+b

ab

ba+b

− ba+b

ba+b

)(229)

It follows that

P(t) =

(1 − a

b1 1

)(1 0

0 e−(a+b)t

)( ba+b

ab

ba+b

− ba+b

ba+b

)=

(b+ae−(a+b)t

a+ba−ae−(a+b)t

a+bb−be−(a+b)t

a+ba+be−(a+b)t

a+b

)

→( b

a+ba

a+bb

a+ba

a+b

) (230)


Example 8.4

Suppose you are given the transition intensity matrix for the permanentdisability model as follows:µ00

x µ01x µ02

x

µ10x µ11

x µ12x

µ20x µ21

x µ22x

=

−0.0508 0.0279 0.02290.0000 −0.0229 0.02290.0000 0.0000 0.0000

(231)

Then

10p0060 = 10p00

60 = e−∫ 10

0 (0.0279+0.0229)ds = 0.60170

10p0160 =

∫ 10

0tp

0060 · µ01

60+t · 10−tp1160+tdt

=

∫ 10

0e−

∫ t0 (0.0279+0.0229)ds · 0.0279 · e−

∫ 10−t0 (0.0229)dsdt

= 0.19363

(232)


MLC Q12 / Nov 2012

A party of scientists arrives at a remote island. Unknown to them, ahungry tyrannosaur lives on the island. You model the future lifetimes ofthe scientists as a three-state model, where:

State 0: no scientists have been eaten.

State 1: exactly one scientist has been eaten.

State 2: at least two scientists have been eaten.

You are given:

(i) Until a scientist is eaten, they suspect nothing, soµ01t = 0.01 + 0.02 · 2t

(ii) Until a scientist is eaten, they suspect nothing, so the tyrannosaurmay come across two together and eat both, with µ02 = 0.5 · µ01

t

(iii) After the first death, scientists become much more careful, soµ12 = 0.01

Calculate the probability that no scientists are eaten in the firstyear.


MLC Q12 / Nov 2012

This is essentially a Permanent Disability model, and so we can computethe transition probabilities accordingly:

tp000 = exp

−∫ t

0

2∑j=1

µ0js ds

= exp

(−∫ t

01.5(0.01 + 0.02 · 2s)ds

)⇒ 1p00

0 = 0.943.

(233)

HW Compute the other transition probabilities using both the integralequation and the matrix exponential method.


Example 8.5 (Forward Euler Method)

Suppose you are given the transition intensity matrix for the disabilityincome insurance model as follows:

µ00x µ01

x µ02x

µ10x µ11

x µ12x

µ20x µ21

x µ22x

=

−µ01x − µ02

x a1 + b1ec1x a2 + b2ec2x

0.1 ·(µ01x

)−µ10

x − µ12x a2 + b2ec2x

0 0 0

(234)

for parameters

(a1 b1 c1a2 b2 c2

)=

[4× 10−4 3.4674× 10−6 0.1381555× 10−4 7.5858× 10−6 0.087498

](235)


Example 8.5

Then Forward Euler applied the Kolmogorov equations leads to

t+hp0060 = tp

0060 − h · tp00

60 ·(µ01

60+t + µ0260+t

)+ h · tp01

60 · µ1060+t + o(h)

t+hp0160 = tp

0160 − h · tp01

60 ·(µ12

60+t + µ1060+t

)+ h · tp00

60 · µ0160+t + o(h)

(236)

Ignoring the o(h) terms, we can iterate forward using, for example, h = 112 .


Example 8.5

In matrix-vector notation, we have(t+hp00

60

t+hp0160

)= [I− hA(t)]

(tp

0060

tp0160

)

I =

(1 00 1

)A(t) =

(µ01

60+t + µ0260+t −µ10

60+t

−µ0160+t µ12

60+t + µ1060+t

) (237)

Keep in mind that A is determined by the given transition intensity matrix.

HW Compute this vector over the interval t ∈ [0, 10] using a time step ofh = 1

12 . Use any numerical solver you like, but please have the valuescomputed into a pair of columns.


Example 8.5


60

t+hp0160

)= [I− hA(t)]

(tp

0060

tp0160

)I =

(1 00 1

)A(t) =

(µ01

60+t + µ0260+t −µ10

60+t

−µ0160+t µ12

60+t + µ1060+t

) (237)





Example 8.5


60

t+hp0160

)= [I− hA(t)]

(tp

0060

tp0160

)I =

(1 00 1

)A(t) =

(µ01

60+t + µ0260+t −µ10

60+t

−µ0160+t µ12

60+t + µ1060+t

) (237)





Permanent Disability Model

Upon finding the right diagonalization, one can show that for a PDM withthe rate matrix

Q =

−(a + b) a b0 −c c0 0 0

(238)

where a, b, c > 0, it follows that for a + b 6= c, P(t) =

e−(a+b)t aa+b−c (e−ct − e−(a+b)t) 1− a

a+b−c e−ct + c−ba+b−c e−(a+b)t

0 e−ct 1− e−ct

0 0 1

(239)


MLC Q16 / Nov 2012

Consider a Modified Disability Model where observed transition intensitiesare (µ01

t , µ10t , µ

12t ) = (0.02, 0.06, 0.10).

Using the Kolmogorov forward equations with step h = 0.5, calculate theprobability that a person currently disabled will be healthy at the end ofone year


MLC Q16 / Nov 2012

Recall that the Forward Euler method can be written explicitly as

f (t + h) ≈ f (t) + h · f ′(t) (240)

and so for the Kolmorgorov Forward ODE system,

t+hpijx = tp

ijx + h ·

( n∑k=0,k 6=j

tpikx µ

kjx+t − tp

ijx

n∑k=0,k 6=j

µjkx+t

)

= tpijx

1− hn∑

k=0,k 6=j

µjkx+t

+n∑

k=0,k 6=j

tpikx µ

kjx+th.

(241)


MLC Q16 / Nov 2012

Again, wlog set x = 0 and fix i , j . For

f (t) := tpij0

f (h) = f (0) + h · f ′(0)(242)

we have the system of equations

hp100 = 0p10

0 + h ·(

(0p110 µ10

0 + 0p120 µ20

0 )− 0p100 · (µ01

0 + µ020 ))

hp110 = 0p11

0 + h ·(

(0p100 µ01

0 + 0p120 µ21

0 )− 0p110 · (µ10

0 + µ120 ))

hp120 = 0p12

0 + h ·(

(0p100 µ02

0 + 0p110 µ12

0 )− 0p120 · (µ20

0 + µ210 )).

(243)


MLC Q16 / Nov 2012

Using the above, and h = 0.5, we obtain for the first iterates:

hp100 = h · µ10

0 = 0.03

hp120 = h · µ12

0 = 0.05

hp110 = 1− h · (µ10

0 + µ120 ) = 0.92.

(244)

Recursively, f (2h) = f (h) + h · f ′(h) and so for f (2h) = 2hpij0 we obtain

2hp100 = hp10

0 + h ·(

(hp110 µ10

h + hp120 µ20

h )− hp100 · (µ01

h + µ02h ))

= 0.03 + 0.5 ·(

(0.92)(0.06) + 0− (0.03)(0.02 + 0))

= 0.0573.

(245)


MLC Q16 / Nov 2012

Note that

Q =

−0.02 0.02 00.06 −0.16 0.10

0 0 0

= UDU−1

where U =

−0.133827 0.92683 0.577350.991005 0.375482 0.57735

0 0 0.57735

D =

−0.168102 0 00 −0.0118975 00 0 0

U−1 =

−0.387597 0.956735 −0.5691381.02298 0.138145 −1.16113

0 0 1.73205

(246)


MLC Q16 / Nov 2012

Note that

Q =

−0.02 0.02 00.06 −0.16 0.10

0 0 0

= UDU−1

where U =

−0.133827 0.92683 0.577350.991005 0.375482 0.57735

0 0 0.57735

D =

−0.168102 0 00 −0.0118975 00 0 0

U−1 =

−0.387597 0.956735 −0.5691381.02298 0.138145 −1.16113

0 0 1.73205

(246)


MLC Q16 / Nov 2012

It follows that

P(t) = UetDU−1

where U =

−0.133827 0.92683 0.577350.991005 0.375482 0.57735

0 0 0.57735

etD =

e−0.168102t 0 00 e−0.0118975t 00 0 1

U−1 =

−0.387597 0.956735 −0.5691381.02298 0.138145 −1.16113

0 0 1.73205

(247)

Compare these with your previously obtained Forward-Euler results.


MLC Q16 / Nov 2012

It follows that

P(t) = UetDU−1

where U =

−0.133827 0.92683 0.577350.991005 0.375482 0.57735

0 0 0.57735

etD =

e−0.168102t 0 00 e−0.0118975t 00 0 1

U−1 =

−0.387597 0.956735 −0.5691381.02298 0.138145 −1.16113

0 0 1.73205

(247)

Compare these with your previously obtained Forward-Euler results.


Premiums

Consider an annuity issued to a life (x) that pays at rate 1 per yearcontinuously while the life is in state j . Then the EPV of this annuity atforce of interest δ per year is

aijx = E[∫ ∞

0e−δt1Y (t)=j |Y (0)=idt

]=

∫ ∞0

e−δtE[1Y (t)=j |Y (0)=i

]dt =

∫ ∞0

e−δt tpijx dt

(248)

If the annuity is payable at the start of each year from the current time,based on the conditional event Y (t) = j | Y (0) = i, then the EPV is

aijx =∞∑k=0

vkkpij

x (249)


Premiums

If a unit benefit is payable to a life (x) on transition to state k , given thatit is currently in state i , then the EPV of this benefit is

Aikx =

∫ ∞0

∑j 6=k

e−δt tpijx µ

jkx+tdt (250)


Example 8.6

An insurer issues a 10−year disability income insurance policy to a healthylife aged 60. Use the model and parameters from i .) Example 8.5 and ii .)Example 8.4. Assume an effective rate of 5% per year and no expenses.Calculate the premiums for the following designs

(a) Premiums are payable continuously while in the healthy state. Abenefit of 20000 per year is payable continuously while in the disabledstate. A death benefit of 50000 is payable immediately upon death.

(b) Premiums are payable monthly in advance conditional on the lifebeing in the healthy state at the premium date. The sickness benefitof 20000 per year is payable monthly in arrear, if the life is in the sickstate at the payment date. A death benefit of 50000 is payableimmediately upon death.


Example 8.6

For case a.), we have via the EPP principle that

20000a0160:10

+ 50000A0260:10

− Pa0060:10

= 0 (251)

and so

P =20000

∫ 100 e−δt tp

0160dt∫ 10

0 e−δt tp0060dt

+50000

∫ 100 e−δt

(tp

0060µ

0260+t + tp

0160µ

1260+t

)dt∫ 10

0 e−δt tp0060dt

.

(252)

For a time step of h = 112 (monthly), we can use the forward-Euler results

from the previous example to calculate P = 3254.65.


Example 8.6

For case b.), we have via the EPP principle that

20000a(12)01

60:10+ 50000A02

60:10− Pa

(12)00

60:10= 0 (253)

and so

P =20000

∑119k=0 v

k12 k

12p01

60∑119k=0 v

k12 k

12p00

60

+50000

∫ 100 e−δt

(tp

0060µ

0260+t + tp

0160µ

1260+t

)dt∑119

k=0 vk12 k

12p00

60

.

(254)

Again, we can use the previously computed values of

k12

p0j60

(1,119)

(j ,k)=(0,0)to

calculate P = 3257.20.


Multiple Decrement Models

Consider the special case for transition matrix

Q(t) =

−∑n

k=1 µ0kx+t µ01

x+t · · · µ0nx+t

0 0 · · · 0...

.... . .

...0 0 · · · 0

(255)

Here, there are multiple exits from state 0, but no further transitions.

0

>>>>>>>>>

''NNNNNNNNNNNNNNNN

1 2 · · · n

Figure: Multiple Decrement Flow Chart


Multiple Decrement Models

tp00x = tp

00x = exp

[−∫ t

0

n∑i=1

µ0ix+sds

]

tp0ix =

∫ t

0sp00

x µ0ix+sds

0pijx = 1i=j

(256)

Note:

Premium is now different when compared to a policy that only allowstransition 0→ 1.

This is because µ00x+t = −

∑nk=1 µ

0kx+t < −µ01

x+t and so tp00x changes

accordingly.

See Example 8.8, where for example an insurer may allow for lowerpremiums via lapse support.


Double Decrement Models

Consider the extra special case for transition matrix

Q(t) =

−(a + b) a b0 0 00 0 0

(257)

Here, there are two possible exits from state 0, but no further transitions.

P(t) =

e−(a+b)t aa+b (1− e−(a+b)t) b

a+b (1− e−(a+b)t)

0 1 00 0 1

(258)

HW For this double decrement model, compute a01x , a

02x as well as A01

x

and A02x . What can you say about these financial instruments as b → 0?




Q(t) =

−(a + b) a b0 0 00 0 0

(257)


P(t) =


a+b (1− e−(a+b)t)

0 1 00 0 1

(258)


02x as well as A01

x





Q(t) =

−(a + b) a b0 0 00 0 0

(257)


P(t) =


a+b (1− e−(a+b)t)

0 1 00 0 1

(258)


02x as well as A01

x



Policy Values

Define

µijy as the transition intensity between states i and j at age y

δt as the force of interest per year at time t

B(i)t as the benefit payment rate while the policyholder is in state i

S(ij)t as the lump sum payment instantaneously at time t on transition

from state i to state j .

Assume the above are all members of C 0[0, n]. Then ∀i ∈ 0, 1, · · · , n,Thiele’s Differential Equation is

d

dttV

i = δt · tV i − B(i)t −

∑j 6=i

µijx+t

(S

(ij)t + tV

j − tVi)

(259)


Joint Life - Last Survivor Benefits

0µ01x+t:y+t//

µ02x+t:y+t

1

µ13x+t

2µ23y+t

// 3

Figure: Joint Model Transition Rates

Define the joint transition matrix via the flow chart above.



Define

tpxy = tp00xy = P[(x) and (y) are both alive in t years]

tqxy = tp01xy + tp

02xy + tp

03xy

= P[(x) and (y) are not both alive in t years]

tpxy = tp00xy + tp

01xy + tp

02xy

= P[ at least one of (x) and (y) is alive in t years]

tqxy = 1− tpxy

µx+t:y+t = µ01x+t:y+t + µ02

x+t:y+t

(260)



Also define

tq1xy = P[(x) dies before (y) and within t years]

=

∫ t

0rp00

xyµ02x+r :y+rdr

6= tp02xy

tq2xy = P[(x) dies after (y) and within t years]

=

∫ t

0rp01

xyµ13x+rdr

(261)


MLC Q1 / Nov 2012

For two lives, (80) and (90), with independent future lifetimes, you aregiven

p80+k = 0.9− 0.1k for k ∈ 0, 1, 2p90+k = 0.6− 0.1k for k ∈ 0, 1, 2

(262)

Calculate the probability that the last survivor will die in the third year.

By definition,

tpxy = tp00xy + tp

01xy + tp

02xy

= P[ at least one of (x) and (y) is alive in t years](263)

is the probability we seek to compute.


MLC Q1 / Nov 2012

For two lives, (80) and (90), with independent future lifetimes, you aregiven

p80+k = 0.9− 0.1k for k ∈ 0, 1, 2p90+k = 0.6− 0.1k for k ∈ 0, 1, 2

(262)

Calculate the probability that the last survivor will die in the third year.By definition,

tpxy = tp00xy + tp

01xy + tp

02xy

= P[ at least one of (x) and (y) is alive in t years](263)

is the probability we seek to compute.


MLC Q1 / Nov 2012

Using the assumption of independence of lives,

2p80:90 − 3p80:90 = P[ at least one of (80) and (90) is alive in 2 years]

− P[ at least one of (80) and (90) is alive in 3 years]

= P[ last survivor dies in the 3rd year]

=(

2p80 + 2p90 − 2p80:90

)−(

3p80 + 3p90 − 3p80:90

)=(

p80p81 + p90p91 − p80p81p90p91

)−(

p80p81p82 + p90p91p92 − p80p81p82p90p91p92

)= 0.24048.

(264)



Insurance Notation

axy = a00xy , the Joint Life Annuity with continuous payment of 1 per

year while both husband and wife are alive.

Axy the Joint Life Insurance with a unit payment immediately uponfirst death.

A1xy , the Contingent Insurance, a unit payment immediately upon

death of the husband given that he dies before his wife.



Insurance Notation

Axy = A03xy , the Last Survivor Insurance with unit payment

immediately upon second death.

ax |y = a02xy the Reversionary Annuity with a continuous payment at

unit rate per year while wife is alive given that husband has died..

axy = a00xy + a01

xy + a02xy , the Last Survivor Annuity, a continuous

payment at rate 1 per year while at least one person is alive.



It can be shown that

axy = ax + ay − axy

ax |y = ay − axy

Axy = Ax + Ay − Axy

axy =1− Axy

δ

(265)

HW

Prove this using explicit integral formulations.

Read over Examples 8.10 and 8.11.


MLC Q21 / Nov 2012

For a fully continuous whole life insurance issued on (x) and (y), you aregiven ∀t ≥ 0:

The death benefit of 100 is payable at the second death.

Premiums are payable until the first death.

The future lifetimes of (x) and (y) are dependent.

tpxy = λe−at + (1− λ)e−bt for some λ ∈ [0, 1].

tpx = e−at

tpy = e−ct for some c < a < b.

The force of interest is constant at δ > 0.

Calculate the annual benefit premium rate P for this insurance


MLC Q21 / Nov 2012

The equation of value here, assuming the EPP, is

0 = 100Axy − Paxy

⇒ P = 100Axy

axy

= 100Ax + Ay − Axy

axy

= 100Ax + Ay − (1− δaxy )

axy

(266)


MLC Q21 / Nov 2012

The equation of value here, assuming the EPP, is

0 = 100Axy − Paxy

⇒ P = 100Axy

axy

= 100Ax + Ay − Axy

axy

= 100Ax + Ay − (1− δaxy )

axy

(266)


MLC Q21 / Nov 2012

However,

axy =

∫ ∞0

e−δt(λe−at + (1− λ)e−bt

)dt

=λ

a + δ+

1− λb + δ

Ax =

∫ ∞0

e−δtae−atdt =a

a + δ

Ay =

∫ ∞0

e−δtce−ctdt =c

c + δ

⇒ P = 100

aa+δ + c

c+δ − 1 + δ(

λa+δ + 1−λ

b+δ

)λ

a+δ + 1−λb+δ

(267)


Example: Joint Life Benefits

For a special whole life insurance policy on (x) and (y) with dependentfuture lifetimes, you are given:

A death benefit of 105, 000 is paid at the end of the year of death ifboth (x) and (y) die within the same year. No death benefits arepayable otherwise.

px+k = 0.85 for all k ∈ 0, 1, 2, ..py+k = 0.8 for all k ∈ 0, 1, 2, ..px+k:y+k = 0.75 for all k ∈ 0, 1, 2, ..The yearly interest rate used is r = 0.05.

Calculate the expected present value of the death benefit.


Example: Joint Life Benefits

First, notice that

1px+k:y+k = px+k + py+k − px+k:y+k = 0.9

1qx+k:y+k = 1− 1px+k:y+k = 0.1

kpxy = Πk−1j=0 px+j :y+j = 0.75k .

(268)

It follows that the

EPV = 105000∞∑k=0

1

(1 + r)k+1 kpxy 1qx+k:y+k

= 105000∞∑k=0

1

(1.05)k+1(0.75k)(0.1)

=10000

1− 57

= 35000.

(269)


Transitions at Specific Ages

Example 8.12 : The employees (0) of a large corporation can leave thecorporation in three ways: they can retire (1), they can withdraw from thecorporation (2), or they can die while they are still employees (3).Consider the model

µ03x ≡ µ13

x ≡ µ23x = µx

µ02x =

µ02, if x < 60

0, if x ≥ 60

(270)

where retirement can only take place only on an employee’s60th, 61st , 62nd , 63rd , 64th, or 65th birthday. Assume that 40% ofemployees reaching exact age 60, 61, 62, 63 or 64 will retire at that ageand that 100% of all employees who reach age 65 retire immediately.



The corporation offers the following benefits to the employees:

For those employees who die while still employed, a lump sum of200000 is payable immediately upon death.

For those employees who retire, a lump sum of 150000 is payableimmediately upon death after retirement.

Theorem

Assuming a constant force of interest of δ per year and the notation of Ax

and nEx from single life computations based on a force of mortality µx , itfollows that the EPV of the Death after retirement benefit of anemployee currently aged 40 is

150000 · 20E40e−20µ02

(0.4 ·

[4∑

k=0

0.6kk|A60

]+ 0.65 · 5|A60

)(271)



To begin our proof, we can compute

E(40)[PV(Benefit) | retire at age 60] = 150000e−20δA60 (272)

20−p0040 = exp

[−∫ 20

0

(µ02 + µ03

40+t

)dt

]= exp

[−∫ 20

0(µ40+t) dt

]e−20µ02

= 20p40e−20µ02

P(40)[retire at age 60] = 0.4 · 20−p0040 = 0.4 · 20p40e−20µ02

20+p0040 = 0.6 · 20−p00

40

21−p0040 = 20+p00

40 · p60

21+p0040 = 0.6 · 21−p00

40 = 0.62 · 21p40e−20µ02

(273)



Also,

P(40)[retire at age 61] = X1X2X3X4

X1 = P(40)[survive in employment to age 60−] = 20−p0040

X2 = P(40)[will not retire at age 60] = 0.6

X3 = P(40)[(60+) will survive to age 61−] = 1p60

X4 = P(40)[will retire at age 61] = 0.4

E(40)[PV(Benefit) | retire at age 61] = 150000e−21δA61

(274)



We can repeat this until

P(40)[retire at age 65]

= 20−p0040 · 0.65 · 1p60 · 1p61 · 1p62 · 1p63 · 1p64

= 25p40e−20µ02 · 0.65

E(40)[PV(Benefit) | retire at age 65] = 150000e−25δA65

(275)

We now have enough information to complete the proof.



Proof.

For benefit after retirement, we have

E(40)[PV(Benefit)] =5∑

k=0

Bk(40)P

k(40)

Bk(40) = E(40)[PV(Benefit) | retire at age 60 + k]

= 150000e−(20+k)δA60+k for k ∈ 0, · · · , 5Pk

(40) := P(40)[retire at age 60 + k]

= 20+kp40e−20µ02 · 0.6k · 0.4 for k ∈ 0, · · · , 4

P5(40) = 25p40e−20µ02 · 0.65

(276)

Substitution and arithmetic lead to the form in the theorem statement.



Also notice that via the Tower property for conditional expectations, wehave a direct version of the proof:

E(40)[PV(Benefit)] = E(40)

[E(60)[PV(Benefit)]

]= 20E40e−20µ02 ·

5∑k=0

Bk(60)P

k(60)

(277)

where

Bk(60) = E(60)[PV(Benefit) | retire at age 60 + k]

= 150000e−kδA60+k for k ∈ 0, · · · , 5Pk

(60) := P(60)[retire at age 60 + k]

= kp60 · 0.6k · 0.4 for k ∈ 0, · · · , 4P5

(60) = 5p60 · 0.65

(278)


Markov Transportation Models

For example, consider a puppy at point 0 of a square lake. The state ofthe puppy is

Y (k) =

0 if the puppy is at corner 0 time k1 if the puppy is at corner 1 time k2 if the puppy is at corner 2 time k3 if the puppy is at corner 3 time k

0 //

2oo

1

OO

// 3

Figure: Joint Model Flow Chart



Now, consider the random variable

Tij = the time it takes to get from i to j . (279)

and the expected values

tij = E[Tij ]. (280)

Consider a transition probability matrix P such that

P =

p00 p01 p02 p03

p10 p11 p12 p13

p20 p21 p22 p23

p30 p31 p32 p33

=

0 1

212 0

12 0 0 1

212 0 0 1

20 0 0 1

. (281)

For this transition model, calculate t03.



For the above matrix, we can see that

t03 =1

2(1 + t13) +

1

2(1 + t23)

t13 =1

2(1 + t03) +

1

2(1)

t23 =1

2(1 + t03) +

1

2(1) .

(282)

It follows that

t03 =1

2

(1 +

1

2(1 + t03) +

1

2(1)

)+

1

2

(1 +

1

2(1 + t03) +

1

2(1)

)= 2 +

1

2t03

∴ t03 = 4.(283)



If the transition matrix is adapted to

P =

p00 p01 p02 p03

p10 p11 p12 p13

p20 p21 p22 p23

p30 p31 p32 p33

=

0 p 1− p 0

1− p 0 0 pp 0 0 1− p0 0 0 1

(284)

then the expected time t03 is computed via

t03 = p (1 + t13) + (1− p) (1 + t23)

t13 = (1− p) (1 + t03) + p (1)

t23 = p (1 + t03) + (1− p) (1) .

(285)



The matrix form of (285) is111

=

1 −p −(1− p)−(1− p) 1 0−p 0 1

t03

t13

t23

(286)

and inversion via WolframAlpha leads to

t03

t13

t23

=

1 −p −(1− p)−(1− p) 1 0−p 0 1

−1111

=

2

2p2−2p+12p2−4p+32p2−2p+1

2p2+12p2−2p+1

. (287)

Note that t03 = 4 when p = 12 .



Now consider an ant at corner 1 of a cube. The state of the ant is

Y (k) = j if the ant is at corner j ∈ 1, 2, ..., 8 at time k . (288)

Consider the general transition probability matrix P

P =

p11 p12 p13 p14 p15 p16 p17 p18

p21 p22 p23 p24 p25 p26 p27 p28

p31 p32 p33 p34 p35 p36 p37 p38

p41 p42 p43 p44 p45 p46 p47 p48

p51 p52 p53 p54 p55 p56 p57 p58

p61 p62 p63 p64 p65 p66 p67 p68

p71 p72 p73 p74 p75 p76 p77 p78

p81 p82 p83 p84 p85 p86 p87 p88

. (289)



For an ant at corner 1 that would like to get to corner 8,

P =

0 13 0 1

3 0 13 0 0

13 0 1

3 0 0 0 13 0

0 13 0 1

3 0 0 0 13

13 0 1

3 0 13 0 0 0

0 0 0 13 0 1

3 0 13

13 0 0 0 1

3 0 13 0

0 13 0 0 0 1

3 0 13

0 0 0 0 0 0 0 1

. (290)

Compute t18 = E[T18].



We can summarize the equations for the expected time to arrival as

t18 = 1 +1

3(t28 + t48 + t68)

t28 = 1 +1

3(t18 + t38 + t78)

t38 = 1 +1

3(t28 + t48 + t88)

t48 = 1 +1

3(t18 + t38 + t58)

t58 = 1 +1

3(t48 + t68 + t88)

t68 = 1 +1

3(t18 + t58 + t78)

t78 = 1 +1

3(t28 + t68 + t88)

t88 = 0.

(291)


Markov Chain Model of Employment

The US government has studied models of employment and has come upwith the following observation:

P[Unemployed finds job by end of the year] = pf ∈ (0, 1)

P[Employed loses job by end of the year] = pl ∈ (0, 1)

Define Wk to be the probability a worker is employed at the beginning ofyear k , Nk the probability she is not working at the beginning of year k .

Then (Wk+1

Nk+1

)=

[1− pl pf

pl 1− pf

](Wk

Nk

)(292)

Finally, assume that Wk + Nk = 1.

Question: Does Wk →W for some W ∈ (0, 1)?



The US government has studied models of employment and has come upwith the following observation:

P[Unemployed finds job by end of the year] = pf ∈ (0, 1)

P[Employed loses job by end of the year] = pl ∈ (0, 1)

Define Wk to be the probability a worker is employed at the beginning ofyear k , Nk the probability she is not working at the beginning of year k .

Then (Wk+1

Nk+1

)=

[1− pl pf

pl 1− pf

](Wk

Nk

)(292)

Finally, assume that Wk + Nk = 1.

Question: Does Wk →W for some W ∈ (0, 1)?



It follows that the matrix can be diagonalized as

(1− pl pf

pl 1− pf

)=

(pfpl−1

1 1

)(1 00 1− pf − pl

)( plpf +pl

plpf +pl

− plpf +pl

pfpf +pl

)(293)

Define

~qk =

(pl

pf +pl

plpf +pl

− plpf +pl

pfpf +pl

)(Wk

Nk

)(294)



It follows that the matrix can be diagonalized as

(1− pl pf

pl 1− pf

)=

(pfpl−1

1 1

)(1 00 1− pf − pl

)( plpf +pl

plpf +pl

− plpf +pl

pfpf +pl

)(293)

Define

~qk =

(pl

pf +pl

plpf +pl

− plpf +pl

pfpf +pl

)(Wk

Nk

)(294)



Hence,

~qk+1 =

(1 00 1− pf − pl

)~qk ⇒ ~qk =

(A

B(1− pf − pl)k

)(295)

Returning to our original notation,

(Wk

Nk

)=

(pfpl−1

1 1

)(1 00 1− pf − pl

)~qk

=

(pfpl−1

1 1

)(1 00 1− pf − pl

)(A

B(1− pf − pl)k

)=

(pfpl−1

1 1

)(A

B(1− pf − pl)k+1

)=

(Apf

pl− B(1− pf − pl)

k+1

A + B(1− pf − pl)k+1

)(296)



Hence,

~qk+1 =

(1 00 1− pf − pl

)~qk ⇒ ~qk =

(A

B(1− pf − pl)k

)(295)

Returning to our original notation,

(Wk

Nk

)=

(pfpl−1

1 1

)(1 00 1− pf − pl

)~qk

=

(pfpl−1

1 1

)(1 00 1− pf − pl

)(A

B(1− pf − pl)k

)=

(pfpl−1

1 1

)(A

B(1− pf − pl)k+1

)=

(Apf


k+1

A + B(1− pf − pl)k+1

)(296)



Solving for our parameters A,B, we see that(W0

N0

)=

(Apf


A + B(1− pf − pl)

)

⇒(

AB

)=

11+

pfplpf

plN0−W0

1−pf−pl

∴

(Wk

Nk

)→

pfpl

1+pfpl

11+

pfpl

=

(pf

pl+pfpl

pl+pf

) (297)



Solving for our parameters A,B, we see that(W0

N0

)=

(Apf


A + B(1− pf − pl)

)

⇒(

AB

)=

11+

pfplpf

plN0−W0

1−pf−pl

∴

(Wk

Nk

)→

pfpl

1+pfpl

11+

pfpl

=

(pf

pl+pfpl

pl+pf

) (297)


Homework Questions

Examples to Read:

8.13, 8.14, 8.15

9.3, 9.4, 9.7, 9.10.

HW Exercises:

8.1, 8.2, 8.3, 8.4, 8.5, 8.8, 8.10, 8.11, 8.12, 8.13, 8.15, 8.16, 8, 18, 8.20

9.1, 9.2, 9.3, 9.4, 9.6, 9.10, 9.13, 9.14, 9.16


Plan Type

We consider two types of retirement plans.

A Defined Contribution plan specifies how much an employer willcontribute, as a percentage of salary, into a plan.

A Defined Benefit plan specifies a level of benefit, most likelyrelated to the employee’s salary near retirement. Here, contributionsmay need to be updated based on the investment returns to ensurethat the benefit is met.


Replacement Ratio

Also, defining the Replacement Ratio

R :=pension income in the year after retirement

salary in the year before retirement(298)

the benefit under DB plans and target under DC plans may aim for

R ∈ (0.5, 0.7). (299)

This of course assumes the member survives the year following retirement.


Salary Scale Function

We can also use a deterministic model to define the salary scale syy≥x0

beginning at some suitiable initial age x0 where the value of sx0 can be setarbitrarily.

The ratio usually given is

sysx

=salary received in year y to y + 1

salary received in year x to x + 1(300)

and, assuming that salaries are increased continuously, the salary rate atage x is defined to be sx− 1

2.


Salary Scale Function

We can also use a deterministic model to define the salary scale syy≥x0

beginning at some suitiable initial age x0 where the value of sx0 can be setarbitrarily.

The ratio usually given is

sysx

=salary received in year y to y + 1

salary received in year x to x + 1(300)

and, assuming that salaries are increased continuously, the salary rate atage x is defined to be sx− 1

2.


Example 10.2

The final average salary for the pension benefit provided by a pension planis defined as the average salary in the three years before retirement.Members’ salaries are increased each year, six months before the valuationdate

A member aged exactly 35 at the valuation date received 75000 insalary in the year to the valuation date. Calculate his predicted finalaverage salary assuming retirement at age 65.

A member aged exactly 55 at the valuation date was paid salary at arate of 100000 per year at that time. Calculate her predicted finalaverage salary assuming retirement at age 65.

Assume a salary scale where sx0+y = 1.04y sx0 .


Example 10.2

For first case

savg = 75000 · 1

3

s62 + s63 + s64

s34

=75000

3·(1.0428 + 1.0429 + 1.0430

)= 234019

(301)

For second case

savg = 100000 · 1

3

s62 + s63 + s64

s54.5

=100000

3·(1.047.5 + 1.048.5 + 1.049.5

)= 139639

(302)

Now read Example 10.3 for more practice and Section 10.4 for setting theDC contribution.


Stochastic Pension Model

We can define a multiple decrement model for a pension plan via states

Y (t) =

0 if (x) is a member at age x + t1 if (x) has withdrawn by time x + t2 if (x) has retired due to disability by age x + t3 if (x) has retired due to age at x + t4 if (x) has died in service by age x + t

0

>>>>>>>>

''NNNNNNNNNNNNNN

1 2 3 4

Figure: Pension Plan Flow Chart. In a DC plan, benefit on exit is the same.However, in a DB plan different benefits may be payable on different forms of exit.


Example 10.5

A pension plan member is entitled to a lump sum benefit on death inservice of four times the salary paid in the year up to death. Assuming themultiple decrement model with

µ01x = µwx

µ02x = µix = 0.001

µ03x = µrx

µ04x = µdx = A + Bcx

= 0.00022 + (2.7× 10−6) · 1.124x

(303)


Example 10.5

Assume

µwx =

0.1, if x < 35

0.05, if 35 ≤ x < 45

0.02, if 45 ≤ x < 60

0, if x ≥ 60

µrx =

0 if x < 60

0.1, if 60 < x < 65

and

P [(x) retires at (60) | survives in employment to (60)] = 0.3

P [(x) retires at (65) | survives in employment to (65)] = 1(304)


Example 10.5

Calculate, for a member aged 35, the probability of retiring at age 65.Notice the similarities to Example 8.12.

For t ∈ (0, 10), we have

tp0035 = exp

[−∫ t

0

(µw35+s + µi35+s + µr35+s + µd35+s

)ds

]= exp

[−∫ t

0

(0.05 + 0.001 + 0 + A + Bc35+s

)ds

]= exp

[−0.05122t +

2.7× 10−6

ln (1.124)1.12435

(1.124t − 1

)](305)

It follows that 10p0035 = 0.597342


Example 10.5

Calculate, for a member aged 35, the probability of retiring at age 65.Notice the similarities to Example 8.12.

For t ∈ (0, 10), we have

tp0035 = exp

[−∫ t

0

(µw35+s + µi35+s + µr35+s + µd35+s

)ds

]= exp

[−∫ t

0

(0.05 + 0.001 + 0 + A + Bc35+s

)ds

]= exp

[−0.05122t +

2.7× 10−6

ln (1.124)1.12435

(1.124t − 1

)](305)

It follows that 10p0035 = 0.597342


Example 10.5

For t ∈ [10, 25), we compute

tp0035

10p0035

= P [ (35, 0)→ (35 + t, 0) | (35, 0)→ (45, 0) ]

= exp

[−∫ t−10

0

(µw45+s + µi45+s + µr45+s + µd45+s

)ds

]= exp

[−∫ t−10

0

(0.02 + 0.001 + 0 + A + Bc45+s

)ds

]= exp

[−0.02122(t − 10) +

2.7× 10−6

ln (1.124)1.12445

(1.124t−10 − 1

)]⇒ 25−p00

35 =25p00

35

10p0035

· 10p0035 = 0.712105 · 0.597342 = 0.425370

and 25+p0035 = 0.7 · 25−p00

35 = 0.297759(306)


Example 10.5

For t ∈ (25, 30), we compute

tp0035

25+p0035

= P[

(35, 0)→ (35 + t, 0) | (35, 0)→ (60+, 0)]

= exp

[−∫ t−25

0

(µw60+s + µi60+s + µr60+s + µd60+s

)ds

]= exp

[−0.10122(t − 25) +

2.7× 10−6

ln (1.124)1.12460

(1.124t−25 − 1

)](307)

It follows that the probability of retirement at exact age 65 is

30−p0035 =

(tp

0035

25+p0035

)(25+p00

35

)= 0.175879 (308)

Now calculate: P35[withdrawal],P35[retirement],P35[disability retirement]and P35[death in service].


Service Table

We can represent the multiple decrement model for pensions in tabularform. Begin by defining a minimum integer entry age x0 and correspondingarbitrary radix (cohort) lx0 . and now organize a table with entries

wx0+k = lx0kp00x0

p01x0+k

ix0+k = lx0kp00x0

p02x0+k

rx0+k = lx0kp00x0

p03x0+k

dx0+k = lx0kp00x0

p04x0+k

lx0+k = lx0kp00x0

(309)

and it follows that

wx0+k + ix0+k + rx0+k + dx0+k = lx0kp00x0

(p01x0+k + p02

x0+k + p03x0+k + p04

x0+k

)= lx0kp00

x0

(1− p00

x0+k

)= lx0

(kp00

x0− k+1p00

x0

)= lx0+k − lx0+k+1.


Service Table

It follows that we can use the service table to answer questions like

P35 [withdraws] =

∑24k=0 w35+k

l35

P35 [retires in ill health] =

∑29k=0 i35+k

l35

P35 [retires for age reasons] =

∑30k=0 r35+k

l35

P35 [dies in service] =

∑29k=0 d35+k

l35

(311)

For long-horizon investments with uncertain returns (forecasts may only bevalid for a small horizon), using tabular methods with UDD approximationis common in pension valuation.


Valuation:Contributions

Employees in a pension plan pay contributions of 6% of their previousmonth’s salary at each month end until age 60. Calculate the EPV atentry of contributions for a new entrant aged 35, with a starting salaryrate of 100000 using the model µ01

x = λ, µ02x = γ, µ03

x = 0 and µ04x = µ

for x ∈ (35, 60). Assume a constant force of interest δ and a salary scalefunction sy = eεy for y ∈ (35, 60).


Valuation

Per month, the contribution amount is a scaling of 0.06 10000012 = 500. It

follows that

E [PV(Contributions)] = 500300∑k=1

k12

p0035eε

k12 e−δ

k12

= 500300∑k=1

e−(µ+γ+λ) k12 eε

k12 e−δ

k12

= 500ex(1− e300x

1− ex

)x =

ε− µ− γ − λ− δ12

(312)


Valuation:Benefits

For a DB plan, the basic annual pension benefit is equal to n · SFin · α,where n is the total number of years of service, SFin is the average salaryin a specified period before retirement (ie. three years preceding exit) andα is the accrual rate, usually between 0.01 and 0.02.


Valuation:Benefits; Example 10.7

Estimate the EPV of the accrued age retirement pension benefit for amember aged 55 with 20 years of service, whose salary in the year prior tothe valuation date was 50000.

Assume that mid-year age retirements happen at exactly halfway intothe year. (Recall Claims Acceleration.)

Assume the final average salary is defined as the earnings in the threeyears before retirement.

Assume α = 0.015.

Calculate this EPV by using elements of a corresponding service table.



Note that for this problem,

n · S · α = (20)(50000)(0.015) = 15000

zy =sy−3 + sy−2 + sy−1

3

(313)

as well as

E [Projected Final Salary | Retirement at age y ] = 50000zys54

=: SDin.

(314)



∴ E [PV(Benefits)] = 15000

(r60−

l55

z60

s54v 5a

(12)60 +

r65−

l55

z65

s54v 10a

(12)65

)+ 15000

r60+

l55

z60.5

s54v 5.5a

(12)60.5

+ 150004∑

k=1

r60+k

l55

z60.5+k

s54v 5.5+k a

(12)60.5+k

zy =sy−3 + sy−2 + sy−1

3

(315)

One can program this using numerical software, using linear interpolationfor mid-year quantities. Read Examples 10.8, 10.9 for a discussion onwithdrawal pension.


Plan Funding

Assuming..

All employer contributions to a fund are paid the start of the year.

There are no employee contributions.

Any benefits payable during the year are paid exactly half-way thoughthe year.

We define the normal contribution due at the start of the year t to t + 1for a member aged x at time t as Ct .

Using reserving principles studied earlier, we have the equation

tV + Ct = E [PV(Benefits for mid-year exits)] + v 1p00x t+1V (316)


Plan Funding

Assuming..

All employer contributions to a fund are paid the start of the year.

There are no employee contributions.

Any benefits payable during the year are paid exactly half-way thoughthe year.

We define the normal contribution due at the start of the year t to t + 1for a member aged x at time t as Ct .

Using reserving principles studied earlier, we have the equation

tV + Ct = E [PV(Benefits for mid-year exits)] + v 1p00x t+1V (316)


Example 10.10

Assume a pension plan with the following valuation methods:

Accrual rate: 1.5%

Final salary plan

Pension based on earnings in the year before age retirement

Normal retirement at age 65

The pension benefit is a life annuity payable monthly in advance

There is no benefit due on death in service

No exits other than by death before normal retirement age


Example 10.10

Calculate the value of the accrued pension benefit and normal contributiondue at the start of the year using a projected unit funding (PUC), whereinterest is set at 5% per year, salaries increase at 4% per year and assumea constant mortality µ before and after retirement.

SFin = 50000s64

s49= 50000(1.04)15 = 90047

0V = 0.015 · 20 · SFin · 15p50 · v 15 · a(12)65

= 27014.10 · e−15µ · v 15 · 1

12

∞∑k=0

vk12 e−µ

k12

=27014.10

12e15µ · 1.0515 ·(

1− 12

√1

1.05eµ

)(317)


Example 10.10

Calculate the value of the accrued pension benefit and normal contributiondue at the start of the year using a projected unit funding (PUC), whereinterest is set at 5% per year, salaries increase at 4% per year and assumea constant mortality µ before and after retirement.

SFin = 50000s64

s49= 50000(1.04)15 = 90047

0V = 0.015 · 20 · SFin · 15p50 · v 15 · a(12)65

= 27014.10 · e−15µ · v 15 · 1

12

∞∑k=0

vk12 e−µ

k12

=27014.10

12e15µ · 1.0515 ·(

1− 12

√1

1.05eµ

)(317)


Example 10.10

It follows that our equation for C is

1V = 0.015 · 21 · SFin · 14p51 · v 14 · a(12)65

=21

20

1

vp50

(0.015 · 20 · SFin · 15p50 · v 15 · a(12)

65

)=

21

20

1

vp50(0V ) .

∴ C = v · p50 · 1V − 0V =21

200V − 0V =

0V

20

(318)

Consider the traditional unit credit funding approach, and see how thisaffects our previous calculation. Also, read over Example 10.11 whichallows for benefits payable on exit during the year.


Homework Questions

HW: 10.1, 10.2, 10.5, 10.7, 10.8, 10.11, 10.12


Law of Total Variance

Recall that for two random variables X ,Y in a probability space (Ω,F ,P)we have the Tower Property

E[E[X | Y ]

]= E[X ]

E[E[Y | X ]

]= E[Y ]

(319)

and so it follows that

V [X ] = E[X 2]−(E[X ]

)2

= E[E[X 2 | Y

] ]−(E[E[X | Y ]

])2

= E[V [X | Y ] + E[X | Y ]2

]−(E[E[X | Y ]

])2

= E[V [X | Y ]

]+ V

[E[X | Y ]

](320)


Deviation as a Risk Measure

Assuming a sequence of i.i.d. Random Variables

Xk

n

k=1, one measure

of the risk associated to the average

Xn :=1

n

n∑k=1

Xk (321)

is the total variance

ρn(X ) := V[Xn

](322)



Correspondingly, we say that such a risk is Diversifiable iflimn→∞ ρn(X ) = 0, and not diversifiable otherwise.

Note that if

Xk

n

k=1are dependent but otherwise identically distributed

with correlation coefficient ρ, mean µ and variance σ2, then

ρn(X ) =nσ2 + n(n − 1)ρσ2

n2→ ρσ2 6= 0 (323)

For a history of variance as a risk measure in Modern Portfolio Theory andthe corresponding use of diversification, click here and references within.


http://en.wikipedia.org/wiki/Modern_portfolio_theory


Recall that for any random variable Y and i.i.d. sequence

Xk

n

k=1with

identical copy X

V[1

n

n∑k=1

Xk

]= E

[V[1

n

n∑k=1

Xk | Y]]

+ V[E[1

n

n∑k=1

Xk | Y]]

=1

nE[V[X | Y

]]+ V

[E[X | Y

]] (324)

It follows that ρn(X )→ 0 as long as V[E[X | Y

]]= 0.




Xk

n

k=1with

identical copy X

V[1

n

n∑k=1

Xk

]= E

[V[1

n

n∑k=1

Xk | Y]]

+ V[E[1

n

n∑k=1

Xk | Y]]

=1

nE[V[X | Y

]]+ V

[E[X | Y

]] (324)


]]= 0.




Xk

n

k=1with

identical copy X

V[1

n

n∑k=1

Xk

]= E

[V[1

n

n∑k=1

Xk | Y]]

+ V[E[1

n

n∑k=1

Xk | Y]]

=1

nE[V[X | Y

]]+ V

[E[X | Y

]] (324)


]]= 0.


Connection with CLT

Note that by the Central Limit Theorem,

limn→∞

P[ ∣∣Xn − µ

∣∣ ≥ k]

= limn→∞

P[ ∣∣∣∣∑n

k=1 Xk − nµ√nσ

∣∣∣∣ ≥ k√

n

σ

]= lim

n→∞2Φ(− k√

n

σ

)= 0.

(325)

This says that for uncorrelated r.v.’s, since the variance of the aggregatemean is linear in n, we have the deviation of the aggregate mean from theindividual mean asymptotically disappears.


Example of Diversifiable Risk

Consider the case where we have an i.i.d. sequence

Xk

n

k=1

Xk ∈ 0, 1P[Xk = 1] = tpx · (1− spx+t).

(326)

It follows that

Xn =1

n

n∑k=1

Xk (327)

models the sample probability of deaths of a population of n alive at age xwhere death occurs between age x + t and age x + t + s. This is of coursea binomial random variable with p = tpx · (1− spx+t).


Example of Diversifiable Risk

We can see that for a copy X of the sequence,

V[Xn

]=

V[∑n

k=1 Xk

]n2

=nV [X ]

n2

=1

n·[tpx · (1− spx+t)

]·[1− tpx · (1− spx+t)

]→ 0

(328)

and so the risk is diversifiable.


Example of Non-Diversifiable Risk

Consider now the case where the Xk model the loss associated with amember of an insured population. If each member has loss function Xk

and the premiums are charged in keeping with the EPP, then we expect

that E[Xk

]= 0 for all k ∈ 1, . . . , n .

If, however, the forecasted yield rate used is a random variable Y , then ifE[Xk | Y ] 6= 0 we have non-diversifiable risk as

V[Xn

]→ V

[E[X | Y

]]6= 0. (329)

Correlated loan losses, such as those in a large portfolio, can be modeledin a similar way. Check out Vasicek’s model, beautifully explained in theseslides by Stephen M. Schaefer of the London Business School.


http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.139.5752&rep=rep1&type=pdf

http://dse.univr.it/safe/documents/SSEFCANAZEI2012/07_correlation_-_modeling.pdf


Example of Non-Diversifiable Risk

Consider now the case where the Xk model the loss associated with amember of an insured population. If each member has loss function Xk

and the premiums are charged in keeping with the EPP, then we expect

that E[Xk

]= 0 for all k ∈ 1, . . . , n .

If, however, the forecasted yield rate used is a random variable Y , then ifE[Xk | Y ] 6= 0 we have non-diversifiable risk as

V[Xn

]→ V

[E[X | Y

]]6= 0. (329)

Correlated loan losses, such as those in a large portfolio, can be modeledin a similar way. Check out Vasicek’s model, beautifully explained in theseslides by Stephen M. Schaefer of the London Business School.


http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.139.5752&rep=rep1&type=pdf



Q : 294 : SOA MLC Study Guide

An insurer issues a number of identical special 1-year term life insurancepolicies.

Each policy has a death benefit of 1000 payable at the end of the year ofdeath, on condition that:

The policyholder dies during the year; and

A stock index ends the year below its value at the start of the year.

Both conditions must be satisfied for the death benefit to be paid.



Furthermore, you are given:

Future lifetimes of the policyholders are independent

qx = 0.05 for all x.

The probability that the stock index ends the year below its value atthe start of the year is 0.1 for all years.

Future lifetimes of the policyholders and the value of the stock indexare independent.

The annual effective rate of interest rate is 3%.

XN denotes the total present value of benefits for N policies.

Calculate

limN→∞

√V [XN ]

N(330)



Define F = Fund drops below current level in the next year . Then

P[F ] = 0.1

E[XN | F ] = N × 1000

1 + iqx = 48.54N

E[XN | F c ] = 0

∴ E[XN ] = E[XN | F ]× P[F ] + E[XN | F c ]× P[F c ]

= 4.854N

(331)



Also,

V [XN | F ] = N ×(1000

1 + i

)2qx(1− qx) = 44773.31N

V [XN | F c ] = 0

E[V [XN | 1F]

]= V [XN | F ]× P[F ] + V [XN | F c ]× P[F c ]

= 4477.331N

V[E[XN | 1F]

]=(E[XN | F ]

)2× P[F ] +

(E[XN | F c ]

)2× P[F c ]

− E[XN

]2

=(

48.54N)2× 0.1− (4.854N)2 = 212.05N2

⇒ limN→∞

√V [XN ]

N= lim

N→∞

√212.05N2 + 4477.331N

N=√

212.05 = 14.56


Homework Questions

HW: 11.1, 11.2, 11.3, 11.5, 11.7


Reserves

Recall the need for policy values when negative future cash flows wereexpected. In this lecture, we cover the idea of reserves, which is the actualamount of money held by the insurer to cover future liabilities associatedwith contracts.


Reserves

The insurer may decide to set aside assets in reserve as equal to the netpremium policy values on a certain (reserve) basis.

For example, consider an n−year term insurance contract issued to a life xwith sum insured S . Since we use the net premium basis to compute fixedpremiums, it follows that

P = SA1x :n

ax :n

⇒ Rt = tV = SA 1x+t:n−t − Pax+t:n−t

= SA1x :n ·

(A 1x+t:n−tA1x :n

−ax+t:n−t

ax :n

) (333)

The cost of setting up, from t − 1 to t, the reserve amount of tV is attime t equal to tV · px+t−1 when valued at time t−

i.e. the proportion of contracts that survive to the end of the year.


Reserves



P = SA1x :n

ax :n


= SA1x :n ·

(A 1x+t:n−tA1x :n

−ax+t:n−t

ax :n

) (333)




Reserves



P = SA1x :n

ax :n


= SA1x :n ·

(A 1x+t:n−tA1x :n

−ax+t:n−t

ax :n

) (333)




Reserves



P = SA1x :n

ax :n


= SA1x :n ·

(A 1x+t:n−tA1x :n

−ax+t:n−t

ax :n

) (333)




Notation

At time t, just before and just after, we have quantities that are assetsand costs.

At time (t − 1)+, we have the cost Et associated from t − 1 to t.

Between (t − 1)+ and t−, we have the payout S settled at time t−with expected value S · qx+t−1.

At time (t − 1)+, we have the asset t−1V which grows at the interestrate i to value (1 + i) · t−1V at time t−

At time (t)−, we have the cost tV · px+t−1.


Profits

Correspondingly, we can set up an equation for the profit at time t,denoted by Prt :

Prt =(t−1V + P − Et

)(1 + i)− Sqx+t−1 − tVpx+t−1 (334)

The Profit Vector

~Pr :=(

Pr0, . . . ,Prn)

(335)

is comprised of elements that represent the expected profit at the end ofthe year given that the policy is in effect at the start of the year.


Profits

The Profit Signature is the the vector ~Π comprised of elements

Πt := t−1pxPrt (336)

that represent the expected profit at the end of the year given that thepolicy was in effect at age x.


Profit Measures

Recall that for any set of cash flows Ct , the internal rate of return IRR (ifit uniquely exists) is the interest rate j such that

n∑t=0

Ct

(1 + j)t= 0. (337)

In accordance with the IRR, the insurer may set a minimum hurdle or riskdiscount rate r such that the contract is satisfiably profitable if IRR > r .


EPV of Future Profit

If the IRR does not exist, the insurer may seek to measure the profitabilityvia the Net Present Value computed using the risk discount rate:

NPV :=n∑

t=0

Πt

(1 + r)t(338)


Profit Margin and DPP

Another measure is the ratio of NPV to E[PV (Premiums)]:

Profit Margin :=NPV

E[PV (Premiums)](339)

as is the discounted payback period DPP:

DPP := min

m :

m∑t=0

Πt

(1 + r)t≥ 0

(340)

which represents the time until the insurer starts to make a profit.

A question naturally arises of how to jointly measure interest risk andprofit. One may even compute the marginal changes in the profit measureswith respect to change in risk discount factor r .


Exercise 12.13 (Profit Testing Basis)

A special 10-year endowment insurance is issued to a healthy life aged 55.The benefits under the policy are

50000 if at the end of a month the life is disabled, having beenhealthy at the start of the month,

100000 if at the end of a month the life is dead, having been healthyat the start of the month,

50000 if at the end of a month the life is dead, having been disabledat the start of the month,

50000 if the life survives as healthy to the end of the term.

On withdrawal at any time, a surrender value equal to 80% of the netpremium policy value is paid, and level monthly premiums are payablethroughout the term while the life is healthy.


Exercise 12.13 (Profit Testing Basis)

Other elements of the profit testing basis are as follows:

Interest: 7% per year.

Expenses: 5% of each gross premium, including the first, togetherwith an additional initial expense of 1000.

The benefit on withdrawal is payable at the end of the month ofwithdrawal and is equal to 80% of the sum of the reserve held at thestart of the month and the premium paid at the start of the month.

Reserves are set equal to the net premium policy values.


Exercise 12.13

Extra modeling assumptions:

The gross premium and net premium policy values arecalculated using the same survival model as for profit testingexcept that withdrawals are ignored, so that µ03

x = 0 for all x .

The net premium policy values are calculated using an interestrate of 5% per year.

The monthly gross premium is calculated using the equivalence principleon the following basis:

Interest: 5.25% per year.

Expenses: 5% of each premium, including the first, together with anadditional initial expense of 1000.


Exercise 12.13

(a) Calculate the monthly premium on the net premium policy valuebasis.

(b) Calculate the reserves at the start of each month for both healthylives and for disabled lives.

(c) Calculate the monthly gross premium.

(d) Calculate the monthly profit earned, using the profit testing basis,

under assumption that the life is healthy (Pr(0)t ) or that the life is

disabled (Pr(1)t ) at the beginning of the month . Hint: Consider

equation (334) for both healthy and disabled lives

(e) Calculate the internal rate of return.

(f) Calculate the NPV, the profit margin (using the EPV of grosspremiums), the NPV as a percentage of the acquisition costs, and thediscounted payback period for the contract, in all cases using a riskdiscount rate of 15 % per year.


Exercise 12.13

The model for state transition in this model follows the flow chart below:

0 (Healthy)

''PPPPPPPPPPP//

1 (Disabled)

3 (Withdrawn) 2 (Dead)

Figure: Multiple State Model


Exercise 12.13

The associated rate matrix for Profit Testing is

Q(t) =

−0.035 0.01 0.015 0.01

0 −0.03 0.03 00 0 0 00 0 0 0

(341)

and the associated rate matrix for gross premium and net premiumpolicy values is

Q(t) =

−0.025 0.01 0.0150 −0.03 0.030 0 0

(342)


Exercise 12.13 - Diagonalization of Q

U =

−1 0.894427 0.245036 0.4380250 0.447214 −0.0439573 0.6345450 0 −0.0439573 0.6345450 0 0.967519 −0.0532764

D =

−0.035 0 0 0

0 −0.03 0 00 0 0 00 0 0 0

U−1 =

−1 2 −1.28571 0.2857140 2.23607 −2.23607 00 0 0.0871109 1.037530 0 1.58197 0.0718735

(343)


Exercise 12.13 - Diagonalization of Q

It follows that if we do allow for withdrawal, our transition probabilitymatrix P(t) = exp (tQ) has the solution

P(t) =

e−0.035t

t p0155 t p

0255 t p

0355

0 e−0.03t 1− e−0.03t 00 0 1 00 0 0 1

(344)

where

t p0155 = 2e−0.03t − 2e−0.035t

t p0255 = 0.71428571 + 1.285714287e−0.035t − 2e−0.03t

=5

7+

9

7e−0.035t − 2e−0.03t

t p0355 = 0.285714287− 0.285714287e−0.035t

=2

7− 2

7e−0.035t

(345)


Exercise 12.13 - No Withdrawal

If we do not allow for withdrawal, however, our transition probabilitymatrix P(t) = exp (tQ) has the simpler solution

P(t) =

e−0.025t 2e−0.025t − 2e−0.03t 1 + 2e−0.03t − 3e−0.025t

0 e−0.03t 1− e−0.03t

0 0 1

.

(346)


Exercise 12.13- Monthly Net Premium

For the equation of value, we determine for the monthly net premium P ′

E[Premium Income] = P ′119∑k=0

k12

p0055v

k12

E[Benefit] = 50000v 1010p00

55

+ 50000119∑k=0

(k12

p0055 1

12p01

55+ k12

+ k12

p0155 1

12p12

55+ k12

)v

k+112

+ 100000119∑k=0

k12

p0055 1

12p02

55+ k12

vk+112

(347)Using our solution for transition probabilities that don’t allow forwithdrawals, a discount rate of v = 1

1.05 and solving the resultinggeometric series for the EPV’s above, we return P ′ = 452.


Exercise 12.13 - Net Policy Values

Recall that tV(i) = E[ Loss | Y (t) = i ]

Given the parameters of our contract, we have the boundary values

10−V (0) = 50000

10−V (1) = 0(348)

and the recursive equations

tV(0) = −P ′ + 1

12p00

55+tv1

12t+ 1

12V (0) + 1

12p01

55+tv1

12 (50000 + t+ 112

V (1))

+ 100000v1

12 112

p0255+t

tV(1) = 1

12p11

55+tv1

12t+ 1

12V (1) + 50000v

112 1

12p12

55+t

(349)A matrix (array) recursion method can be encoded via spreadsheet orother numerical software to iterate backwards from t = 10.


Exercise 12.13 - Monthly Gross Premium

For the equation of value, we determine for the monthly gross premium P

E[Premium Income] = 0.95P119∑k=0

k12

p0055v

k12

E[Benefit] = 50000v 1010p00

55

+ 50000119∑k=0

(k12

p0055 1

12p01

55+ k12

+ k12

p0155 1

12p12

55+ k12

)v

k+112

+ 100000119∑k=0

k12

p0055 1

12p02

55+ k12

vk+112

+ 1000.(350)

Using our solution for transition probabilities that don’t allow forwithdrawals, a new discount rate of v = 1

1.0525 , and solving the resultinggeometric series for the EPV’s above, we return P = 484.27.


Exercise 12.13

Once we have the arrays(

Pr(0)t ,Pr

(1)t

)from part (d), we can further

compute

Πt = t− 112

p0055Pr

(0)t + t− 1

12p01

55Pr(1)t .

An IRR j such that∑120

k=0

Π k12

(1+j)k12

= 0.

The net present value NPV =∑120

k=0

Π k12

(1.15)k12

.

The discounted payback period m, where m is the smallest value such

that∑m

k=0

Π k12

(1.15)k12≥ 0.

The profit margin

∑120k=0

Π k12

(1.15)k12

P∑119

k=0

k12

p0055

(1.15)k12

.


Homework Questions

Finish Exercise 12.13 by filling in the numbers in the previous slide.

Read Example 12.1.

HW: 12.1, 12.2, 12.3, 12.4, 12.6


Equity Linked Insurance

Modern insurance contracts can include some form of guarantee.

These are known in America as Variable Annuities and SegregatedFunds in Canada.

The accumulating premiums the policyholder pays is invested on thepolicyholder’s behalf.

These premiums form the policyholder’s fund, from which regularmanagement charges are deducted by the insurer and paid into theinsurer’s fund to cover expenses and insurance charges.



On survival to the end of the contract term the benefit may be just thepolicyholder’s fund and no more, or there may be a guaranteed minimummaturity benefit (GMMB). There may also be a guaranteed minimumdeath benefit (GMDB).

There are very real consequences to the differences between financialpricing and actuarial reserving. A short but excellent analysis can be foundin the paper by Bangwon Ko and Elias S. W. Shiu on Financial Pricingand Actuarial Reserving.

Also consider A Heavy Traffic Approach to Modeling Large LifeInsurance Portfolios (Stochastic modeling of actuarial reserve, with Itointegration of a time-changed Brownian Bridge.)

We follow the example set by Shiu and Ko now.


https://www.iabe.be/sites/default/files/bijlagen/vol7.7-ko_0.pdf


http://www.columbia.edu/~jb2814/papers/Collective_insurance4.pdf



On survival to the end of the contract term the benefit may be just thepolicyholder’s fund and no more, or there may be a guaranteed minimummaturity benefit (GMMB). There may also be a guaranteed minimumdeath benefit (GMDB).

There are very real consequences to the differences between financialpricing and actuarial reserving. A short but excellent analysis can be foundin the paper by Bangwon Ko and Elias S. W. Shiu on Financial Pricingand Actuarial Reserving.

Also consider A Heavy Traffic Approach to Modeling Large LifeInsurance Portfolios (Stochastic modeling of actuarial reserve, with Itointegration of a time-changed Brownian Bridge.)

We follow the example set by Shiu and Ko now.






Stochastic Actuarial Reserving

Fix a probability space(

Ω,F ,P)

and a standard Brownian motion W

that lives on this space.

Consider now a term contact with term T and let α denote themanagement charges factor along with β representing the policyholder’sparticipation factor.

Furthermore, assume mean and standard deviation parameters (µ, σ)respectively and the corresponding Geometric Brownian Mutual Fund Asset

St = S0eµt+σWt . (351)



Using this as the model of the asset returns upon which premiums areinvested, the policyholder wishes to purchase a contract that pays amaturity benefit credited at a rate of return which is the greater of

the customer’s risk discount rate r , where r < µ or

the participation rate of the stock index returns of S .

Symbolically, for a current premium P invested in the , the contractpayout value at maturity is

V (T ) = (1− α)P max

erT , 1 + β

(ST

S0− 1)

. (352)



Assume that the policyholder is able to fully participate in the returns fromthe fund (i.e. β = 1.)Then

V (T ) = (1− α)PST

S0+ (1− α)P max

erT −

(ST

S0

), 0

:= V1(T ) + V2(T ).

(353)

Here, V1(T ) is the net premium, or payoff, for investing in the index fundand V2(T ) is the guaranteed option payoff if the index fundunder-performs relative to the risk discount rate r .

How does one reserve to meet the obligations of V2(T ).



One can see that the probability of a payout, that V2(T ) 6= 0 is for large T

P[V2(T ) 6= 0] = P[rT > µT + σWT ] = Φ( r − µ

σ

√T)≈ 0. (354)

Since it is a low probability event that we have to prepare for a payoutV2(T ) and since we can directly replicate the payoff V1(T ) by initially

purchasing (1−α)PS0

units of the index fund, an actuary may be tempted tonot reserve for the uncertain portion of the guarantee, V2(T ), if thecontract has a relatively long term T .

Is this a wise decision?




P[V2(T ) 6= 0] = P[rT > µT + σWT ] = Φ( r − µ

σ

√T)≈ 0. (354)








P[V2(T ) 6= 0] = P[rT > µT + σWT ] = Φ( r − µ

σ

√T)≈ 0. (354)






Stochastic Reserving for non-diversifiable risk

Given a random loss L, we define the quantile reserve, also known as theValue at Risk with parameter α, as the amount which with probability αwill not be exceeded by the loss.

Symbolically, if L has a continuous distribution function FL, then theα−quantile reserve is Qα where

P[L ≤ Qα] = α. (355)


Stochastic Reserving for non-diversifiable risk

One feature that is missing in VaR is the description of what the loss couldbe if it does exceed the quantile Qα. In this case, the Conditional TailExpectation (CTEα) is defined as

CTEα = E[L | L ≥ Qα]. (356)

A risk manager should not rely on static measures of risk involved with aportfolio of liabilities. Rather, the CTE or VaR reserve should be regularlyupdated to incorporate market information as it arrives. This allowsreserves which are held in less-risky (and possibly more liquid) funds to beinvested higher return and higher risk assets if current market informationdictates that CTE reserves can be reduced.


Hedging Retirement Risks? Ideas from a Nobel Laureate..

Another set of problems that actuaries share with retirement plannersand financial economists alike is the lack of planning many folks dofor their retirement.

Consider the (2014) article 1 The Crisis in Retirement Planning byRobert C. Merton, a Nobel Laureate and a founding father ofmathematical finance.

In this work, Prof. Merton discusses the move from DB to DC plans,and the weight of managing one’s retirement account in the light ofthis cultural shift.

1Merton, Robert C. ”The Crisis in Retirement Planning.” Harvard Business Review92, nos. 7/8 (July–August 2014): 43–50.


http://robertcmerton.com/publication/the-crisis-in-retirement-planning-2/

Hedging Retirement Risks? Ideas from a Nobel Laureate..

Defining the strength of one’s DC pension, seen as an investmentaccount, Merton (2014) argues for maximizing the probability ofhitting a desired income level in retirement, as opposed to maximizingtotal capital in savings.

In more recent work, Merton has proposed the development of newfinancial instruments to hedge more of this DC risk. One suchinstrument 2,are known as Standard of Living indexed,Forward-starting, Income-only Securities.

These are government issued default-free bonds which offer ”...certainty about two characteristics critical for DC retirementportfolios: (i) a commitment to pay over a particular time horizon(how/when one is paid); and (ii) a specific cash flow (what is paid).DC investors require a guaranteed cash flow that protects their realpurchasing power in retirement.”

2Muralidhar, Arun and Merton, Robert. (2017). Time for retirement ’SeLFIES’?.Investments and Pensions Europe.


https://www.ipe.com/reports/special-reports/ipe-at-20/ipe-at-20-time-for-retirement-selfies/10018263.article

https://www.ipe.com/reports/special-reports/ipe-at-20/ipe-at-20-time-for-retirement-selfies/10018263.article

Homework Questions

Read Example 14.1 and Section 14.4

HW: 14.1− 14.5


Documents

STT 455-6: Actuarial ModelsSTT 455-6: Actuarial Models Albert Cohen Actuarial Sciences Program Department of Mathematics Department of Statistics and Probability C336 Wells Hall Michigan