10_FM_2_tn_1pp

Financial Mathematics


Jonathan Ziveyi1

1University of New South Wales

Actuarial Studies, Australian School of Business

[email protected]

Module 2 Topic Notes

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Plan

Module 2: Valuation of Contingent Cash FlowsIntroductionA Continuous Model for Survival AnalysisA Discrete Model for Survival AnalysisExpected Present Value of Basic Life Insurance Products

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Module 2: Valuation of Contingent Cash Flows

Introduction

Plan


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Introduction

Contingent Cash Flows

a contingent cash flow is a cash flow that happens if. . .

Contingent cash flows happen in a variety of situations:

depending on an event that will happen, and only once related to life insurance mathematics (death is such an event) a probability model is necessary for that event we can calculate standard expected present values this module

depending on other securities: derivatives use probability model or replicating portfolio (arbitrage-free) another module of this course

depending on an event that may not happen at all, or indeedseveral times, and for random claim amounts this is the case of any GI insurance product pricing techniques are discussed in other courses

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Introduction

History of life insurance mathematics Several steps to get to lifeinsurance mathematics

Simon Stevin (1548-1620), a Dutch mathematician, developedthe first compound interest tables

Blaise Pascal (1623-1662), a French mathematician andphilosopher, gives birth to probability calculus

Edmund Halley (1656-1742), a British mathematician andastronomer, builds the first mortality table

James Dodson (1705-1757), a British mathematician, was thefirst to put all components together. In a 1756 lecture, heshowed how a life insurance plan should be set up premium rates should be calculated reserves would build up

The discipline of life insurance mathematics was born!

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Introduction

Plan of this module

1. Introduction

2. A Continuous Model for Survival Analysis

3. A Discrete Model for Survival Analysis

4. Expected Present Value of Basic Life Insurance Products

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A Continuous Model for Survival Analysis

Plan


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Survival Probabilities Let an individuals age-at-death, Z , be acontinuous r.v. with distribution function

FZ (z) = Pr(Z z), x 0.

Its complement, the survival function S(x), is defined as

S(x) = 1 FZ (x)

= Pr(Z > x), x 0,

where S(x) is seen to be the probability that a newborn will attainage x .

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Probabilities of dying (surviving) over the next t years Convention:

a life aged x will be denoted by (x)

x for men, y for women

The future lifetime

T (x) = Z x |Z > x

of (x) is a random variable with distribution function

tqx = 1 tpx

= Pr(Z x t | Z > x)

=FZ (x + t) FZ (x)

1 FZ (x)

=S(x) S(x + t)

S(x)

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Probability of dying over the next t years, in s years The probability(x) will die between ages x + s and x + s + t is

s|tqx = Pr(s < T (x) s + t)

= Pr(s < Z x s + t | Z > x)

= Pr(x + s < Z x + s + t | Z > x)

=S(x + s) S(x + s + t)

S(x).

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Force of mortality Let s = 0 and t be infinitesimal (a dt):

dtqx is the probability that (x) will die in the next instant

this is equal to the pdf of T (x) at s = 0, times dt

We have

dtqx =S(x) S(x + dt)

S(x)=

S(x + dt) S(x)

dt

dt

S(x)=

S (x)

S(x)dt.

The coefficient of dt is called the force of mortality, and representsthe likelihood for the individual (x) to die in the next instant dt:

x = S (x)

S(x)= [lnS(x)] =

fZ (x)

S(x)=

fZ (x)

1 FZ (x).

Note thatS(x) = e

Rx

0tdt .

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One more step to obtain the pdf of T (x) What if s > 0?

we have to account for the probability of surviving s years,before dying between s and s + dt

We have then

s|dtqx = S(x + s)

S(x)

S(x + s + dt) S(x + s)

S(x + s)dtdt = spxx+sdt

This integrates to 1. We can now calculate the expected futurelifetime of (x):

ex = E [T (x)] =

0

t tpxx+tdt.

Note that

tpx = e

Rt

0x+sds .

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Example The lifetime of a light bulb is sometimes modeled using anexponential distribution:

fZ (z) = ez .

Find

1. the probability that the bulb will fail before x hours of usage;

2. the probability that it will function more than x hours;

3. its failure rate;

4. its expected lifetime;

5. the probability that it will function for another t years if it hasalready functioned for x hours.

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Example

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Analytical distributions of T (x) de Moivre (1724):

maximum age

T (x) is uniformly distributed between 0 and x(the remaining years)

x+t =1

x t

Gompertz (1824)

the force of mortality grows exponentially

x+t = Bcx+t

Makeham (1860)

generalises Gompertz

x+t = A + Bcx+t

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Analytical distributions of T (x) Lee-Carter (1992):

one of the models that are most used now (severalmodifications were (and still are) developed to improve it)

x ,t = ex+xt + x ,t = Ax B

tx + x ,t

where

x ,t is the force of mortality of (x) at time t

x and x depend on x

t depends on t

x ,t are iid N(0, 2) random variables

This is the continuous model for longitudinal or generation lifetables, that relate to the generation of persons born at time t x .

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A Discrete Model for Survival Analysis

Plan


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Curtate Future Lifetime Let K (x) T (x) be the random numberof completed years lived by (x), or the curtate future lifetime of(x). This is a discrete random variable.

Pr[K (x) = k ] = Pr[k < T (x) k + 1]

= k|1qx

=FZ (x + k + 1) FZ (x + k)

1 FZ (x)

=1 FZ (x + k)

1 FZ (x)FZ (x + k + 1) FZ (x + k)

1 FZ (x + k)

=S(x + k)

S(x)

S(x + k) S(x + k + 1)

S(x + k)= kpx qx+k

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Expected life in discrete time The expected curtate future lifetimeof (x) is denoted by

E [K (x)] = ex ex 1

2.

It can be calculated as

ex =

k=1

kkpxqx+k

or

ex =

k=1

Pr[K k ] =

k=1

kpx .

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Numerical Example For a newborn (0):

t qt 1 qt = pt S(t) = tp0 Pr[K (0) = t]

0 0.3 0.7 1.00000 0.300001 0.1 0.9 0.70000 0.070002 0.2 0.8 0.63000 0.126003 0.5 0.5 0.50400 0.252004 0.7 0.3 0.25200 0.176405 0.9 0.1 0.07560 0.068046 1.0 0.0 0.00756 0.00756

e0 = 2.16916 1.00000

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Numerical Example For (2):

t qt 1 qt = pt tq2 tp2 Pr[K (2) = t]

0 0.3 0.7 0.000 1.000 0.2001 0.1 0.9 0.200 0.800 0.4002 0.2 0.8 0.600 0.400 0.2803 0.5 0.5 0.880 0.120 0.1084 0.7 0.3 0.988 0.012 0.0125 0.9 0.1 1.000 0.000 0.0006 1.0 0.0

e2 = 1.332 1.000

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Expected Present Value of Basic Life Insurance Products

Plan


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Insurance products: conventions

historically (and practically) priced in discrete time

premiums are paid at the beginning of the year, benefits at theend of the year(have you ever wondered why all insurance premiums had tobe paid in advance?)

the net single premium, or actuarial present value, or riskpremium of the policy is the expected present value of futurebenefits.

the actual premium paid by the customer is loaded; this is notdiscussed here

let i be the technical rate of interest

In life insurance, benefit payments are contingent on the deathand/or survival of the person insured.

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How to interpret this section Some standard EPV related to anevent that will happen only once:

an annuity until it happens

an annuity until it happens, but for a maximum of n years

a capital when it happens

a capital when it happens, if it happens before n years

a capital after n years, if it hasnt happened yet

a capital when it happens if it happens before n years after n years if it hasnt happened yet

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Insurance in case of death Life insurance

single payment, the sum insured

net single premium is paid at time 0, for a sum insured of 1

the benefit isB = vK(x)+1,

a random variable with distribution

Pr[vK(x)+1 = vk+1] = Pr[K (x) = k ],

= kpx qx+k , k = 0, 1, . . . .

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Insurance in case of death Whole life insurance

Ax = E [B ] = E[vK(x)+1

]=

k=0

vk+1kpxqx+k ,

andVar(B) = 2Ax (Ax)

2

where

2Ax = E[B2]

= E[v2(K(x)+1)

]

= E[v

K(x)+1j

]

where j = (1+ i)2 1, i.e.,the effective two yearly rate of interest.

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Insurance in case of death Term insurance

A1x :n =

n1k=0

vk+1kpx qx+k .

In continuous timeIf B = vT (x), then

Ax =

0

v t tpx x+tdt, and

A1x :n =

n0

v t tpx x+tdt.

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Insurance in case of survival Pure endowment

A 1x :n = vnnpx .

Whole life annuity-dueThe present value of benefits is

B = 1+ v + v2 + + vK(x) = aK(x)+1

with pmf

Pr[B = ak+1 ] = Pr[K (x) = k ] = kpx qx+k .

We have then

ax =

k=0

ak+1 kpx qx+k

=

k=0

vkkpx .

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Insurance in case of survival Whole life annuity-immediateThis is simply

ax = ax 1.

Temporary life annuity-due for n yearsThe present value of benefits is

B =

{aK(x)+1 K (x) = 0, 1, 2, . . . , n 1

an K (x) = n, n + 1, n + 2, . . .

We have then

ax :n =

n1k=0

ak+1 kpx qx+k + an npx

=n1k=0

vkkpx .

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Insurance in case of death and survival EndowmentPayment of 1

if death before n, or

after n years if survival.

The present value of benefits is

B =

{vK(x)+1 K (x) = 0, 1, 2, . . . , n 1vn K (x) = n, n + 1, n + 2, . . .

with present valueAx :n = A

1x :n + A

1x :n .

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Numerical example (continued), for (2) and n = 3Whole life insurance

A2 = 0.9132,2A2 = 0.8351, and thus the variance is (0.0339)

2

Endowment = Term insurance + Pure endowment

A2:3 = A12:3

+ A 12:3

0.9177 = 0.8110 + 0.1067

Life annuities

a2 = 2.2560 a2 = 1.2560

Temporary life annuity-due

a2:3 = 2.1391

Note: the technical rate of interest is 4%27/28




Important relations There are many.... Here are two important ones:

dax + Ax = 1,

for whole life insurance and

dax :n + Ax :n = 1,

for term insurance.

Both are similar todan + v

n = 1.

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