Actuarial mathematics 3280 - Department of Mathematics …efurman/actmath2009/lecture5.pdf · Premium equations Actuarial mathematics 3280 Beneﬁt premiums Edward Furman Department

Premium equations

Actuarial mathematics 3280Benefit premiums

Edward Furman

Department of Mathematics and StatisticsYork University

January 14, 2010

Edward Furman Actuarial mathematics 3280 1 / 46

Premium equations

Insurance businessAn insurance system is a mechanism for reducing the adversefinancial impact of random events that prevent the fulfillment ofreasonable expectations. (see, Bowers et al., 1989)

Definition 1.1Conditional state independence Let X be a set of insurancerisks X1,X2, . . ., then a premium calculation principle (pcp) isdefined as the map

π : X → [0,∞],

such thatπ[X ] ≥ E[X ],

for every X ∈ X .

Note that π implies ordering for each X ,Y ∈ X .


Premium equations



π : X → [0,∞],

such that

π[X ] ≥ E[X ],

for every X ∈ X .



Premium equations



π : X → [0,∞],


for every X ∈ X .



Premium equations



π : X → [0,∞],


for every X ∈ X .



Premium equations

What you pay is what you get

Letw1 denote the capital an insurance buyer has;X be the risk the person faces;π represent the premium the buyer is ready to pay.

Then π may be determined from the equation

w1 − π = E[w1 − X ] = w1 − E[X ],

thus yieldingπ = E[X ].


Premium equations




w1 − π = E[w1 − X ] = w1 − E[X ],

thus yielding

π = E[X ].


Premium equations




w1 − π = E[w1 − X ] = w1 − E[X ],

thus yieldingπ = E[X ].


Premium equations

Insured side

Let us now assume the above mentioned person has autility function that describes his/her preferences. Moreprecisely, u : Y → R, a function from e.g. a consumptionset to reals.

The premium defining equation from the buyer’s point ofview becomes:

u(w1 − π) = E[u(w1 − X )]

We will further assume that u : R→ R and the person1 prefers more to less, and2 enjoys much more when is poor and much less when is

rich.


Premium equations

Insured side

Let us now assume the above mentioned person has autility function that describes his/her preferences. Moreprecisely, u : Y → R, a function from e.g. a consumptionset to reals.The premium defining equation from the buyer’s point ofview becomes:

u(w1 − π) = E[u(w1 − X )]


rich.


Premium equations

Insured side

Let us now assume the above mentioned person has autility function that describes his/her preferences. Moreprecisely, u : Y → R, a function from e.g. a consumptionset to reals.The premium defining equation from the buyer’s point ofview becomes:

u(w1 − π) = E[u(w1 − X )]


rich.


Premium equations

The latter two assumption are reformulated as

u′(x) > 0

and

u′′(x) < 0.

Using the above and Jensen’s inequality, we arrive at

u(w1 − π) = E[u(w1 − X )] ≤ u(w1 − E[X ]),

leading to the requirement

π ≥ E[X ],

or alternatively0 ≥ E[X ]− π.


Premium equations


u′(x) > 0

andu′′(x) < 0.


u(w1 − π) = E[u(w1 − X )] ≤ u(w1 − E[X ]),


π ≥ E[X ],



Premium equations


u′(x) > 0

andu′′(x) < 0.


u(w1 − π) = E[u(w1 − X )] ≤

u(w1 − E[X ]),


π ≥ E[X ],



Premium equations


u′(x) > 0

andu′′(x) < 0.


u(w1 − π) = E[u(w1 − X )] ≤ u(w1 − E[X ]),


π ≥ E[X ],



Premium equations


u′(x) > 0

andu′′(x) < 0.


u(w1 − π) = E[u(w1 − X )] ≤ u(w1 − E[X ]),


π ≥ E[X ],



Premium equations

Insurer’s side

Let us now look at the insurer side. Letw2 denote the capital the insurer has;X be the risk the insurer takes (transferred to him by thebuyer) ;θ represent the premium the insurer intends to charge.

Then the premium defining equation from the insurer’s side is:

u(w2) = E[u(w2 + θ − X )] ≤ u(E[w2 + θ − X ])

thus yieldingE[θ − X ] ≥ 0⇔ θ ≥ E[X ],

or alternatively demanding non-positive loss, i.e.,

E[X ]− θ ≤ 0.


Premium equations

Insurer’s side



u(w2) = E[u(w2 + θ − X )] ≤ u(E[w2 + θ − X ])



E[X ]− θ ≤ 0.


Premium equations

Insurer’s side



u(w2) = E[u(w2 + θ − X )] ≤

u(E[w2 + θ − X ])



E[X ]− θ ≤ 0.


Premium equations

Insurer’s side



u(w2) = E[u(w2 + θ − X )] ≤ u(E[w2 + θ − X ])

thus yielding

E[θ − X ] ≥ 0⇔ θ ≥ E[X ],


E[X ]− θ ≤ 0.


Premium equations

Insurer’s side



u(w2) = E[u(w2 + θ − X )] ≤ u(E[w2 + θ − X ])



E[X ]− θ ≤ 0.


Premium equations

Feasibility

Definition 1.2An insurance contract is called feasible whenever

π ≥ θ ≥ E[X ].

There are a lot of various pricing principles in the literature.In this course we will mostly assume linear utility(u′′(x) = 0) that is

π = θ = E[X ],

and thus the insurance contract is always feasible.Moreover, the loss has then to be zero (equivalenceprinciple), i.e., e.g.,

E[X ]− π = 0.


Premium equations

Feasibility


π ≥ θ ≥ E[X ].

There are a lot of various pricing principles in the literature.

In this course we will mostly assume linear utility(u′′(x) = 0) that is

π = θ = E[X ],


E[X ]− π = 0.


Premium equations

Feasibility


π ≥ θ ≥ E[X ].

There are a lot of various pricing principles in the literature.In this course we will mostly assume linear utility(u′′(x) = 0) that is

π = θ = E[X ],


E[X ]− π = 0.


Premium equations

Example 1.1

A person age (x) enters a contract that pays $1 at the end ofthe year of death of the buyer. In return, the buyer is required topay premiums at the beginning of every year while he is alive.State the formula for the premium using the equivalenceprinciple.

SolutionWe observe that

E[π] = E[P..aK+1 ] = P

..ax

andE[X ] = E[vK+1] = Ax .


Premium equations

Example 1.1



E[π] = E[P..aK+1 ] = P

..ax



Premium equations

Example 1.1



E[π] = E[P..aK+1 ] =

P..ax



Premium equations

Example 1.1



E[π] = E[P..aK+1 ] = P

..ax

andE[X ] = E[vK+1] =

Ax .


Premium equations

Example 1.1



E[π] = E[P..aK+1 ] = P

..ax



Premium equations

SolutionThus, due to the equivalence principle, we require that

Ax − P..ax = 0⇔ P = Ax/

..ax .

The loss random variable is given by

L = vK+1 − P..aK+1 ,

we can find its variance:

Var[L] = Var[vK+1 − P..aK+1 ]

= Var[vK+1 − P

1− vK+1

d

]= Var

[vK+1 − P

d+ P

vK+1

d

]= Var

[vK+1

(1 +

Pd

)− P

d

].


Premium equations


Ax − P..ax = 0⇔ P = Ax/

..ax .


L = vK+1 − P..aK+1 ,



= Var[vK+1 − P

1− vK+1

d

]= Var

[vK+1 − P

d+ P

vK+1

d

]= Var

[vK+1

(1 +

Pd

)− P

d

].


Premium equations


Ax − P..ax = 0⇔ P = Ax/

..ax .


L = vK+1 − P..aK+1 ,


Var[L] =

Var[vK+1 − P..aK+1 ]

= Var[vK+1 − P

1− vK+1

d

]= Var

[vK+1 − P

d+ P

vK+1

d

]= Var

[vK+1

(1 +

Pd

)− P

d

].


Premium equations


Ax − P..ax = 0⇔ P = Ax/

..ax .


L = vK+1 − P..aK+1 ,



=

Var[vK+1 − P

1− vK+1

d

]= Var

[vK+1 − P

d+ P

vK+1

d

]= Var

[vK+1

(1 +

Pd

)− P

d

].


Premium equations


Ax − P..ax = 0⇔ P = Ax/

..ax .


L = vK+1 − P..aK+1 ,



= Var[vK+1 − P

1− vK+1

d

]=

Var[vK+1 − P

d+ P

vK+1

d

]= Var

[vK+1

(1 +

Pd

)− P

d

].


Premium equations


Ax − P..ax = 0⇔ P = Ax/

..ax .


L = vK+1 − P..aK+1 ,



= Var[vK+1 − P

1− vK+1

d

]= Var

[vK+1 − P

d+ P

vK+1

d

]=

Var[vK+1

(1 +

Pd

)− P

d

].


Premium equations


Ax − P..ax = 0⇔ P = Ax/

..ax .


L = vK+1 − P..aK+1 ,



= Var[vK+1 − P

1− vK+1

d

]= Var

[vK+1 − P

d+ P

vK+1

d

]= Var

[vK+1

(1 +

Pd

)− P

d

].


Premium equations

SolutionThen,

Var[L] =

Var[vK+1

(1 +

Pd

)]= Var

[vK+1

](1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Ax..axd

)2

=2Ax − (Ax )2

(..axd)2

,

since..axd + Ax = 1.


Premium equations

SolutionThen,

Var[L] = Var[vK+1

(1 +

Pd

)]=

Var[vK+1

](1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Ax..axd

)2

=2Ax − (Ax )2

(..axd)2

,



Premium equations

SolutionThen,

Var[L] = Var[vK+1

(1 +

Pd

)]= Var

[vK+1

](1 +

Pd

)2

=

(2Ax − (Ax )2)

(1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Ax..axd

)2

=2Ax − (Ax )2

(..axd)2

,



Premium equations

SolutionThen,

Var[L] = Var[vK+1

(1 +

Pd

)]= Var

[vK+1

](1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Pd

)2

=

(2Ax − (Ax )2)

(1 +

Ax..axd

)2

=2Ax − (Ax )2

(..axd)2

,



Premium equations

SolutionThen,

Var[L] = Var[vK+1

(1 +

Pd

)]= Var

[vK+1

](1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Ax..axd

)2

=

2Ax − (Ax )2

(..axd)2

,



Premium equations

SolutionThen,

Var[L] = Var[vK+1

(1 +

Pd

)]= Var

[vK+1

](1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Pd

)2

= (2Ax − (Ax )2)

(1 +

Ax..axd

)2

=2Ax − (Ax )2

(..axd)2

,



Premium equations

We can easily generalize the ideas in Example 1. Indeed,denote by

b(K + 1) the benefit function,

v(K + 1) the discount function,P the general symbol for the premium,Z the discrete annuity random variable (e.g.

..aK+1 ).

Then using the already familiar ‘equivalence principle’, wearrive at:

P =E[b(K + 1)v(K + 1)]

E[Z ].

Putting b(K + 1) ≡ 1, v(K + 1) = vk+1 and Z =..aK+1 , we

arrive at the case of Example 1.


Premium equations


b(K + 1) the benefit function,v(K + 1) the discount function,

P the general symbol for the premium,Z the discrete annuity random variable (e.g.

..aK+1 ).


P =E[b(K + 1)v(K + 1)]

E[Z ].




Premium equations


b(K + 1) the benefit function,v(K + 1) the discount function,P the general symbol for the premium,

Z the discrete annuity random variable (e.g...aK+1 ).


P =E[b(K + 1)v(K + 1)]

E[Z ].




Premium equations


b(K + 1) the benefit function,v(K + 1) the discount function,P the general symbol for the premium,Z the discrete annuity random variable (e.g.

..aK+1 ).


P =E[b(K + 1)v(K + 1)]

E[Z ].




Premium equations


b(K + 1) the benefit function,v(K + 1) the discount function,P the general symbol for the premium,Z the discrete annuity random variable (e.g.

..aK+1 ).


P =E[b(K + 1)v(K + 1)]

E[Z ].




Premium equations

Example 1.2Let us now consider the case of the n-year term insurance withpremiums payable at the beginning of each year until end of thecontract. Derive the equation for the premium and calculate thevariance of the loss.

Solution

We have b(k + 1) ≡ 1, v(k + 1) = vk+1 and for the premiums

Z =

{ ..aK+1 , 0 ≤ K < n

..an , K ≥ n

.

As to the loss, let

Z ∗ =

{vK+1, K = 0,1,2, . . . ,n − 1vn, elsewhere

.


Premium equations


Solution


Z =

{ ..aK+1 , 0 ≤ K < n

..an , K ≥ n

.

As to the loss, let

Z ∗ =


.


Premium equations


Solution


Z =

{ ..aK+1 , 0 ≤ K < n

..an , K ≥ n

.

As to the loss, let

Z ∗ =


.


Premium equations


Solution


Z =

{ ..aK+1 , 0 ≤ K < n

..an , K ≥ n

.

As to the loss, let

Z ∗ =


.


Premium equations


Solution


Z =

{ ..aK+1 , 0 ≤ K < n

..an , K ≥ n

.

As to the loss, let

Z ∗ =

{vK+1, K = 0,1,2, . . . ,n − 1

vn, elsewhere.


Premium equations


Solution


Z =

{ ..aK+1 , 0 ≤ K < n

..an , K ≥ n

.

As to the loss, let

Z ∗ =


.


Premium equations

SolutionThus the premium is formulated as

Px :n =E[Z ∗]E[Z ]

=

Ax :n..ax :n

.

The loss random variable is

L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗

d.

Thus the variance is

Var[L] = Var[Z ∗ − Px :n

1− Z ∗

d

]= Var

[Z ∗(

1 +Px :n

d

)− Px :n

d

]= Var

[Z ∗(

1 +Px :n

d

)].


Premium equations


Px :n =E[Z ∗]E[Z ]

=Ax :n..ax :n

.


L =

Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗

d.



1− Z ∗

d

]= Var

[Z ∗(

1 +Px :n

d

)− Px :n

d

]= Var

[Z ∗(

1 +Px :n

d

)].


Premium equations


Px :n =E[Z ∗]E[Z ]

=Ax :n..ax :n

.


L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗

d.


Var[L] =

Var[Z ∗ − Px :n

1− Z ∗

d

]= Var

[Z ∗(

1 +Px :n

d

)− Px :n

d

]= Var

[Z ∗(

1 +Px :n

d

)].


Premium equations


Px :n =E[Z ∗]E[Z ]

=Ax :n..ax :n

.


L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗

d.



1− Z ∗

d

]=

Var[Z ∗(

1 +Px :n

d

)− Px :n

d

]= Var

[Z ∗(

1 +Px :n

d

)].


Premium equations


Px :n =E[Z ∗]E[Z ]

=Ax :n..ax :n

.


L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗

d.



1− Z ∗

d

]= Var

[Z ∗(

1 +Px :n

d

)− Px :n

d

]=

Var[Z ∗(

1 +Px :n

d

)].


Premium equations


Px :n =E[Z ∗]E[Z ]

=Ax :n..ax :n

.


L = Z ∗ − Px :n Z = Z ∗ − Px :n1− Z ∗

d.



1− Z ∗

d

]= Var

[Z ∗(

1 +Px :n

d

)− Px :n

d

]= Var

[Z ∗(

1 +Px :n

d

)].


Premium equations

SolutionFurther because of the equivalence principle, we have that

Var[L] =

(1 +

Px :n

d

)2 (2Ax :n − (Ax :n )2

)=

(2Ax :n − (Ax :n )2)(d

..ax :n )2

,

recalling that1 = d

..ax :n + Ax :n

that leads to

1 + Px :n /d = 1 +Ax :n

d..ax :n

=1

d..ax :n

.


Premium equations


Var[L] =

(1 +

Px :n

d

)2 (2Ax :n − (Ax :n )2

)=

(2Ax :n − (Ax :n )2)(d

..ax :n )2

,

recalling that1 = d

..ax :n + Ax :n

that leads to

1 + Px :n /d = 1 +Ax :n

d..ax :n

=1

d..ax :n

.


Premium equations


Var[L] =

(1 +

Px :n

d

)2 (2Ax :n − (Ax :n )2

)=

(2Ax :n − (Ax :n )2)(d

..ax :n )2

,

recalling that1 = d

..ax :n + Ax :n

that leads to

1 + Px :n /d = 1 +Ax :n

d..ax :n

=1

d..ax :n

.


Premium equations

Quantile premium

Besides the equivalence principle, there are a lot of other usefulways of pricing insurance contracts. We shall now define the socalled quantile based premium.

Definition 1.3 (Quantile or Value-at-Risk premium)The quantile based premium for a continuous random variableL is the solution in π of the following equation:

P(L(π) ≥ 0) = q,

where L represents the loss random variable and q ∈ (0,1).

It is useful to recall the following definition of the q-th quantile ofX v F ,

xq = inf{x : F (x) ≥ q} = sup{x : F (x) ≤ q}.


Premium equations

Example 1.3Consider a 10000 fully discrete whole life insurance. Use ELTand interest rate equal to 6% and issue age 35. Calculate therequired premium such that

the expectation of the distribution of loss is zero,the probability is less than 0.5 that the loss is positive(approximate the smallest premium).the probability of a positive total loss is 0.05 using thenormal. approximation

Find the variance of the loss distribution in these three cases.


Premium equations

Solutiona.

Due to the equivalence principle

P = 10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.

The variance is:

Var[L] = 1000022A35 − (A35)2

d..a35

= 100002 0.0348843− 0.12871942

((0.06/1.06)(15.39262))2

= 2412713

and √Var[L] = 1553.3.


Premium equations

Solutiona. Due to the equivalence principle

P = 10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.

The variance is:

Var[L] = 1000022A35 − (A35)2

d..a35

= 100002 0.0348843− 0.12871942

((0.06/1.06)(15.39262))2

= 2412713

and √Var[L] = 1553.3.


Premium equations


P =

10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.

The variance is:

Var[L] = 1000022A35 − (A35)2

d..a35

= 100002 0.0348843− 0.12871942

((0.06/1.06)(15.39262))2

= 2412713

and √Var[L] = 1553.3.


Premium equations


P = 10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.

The variance is:

Var[L] = 1000022A35 − (A35)2

d..a35

= 100002 0.0348843− 0.12871942

((0.06/1.06)(15.39262))2

= 2412713

and √Var[L] = 1553.3.


Premium equations


P = 10000A35/..a35 = 10000 · 0.1287194/15.39262 = 83.62.

The variance is:

Var[L] = 1000022A35 − (A35)2

d..a35

= 100002 0.0348843− 0.12871942

((0.06/1.06)(15.39262))2

= 2412713

and √Var[L] = 1553.3.


Premium equations

Solutionb. We are interested in

P(L(π) > 0) < 0.5⇔ P(

10000vK+1 − π1− vK+1

d> 0

)< 0.5.

Thus we arrive at

P(

10000vK+1 − π

d+ π

vK+1

d> 0

)< 0.5

⇔ P(

vK+1(

10000 +π

d

)− π

d> 0

)< 0.5

⇔ P(

vK+1 >π

d10000 + π

)< 0.5

⇔ P(

(K + 1) log(v) > log(

π

d10000 + π

))< 0.5.


Premium equations


P(L(π) > 0) < 0.5⇔

P(

10000vK+1 − π1− vK+1

d> 0

)< 0.5.

Thus we arrive at

P(

10000vK+1 − π

d+ π

vK+1

d> 0

)< 0.5

⇔ P(

vK+1(

10000 +π

d

)− π

d> 0

)< 0.5

⇔ P(

vK+1 >π

d10000 + π

)< 0.5

⇔ P(

(K + 1) log(v) > log(

π

d10000 + π

))< 0.5.


Premium equations


P(L(π) > 0) < 0.5⇔ P(

10000vK+1 − π1− vK+1

d> 0

)< 0.5.

Thus we arrive at

P(

10000vK+1 − π

d+ π

vK+1

d> 0

)< 0.5

⇔ P(

vK+1(

10000 +π

d

)− π

d> 0

)< 0.5

⇔ P(

vK+1 >π

d10000 + π

)< 0.5

⇔ P(

(K + 1) log(v) > log(

π

d10000 + π

))< 0.5.


Premium equations


P(L(π) > 0) < 0.5⇔ P(

10000vK+1 − π1− vK+1

d> 0

)< 0.5.

Thus we arrive at

P(

10000vK+1 − π

d+ π

vK+1

d> 0

)< 0.5

⇔ P(

vK+1(

10000 +π

d

)− π

d> 0

)< 0.5

⇔ P(

vK+1 >π

d10000 + π

)< 0.5

⇔ P(

(K + 1) log(v) > log(

π

d10000 + π

))< 0.5.


Premium equations


P(L(π) > 0) < 0.5⇔ P(

10000vK+1 − π1− vK+1

d> 0

)< 0.5.

Thus we arrive at

P(

10000vK+1 − π

d+ π

vK+1

d> 0

)< 0.5

⇔ P(

vK+1(

10000 +π

d

)− π

d> 0

)< 0.5

⇔ P(

vK+1 >π

d10000 + π

)< 0.5

⇔ P(

(K + 1) log(v) > log(

π

d10000 + π

))< 0.5.


Premium equations


P(L(π) > 0) < 0.5⇔ P(

10000vK+1 − π1− vK+1

d> 0

)< 0.5.

Thus we arrive at

P(

10000vK+1 − π

d+ π

vK+1

d> 0

)< 0.5

⇔ P(

vK+1(

10000 +π

d

)− π

d> 0

)< 0.5

⇔ P(

vK+1 >π

d10000 + π

)< 0.5

⇔ P(

(K + 1) log(v) > log(

π

d10000 + π

))< 0.5.


Premium equations

SolutionAs log(v) < 0, we have that

⇔ P(

K + 1 < log(

π

d10000 + π

)/ log(v)

)< 0.5

⇔ P(

K < log(

π

d10000 + π

)/ log(v)− 1

)< 0.5

⇔ P(

K (x) ≥ log(

π

d10000 + π

)/ log(v)− 1

)> 0.5

We thus look for the highest value k for which P(K (x) > k − 1)=k px > 0.5. Take k − 1 = 41, then 42p35 = 0.5125. Go further,take k − 1 = 42, then 43p35 = 0.4808 - less than 0.5. Thus thenecessary age is k − 1 = 41. Solving

log(

π

d10000 + π

)/ log(v)− 1 = 42


Premium equations


⇔ P(

K + 1 < log(

π

d10000 + π

)/ log(v)

)< 0.5

⇔ P(

K < log(

π

d10000 + π

)/ log(v)− 1

)< 0.5

⇔ P(

K (x) ≥ log(

π

d10000 + π

)/ log(v)− 1

)> 0.5


log(

π

d10000 + π

)/ log(v)− 1 = 42


Premium equations


⇔ P(

K + 1 < log(

π

d10000 + π

)/ log(v)

)< 0.5

⇔ P(

K < log(

π

d10000 + π

)/ log(v)− 1

)< 0.5

⇔ P(

K (x) ≥ log(

π

d10000 + π

)/ log(v)− 1

)> 0.5

We thus look for the highest value k for which P(K (x) > k − 1)=k px > 0.5. Take k − 1 = 41, then

42p35 = 0.5125. Go further,take k − 1 = 42, then 43p35 = 0.4808 - less than 0.5. Thus thenecessary age is k − 1 = 41. Solving

log(

π

d10000 + π

)/ log(v)− 1 = 42


Premium equations


⇔ P(

K + 1 < log(

π

d10000 + π

)/ log(v)

)< 0.5

⇔ P(

K < log(

π

d10000 + π

)/ log(v)− 1

)< 0.5

⇔ P(

K (x) ≥ log(

π

d10000 + π

)/ log(v)− 1

)> 0.5


log(

π

d10000 + π

)/ log(v)− 1 = 42


Premium equations


⇔ P(

K + 1 < log(

π

d10000 + π

)/ log(v)

)< 0.5

⇔ P(

K < log(

π

d10000 + π

)/ log(v)− 1

)< 0.5

⇔ P(

K (x) ≥ log(

π

d10000 + π

)/ log(v)− 1

)> 0.5


log(

π

d10000 + π

)/ log(v)− 1 = 42


Premium equations

SolutionOr equivalently

log(

π

d10000 + π

)= 43 log(v),

that is

π

d10000 + π= v43

leads to

π =d10000v43

1− v43 =46.20240.9184

= 50.31.

The variance is given by

Var[L(π)] = 100002(2A35 − (A35)2)(

1 +π

d10000

)2

Thus√

Var[L(π)] = 1473.65.


Premium equations


log(

π

d10000 + π

)= 43 log(v),

that isπ

d10000 + π= v43

leads to

π =d10000v43

1− v43 =46.20240.9184

= 50.31.


Var[L(π)] = 100002(2A35 − (A35)2)(

1 +π

d10000

)2

Thus√

Var[L(π)] = 1473.65.


Premium equations


log(

π

d10000 + π

)= 43 log(v),

that isπ

d10000 + π= v43

leads to

π =d10000v43

1− v43 =46.20240.9184

= 50.31.


Var[L(π)] = 100002(2A35 − (A35)2)(

1 +π

d10000

)2

Thus√

Var[L(π)] = 1473.65.


Premium equations

Solutionc. This time we are interested in

P(L(π) > 0) = 0.05

using normal approximation. Thus,

P((L(π)−E[L(π)])/√

Var[L(π)] > −E[L(π)]/√

Var[L(π)]) = 0.05,

leading to

P((L(π)−E[L(π)])/√

Var[L(π)] ≤ −E[L(π)]/√

Var[L(π)]) = 0.95.

We know that in the normal case, Φ−1(0.95) = 1.6449, thus tofind the premium the following equation in π has to be solved:

−E[L(π)]/√

Var[L(π)] = 1.6449


Premium equations


P(L(π) > 0) = 0.05


P((L(π)−E[L(π)])/√

Var[L(π)] > −E[L(π)]/√

Var[L(π)]) = 0.05,

leading to

P((L(π)−E[L(π)])/√

Var[L(π)] ≤ −E[L(π)]/√

Var[L(π)]) = 0.95.


−E[L(π)]/√

Var[L(π)] = 1.6449


Premium equations


P(L(π) > 0) = 0.05


P((L(π)−E[L(π)])/√

Var[L(π)] > −E[L(π)]/√

Var[L(π)]) = 0.05,

leading to

P((L(π)−E[L(π)])/√

Var[L(π)] ≤ −E[L(π)]/√

Var[L(π)]) = 0.95.


−E[L(π)]/√

Var[L(π)] = 1.6449


Premium equations


P(L(π) > 0) = 0.05


P((L(π)−E[L(π)])/√

Var[L(π)] > −E[L(π)]/√

Var[L(π)]) = 0.05,

leading to

P((L(π)−E[L(π)])/√

Var[L(π)] ≤ −E[L(π)]/√

Var[L(π)]) = 0.95.


−E[L(π)]/√

Var[L(π)] = 1.6449


Premium equations


P(L(π) > 0) = 0.05


P((L(π)−E[L(π)])/√

Var[L(π)] > −E[L(π)]/√

Var[L(π)]) = 0.05,

leading to

P((L(π)−E[L(π)])/√

Var[L(π)] ≤ −E[L(π)]/√

Var[L(π)]) = 0.95.


−E[L(π)]/√

Var[L(π)] = 1.6449


Premium equations

SolutionNote that

E[L(π)] =

(10000− π

d

)A35 −

π

d.

andVar[L(π)] =

(10000− π

d

)2(2A35 − (A35)2).

Taking into account the above, the premium is 100.66.

Under the equivalence principle we have that

..a−1x = πx + d and πx =

dAx

1− Ax.

Check the above for term insurances.


Premium equations

SolutionNote that

E[L(π)] =(

10000− π

d

)A35 −

π

d.

andVar[L(π)] =

(10000− π

d

)2(2A35 − (A35)2).




dAx

1− Ax.



Premium equations

SolutionNote that

E[L(π)] =(

10000− π

d

)A35 −

π

d.

andVar[L(π)] =

(10000− π

d

)2(2A35 − (A35)2).




dAx

1− Ax.



Premium equations

It is interesting to note that both sides of

..a−1x = πx + d

stand for the annual amount paid by a life annuityproduced by one monetary unit at time zero. Indeed

1 =..a−1x

..ax = (πx + d)

..ax .


Premium equations

It is interesting to note that both sides of

..a−1x = πx + d

stand for the annual amount paid by a life annuityproduced by one monetary unit at time zero. Indeed

1 =..a−1x

..ax = (πx + d)

..ax .


Premium equations

Problem 1.1A person age (x) borrows an amount of money equal to Ax inorder to buy a whole life insurance with unit benefit and a singlepremium payed at the beginning of the contract. He repays theinterest on Ax in advance at the beginning of each year ifsurvives and agrees to repay the loan itself from the unit deathbenefit at the end of the year of death. What is the annualpremium for a full unit of this insurance?

SolutionThe equation is actually

dAx..ax + Ax · Ax = Ax

ordAx

..ax = (1− Ax )Ax .


Premium equations



dAx..ax

+ Ax · Ax = Ax

ordAx

..ax = (1− Ax )Ax .


Premium equations



dAx..ax + Ax

· Ax = Ax

ordAx

..ax = (1− Ax )Ax .


Premium equations



dAx..ax + Ax ·

Ax = Ax

ordAx

..ax = (1− Ax )Ax .


Premium equations



dAx..ax + Ax · Ax =

Ax

ordAx

..ax = (1− Ax )Ax .


Premium equations




or

dAx..ax = (1− Ax )Ax .


Premium equations




ordAx

..ax = (1− Ax )Ax .


Premium equations

Figure: Discrete premiums


Premium equations

Continuous premiums

The same reasoning applies to continuous premiums. Forinstance, when we have a whole life continuous insurance Axand premiums are payable continuously then due to theequivalence principle, we obtain that:

P(Ax ) =Ax

ax

Also

Var[L] = Var[vT − P

1− vT

δ

]=

2Ax − (Ax )2

(δax )2 .


Premium equations

Continuous premiums


P(Ax ) =Ax

ax

Also

Var[L] =

Var[vT − P

1− vT

δ

]=

2Ax − (Ax )2

(δax )2 .


Premium equations

Continuous premiums


P(Ax ) =Ax

ax

Also

Var[L] = Var[vT − P

1− vT

δ

]=

2Ax − (Ax )2

(δax )2 .


Premium equations

Example 1.4Calculate the premium when the force of mortality is constantµ, and the force of interest is δ.

SolutionWith the above assumptions

Ax =

∫ ∞0

e−δtµe−µtdt

=µ

µ+ δ

∫ ∞0

(µ+ δ)e−(δ+µ)tdt

=µ

µ+ δ.

In our caseAx = 0.04/0.1 = 0.4.

Also,


Premium equations



Ax =

∫ ∞0

e−δtµe−µtdt

=

µ

µ+ δ

∫ ∞0


=µ

µ+ δ.

In our caseAx = 0.04/0.1 = 0.4.

Also,


Premium equations



Ax =

∫ ∞0

e−δtµe−µtdt

=µ

µ+ δ

∫ ∞0


=

µ

µ+ δ.

In our caseAx = 0.04/0.1 = 0.4.

Also,


Premium equations



Ax =

∫ ∞0

e−δtµe−µtdt

=µ

µ+ δ

∫ ∞0


=µ

µ+ δ.

In our caseAx = 0.04/0.1 = 0.4.

Also,


Premium equations

Solution

ax =

1− Ax

δ=

1− 0.40.06

= 10.

The premium is then:

P(Ax ) = 0.4/10 = 0.04 = µ,

a coincidence? As to the variance, we have that:

2Ax =

∫ ∞0

e−2δtµe−µtdt

=µ

2δ + µ= 0.25.

Thus

Var[L] =25− 0.42

(0.06 · 10)2 = 0.25.


Premium equations

Solution

ax =1− Ax

δ=

1− 0.40.06

= 10.


P(Ax ) = 0.4/10 = 0.04 = µ,


2Ax =

∫ ∞0

e−2δtµe−µtdt

=µ

2δ + µ= 0.25.

Thus

Var[L] =25− 0.42

(0.06 · 10)2 = 0.25.


Premium equations

Solution

ax =1− Ax

δ=

1− 0.40.06

= 10.


P(Ax ) = 0.4/10 = 0.04 = µ,


2Ax =

∫ ∞0

e−2δtµe−µtdt

=µ

2δ + µ= 0.25.

Thus

Var[L] =25− 0.42

(0.06 · 10)2 = 0.25.


Premium equations

Solution

ax =1− Ax

δ=

1− 0.40.06

= 10.


P(Ax ) = 0.4/10 = 0.04 = µ,


2Ax =

∫ ∞0

e−2δtµe−µtdt

=µ

2δ + µ= 0.25.

Thus

Var[L] =25− 0.42

(0.06 · 10)2 = 0.25.


Premium equations

Solution

ax =1− Ax

δ=

1− 0.40.06

= 10.


P(Ax ) = 0.4/10 = 0.04 = µ,


2Ax =

∫ ∞0

e−2δtµe−µtdt

=

µ

2δ + µ= 0.25.

Thus

Var[L] =25− 0.42

(0.06 · 10)2 = 0.25.


Premium equations

Solution

ax =1− Ax

δ=

1− 0.40.06

= 10.


P(Ax ) = 0.4/10 = 0.04 = µ,


2Ax =

∫ ∞0

e−2δtµe−µtdt

=µ

2δ + µ= 0.25.

Thus

Var[L] =

25− 0.42

(0.06 · 10)2 = 0.25.


Premium equations

Solution

ax =1− Ax

δ=

1− 0.40.06

= 10.


P(Ax ) = 0.4/10 = 0.04 = µ,


2Ax =

∫ ∞0

e−2δtµe−µtdt

=µ

2δ + µ= 0.25.

Thus

Var[L] =25− 0.42

(0.06 · 10)2 = 0.25.


Premium equations

Let us have a closer look at the loss random variable forthe whole insurance contract.

Figure: Loss rv L is a non-increasing function since log(v) < 0

Note that the events {L > 0} and {T < c} are equivalent.


Premium equations





Premium equations



Note that the events {L > 0} and {T < c} are

equivalent.


Premium equations





Premium equations

If the premiums is chosen such that P[L > 0] = q, thenbecause of the equivalence of {L > 0} and {T < c}, itholds that P[T < c] = q, where c = xq of T .

Thusvxq − P

..axq = 0,

from which

P =vxq

..axq

=1

..axq (1 + i)xq

=1..sxq

.

Intuitive?Of course! With probability q, we have that

..sT /

..sxq < 1,

and with probability 1− q, we have that..sT /

..sxq > 1


Premium equations

If the premiums is chosen such that P[L > 0] = q, thenbecause of the equivalence of {L > 0} and {T < c}, itholds that P[T < c] = q, where c = xq of T .Thus

vxq − P..axq = 0,

from which

P =vxq

..axq

=

1..axq (1 + i)xq

=1..sxq

.


..sT /

..sxq < 1,


..sxq > 1


Premium equations


vxq − P..axq = 0,

from which

P =vxq

..axq

=1

..axq (1 + i)xq

=

1..sxq

.


..sT /

..sxq < 1,


..sxq > 1


Premium equations


vxq − P..axq = 0,

from which

P =vxq

..axq

=1

..axq (1 + i)xq

=1..sxq

.

Intuitive?

Of course! With probability q, we have that..sT /

..sxq < 1,


..sxq > 1


Premium equations


vxq − P..axq = 0,

from which

P =vxq

..axq

=1

..axq (1 + i)xq

=1..sxq

.

Intuitive?Of course!

With probability q, we have that..sT /

..sxq < 1,


..sxq > 1


Premium equations


vxq − P..axq = 0,

from which

P =vxq

..axq

=1

..axq (1 + i)xq

=1..sxq

.


..sT /

..sxq < 1,


..sxq > 1


Premium equations

Figure: The cdf of the loss rv associated with a whole life insurance.The cdf is a function of two variables, i.e., the premium and the loss.


Premium equations

Example 1.5

Let fT (55)(t) =t p55µ(55 + t) = 1/45, 0 < t < 45. Display thecdf of L and determine the amount P as the smallestnon-negative number for which P[L(P) > 0] ≤ 0.25, in thecontext of

a. 20 year endowment insuranceb. 20 year term insurancec. 10 year term insurance.

Also the force of interest is 0.06.

Solution

We find FL first. Note that FL(u) = 0 for all u < v20 − Pa20 . Theloss is thus bounded from below, i.e., the profit cannot exceedcertain value.

Further for v20 − Pa20 ≤ u ≤ 1, we have that


Premium equations

Example 1.5

Let fT (55)(t) =t p55µ(55 + t) = 1/45, 0 < t < 45. Display thecdf of L and determine the amount P as the smallestnon-negative number for which P[L(P) > 0] ≤ 0.25, in thecontext of

a. 20 year endowment insuranceb. 20 year term insurancec. 10 year term insurance.

Also the force of interest is 0.06.

Solution

We find FL first. Note that FL(u) = 0 for all u < v20 − Pa20 . Theloss is thus bounded from below, i.e., the profit cannot exceedcertain value. Further for v20 − Pa20 ≤ u ≤ 1, we have that


Premium equations

Solution

FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]

= P[vT − P

1− vT

δ≤ u

]= P

[vT ≤ δu + P

δ + P

]

= P

[T ≥ −1

δlog

(δu + Pδ + P

)]

= 1− FT

(−1δ

log

(δu + Pδ + P

)),

for all u > −P/δ. Surprised?


Premium equations

Solution

FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]

= P[vT − P

1− vT

δ≤ u

]

= P

[vT ≤ δu + P

δ + P

]

= P

[T ≥ −1

δlog

(δu + Pδ + P

)]

= 1− FT

(−1δ

log

(δu + Pδ + P

)),



Premium equations

Solution

FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]

= P[vT − P

1− vT

δ≤ u

]= P

[vT ≤ δu + P

δ + P

]

= P

[T ≥ −1

δlog

(δu + Pδ + P

)]

= 1− FT

(−1δ

log

(δu + Pδ + P

)),



Premium equations

Solution

FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]

= P[vT − P

1− vT

δ≤ u

]= P

[vT ≤ δu + P

δ + P

]

= P

[T ≥ −1

δlog

(δu + Pδ + P

)]

= 1− FT

(−1δ

log

(δu + Pδ + P

)),



Premium equations

Solution

FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]

= P[vT − P

1− vT

δ≤ u

]= P

[vT ≤ δu + P

δ + P

]

= P

[T ≥ −1

δlog

(δu + Pδ + P

)]

= 1− FT

(−1δ

log

(δu + Pδ + P

)),



Premium equations

Solution

FL(u) = P[L ≤ u] = P[vT − PaT ≤ u]

= P[vT − P

1− vT

δ≤ u

]= P

[vT ≤ δu + P

δ + P

]

= P

[T ≥ −1

δlog

(δu + Pδ + P

)]

= 1− FT

(−1δ

log

(δu + Pδ + P

)),



Premium equations

SolutionWe know the cdf of T , i.e., we know that

FT (u) =145

u, 0 < u < 45.

Hence

P[L > 0] =

1− P[L ≤ 0] = FT

(−1δ

log

(P

δ + P

))= 0.25.

Equivalently

1 +1

45

(1

0.06log

(P

0.06 + P

))= 0.75.

Solving for P, we find thatP = 0.06224.


Premium equations


FT (u) =145

u, 0 < u < 45.

Hence

P[L > 0] = 1− P[L ≤ 0] =

FT

(−1δ

log

(P

δ + P

))= 0.25.

Equivalently

1 +1

45

(1

0.06log

(P

0.06 + P

))= 0.75.



Premium equations


FT (u) =145

u, 0 < u < 45.

Hence

P[L > 0] = 1− P[L ≤ 0] = FT

(−1δ

log

(P

δ + P

))= 0.25.

Equivalently

1 +1

45

(1

0.06log

(P

0.06 + P

))= 0.75.



Premium equations


FT (u) =145

u, 0 < u < 45.

Hence

P[L > 0] = 1− P[L ≤ 0] = FT

(−1δ

log

(P

δ + P

))= 0.25.

Equivalently

1 +1

45

(1

0.06log

(P

0.06 + P

))= 0.75.



Premium equations

SolutionAnother way would be to recall that {L > 0} is equivalent to{T < c}. Thus the premium is

P =1

..sx0.25

,

wherex0.25 = F−1

T (0.25) = 0.25 · 45 = 11.25.

The premium is then again

P = 0.06224.

Going back to the cdf of L, we note that FL(u) for u > 1 is 1.


Premium equations

SolutionAnother way would be to recall that {L > 0} is equivalent to{T < c}. Thus the premium is

P =1

..sx0.25

,

wherex0.25 = F−1

T (0.25) = 0.25 · 45 = 11.25.

The premium is then again

P = 0.06224.

Going back to the cdf of L, we note that FL(u) for u > 1 is 1.


Premium equations

Solution

Figure: The cdf of L when the insurance is a 20 year endowment one.Edward Furman Actuarial mathematics 3280 36 / 46

Premium equations

Solutionb. Now we have a 20 year term insurance. This means that theloss rv L can take on values in the interval

[−Pa20 ,1

].

Thus FL(u) = 0 for all u < −Pa20 .We also know that for u > 1 we have that FL(u) = 1.Another interesting point is v20 − Pa20 . However in theinterval

[v20 − Pa20 ,1

]the cdf behaves as the cdf of a

loss of the whole life insurance. So

FL(u) = 1 +145

(1

0.06log

(0.06u + P0.06 + P

))

for u ∈[v20 − Pa20 ,1

].


Premium equations


[−Pa20 ,1

].

Thus FL(u) = 0 for all u

< −Pa20 .We also know that for u > 1 we have that FL(u) = 1.Another interesting point is v20 − Pa20 . However in theinterval

[v20 − Pa20 ,1



FL(u) = 1 +145

(1

0.06log

(0.06u + P0.06 + P

))

for u ∈[v20 − Pa20 ,1

].


Premium equations


[−Pa20 ,1

].

Thus FL(u) = 0 for all u < −Pa20 .We also know that for u > 1 we have that FL(u) =

1.Another interesting point is v20 − Pa20 . However in theinterval

[v20 − Pa20 ,1



FL(u) = 1 +145

(1

0.06log

(0.06u + P0.06 + P

))

for u ∈[v20 − Pa20 ,1

].


Premium equations


[−Pa20 ,1

].

Thus FL(u) = 0 for all u < −Pa20 .We also know that for u > 1 we have that FL(u) = 1.Another interesting point is v20 − Pa20 . However in theinterval

[v20 − Pa20 ,1



FL(u) = 1 +145

(1

0.06log

(0.06u + P0.06 + P

))

for u ∈[v20 − Pa20 ,1

].


Premium equations

Solution

In the interval u ∈[−Pa20 , v

20 − Pa20

], we have that

FL(u) = P[T > 20] = 1− 20/45 = 25/45.

The premium is the same as in a. since we again have tofind the 0.25 quantile of L, and it again ‘sits’ on theincreasing part of FL(u).


Premium equations

Solution


20 − Pa20

], we have that

FL(u) = P[T > 20] = 1− 20/45 = 25/45.The premium is the same as in a.

since we again have tofind the 0.25 quantile of L, and it again ‘sits’ on theincreasing part of FL(u).


Premium equations

Solution


20 − Pa20

], we have that

FL(u) = P[T > 20] = 1− 20/45 = 25/45.The premium is the same as in a. since we again have tofind the 0.25 quantile of L, and it again ‘sits’ on theincreasing part of FL(u).


Premium equations

Solution

Figure: The cdf of L when the insurance is a 20 year term one.


Premium equations

Solutionc. We now have a 10 year term insurance. So

FL(u) = 0 for all u < −Pa10 ,

Also, FL(u) = 1 for all u > 1,


10 − Pa10

], we can calculate

the cdf from the probabilityP[T > 10] = 1− P[T ≤ 10] = 1− 10/45 = 35/45 > 0.75.So?As in part b., we have that the cdf is zero first and then wehave a jump to 35/45. This time however the premium isnot P = 0.06224 anymore.


Premium equations


FL(u) = 0 for all u < −Pa10 ,Also, FL(u) = 1 for all u > 1,


10 − Pa10

],

we can calculatethe cdf from the probabilityP[T > 10] = 1− P[T ≤ 10] = 1− 10/45 = 35/45 > 0.75.So?As in part b., we have that the cdf is zero first and then wehave a jump to 35/45. This time however the premium isnot P = 0.06224 anymore.


Premium equations




10 − Pa10

], we can calculate

the cdf from the probabilityP[T > 10] = 1− P[T ≤ 10] = 1− 10/45 = 35/45 > 0.75.So?

As in part b., we have that the cdf is zero first and then wehave a jump to 35/45. This time however the premium isnot P = 0.06224 anymore.


Premium equations




10 − Pa10

], we can calculate

the cdf from the probabilityP[T > 10] = 1− P[T ≤ 10] = 1− 10/45 = 35/45 > 0.75.So?As in part b., we have that the cdf is zero first and then wehave a jump to 35/45. This time however the premium isnot P = 0.06224 anymore.


Premium equations

Solution

Remember that we choose P as the smallest value suchthat P[L(P) > 0] ≤ 0.25.

Try the smallest possible premiumP = 0. Then

P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.

Zero premium and non zero benefit. Reasonable? No.Does it mean that the quantile based premium is “bad”?Not at all. Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.


Premium equations

Solution

Remember that we choose P as the smallest value suchthat P[L(P) > 0] ≤ 0.25. Try the smallest possible premiumP = 0. Then

P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.

Zero premium and non zero benefit. Reasonable?

No.Does it mean that the quantile based premium is “bad”?Not at all. Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.


Premium equations

Solution


P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.

Zero premium and non zero benefit. Reasonable? No.Does it mean that the quantile based premium is “bad”?

Not at all. Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.


Premium equations

Solution


P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.

Zero premium and non zero benefit. Reasonable? No.Does it mean that the quantile based premium is “bad”?Not at all.

Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.


Premium equations

Solution


P[L(0) > 0] = 1− P[L ≤ 0] = 1− 35/45 = 10/45 < 0.25.

Zero premium and non zero benefit. Reasonable? No.Does it mean that the quantile based premium is “bad”?Not at all. Note that in this particular case P < E[L], so P isnot a legitimate pricing principle.


Premium equations

Solution

Figure: The cdf of L when the insurance is a 10 year term one.


Premium equations

Figure: Continuous premiums


Premium equations

Example 1.6Let the utility function of an insurer be

u(x) = x − 0.01x2.

Also, the insurer is known to have the initial wealth w .Determine the annual premium that maximizes the utility above.

Solution

We must find π such that

π = argmaxP

E[u(w + P − X )].

Thus we have a closer look at

∂

∂PE[u(w + P − X )] = E

[∂

∂Pu(w + P − X )

]= 0


Premium equations


u(x) = x − 0.01x2.


SolutionWe must find π such that

π = argmaxP

E[u(w + P − X )].


∂

∂PE[u(w + P − X )] = E

[∂

∂Pu(w + P − X )

]= 0


Premium equations


u(x) = x − 0.01x2.



π = argmaxP

E[u(w + P − X )].


∂

∂PE[u(w + P − X )]

= E[∂

∂Pu(w + P − X )

]= 0


Premium equations


u(x) = x − 0.01x2.



π = argmaxP

E[u(w + P − X )].


∂

∂PE[u(w + P − X )] = E

[∂

∂Pu(w + P − X )

]= 0


Premium equations

SolutionSubstituting u as in the question leads to

E[∂

∂Pu(w + P − X )

]

= E[∂

∂P

((w + P − X )− 0.01(w + P − X )2

)]= E [1− 0.02(w + P − X )] = 0.

which is equivalent to

0.02E [(w + P − X )] = 1

orE [(w + P − X )] = 50


Premium equations


E[∂

∂Pu(w + P − X )

]= E

[∂

∂P

((w + P − X )− 0.01(w + P − X )2

)]

= E [1− 0.02(w + P − X )] = 0.


0.02E [(w + P − X )] = 1

orE [(w + P − X )] = 50


Premium equations


E[∂

∂Pu(w + P − X )

]= E

[∂

∂P

((w + P − X )− 0.01(w + P − X )2

)]= E [1− 0.02(w + P − X )] = 0.


0.02E [(w + P − X )] = 1

orE [(w + P − X )] = 50


Premium equations


E[∂

∂Pu(w + P − X )

]= E

[∂

∂P

((w + P − X )− 0.01(w + P − X )2

)]= E [1− 0.02(w + P − X )] = 0.


0.02E [(w + P − X )] = 1

orE [(w + P − X )] = 50


Premium equations

SolutionIn other words

π = 50− w + E[X ].


Documents

Actuarial mathematics 3280 - Department of Mathematics …efurman/actmath2009/lecture5.pdf · Premium equations Actuarial mathematics 3280 Beneﬁt premiums Edward Furman Department