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Actuarial mathematics 2Life insurance contracts

Edward Furman

Department of Mathematics and StatisticsYork University

January 31, 2011

Edward Furman Actuarial mathematics MATH 3280 1 / 4 5
http://goforward/http://find/http://goback/


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Definition 0.1 (Life insurance.)

Life insurance is a contract that is designed to reduce the

financial impact of an untimely death.

The payment is a single one.

The payment can be made either upon death or thereafter.

Question

Given a life status (u), what is its expected future lifetime? If theinsurance amount is one dollar: 1.) what is the r.v. representing

the payment upon death of (u)? 2.) what is the r.v. representing

the payment at the end of the year of death of (u)?

Recall that, e.g., T(u) : Ru [0, ].



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Solution

The expected lifetime of (u) is E[T(u)].

The payment upon death is vT(u).

The payment at the end of the year of death is vK(u)+1.

Recall that the price is the expected loss, for identity utility =

fairness principle (or equivalence principle).

Net premium for an insurance contract.

The net premium for an immediately payable insurance is

E[vT(u)] for a life status (u). Also, the net premium for an

insurance payable at the end of the year of death isE

[vK(u)+1

].I.e.,

E[vT(u)] =

Ru

vttpu(u+ t)dt =

Ru

vtdtpu := Au.



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Also

E[v

K(u)+1

] =

kRuv

k+1

kpuqu+k =

kRuv

k+1

kpu := Au.

Example 0.1 (Whole life insurance.)

Let (u) = (x). Then Ru = [0, ), and

Ax := E[vT(u)] =

0

vttpx(x + t)dt,

as well as

Ax := E[vK(u)+1] =

k=0

vk+1kpxqx+k.



5/45Edward Furman Actuarial mathematics MATH 3280 5 / 4 5


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Example 0.2 (n-year term insurance.)

Let (u) = (1x : n). Thus Ru = [0, n) {}, and

A1x:n

:= E[vT(u)] =

n0

vttpx(x + t)dt + 0,

as well as

A1x:n

:= E[vK(u)+1] =n1k=0

vk+1kpxqx+k.

Proposition 0.1

We have that, for A 1x:0

= 0 (why?), and for all x,

A1x:n

= vqx + vpxA 1x+1:n1

.



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Proof.

A1x:n

=n1k=0

vk+1kpxqx+k = vqx +n1k=1

vk+1kpxqx+k

= vqx + vpx

n1

k=1

vkk1px+1qx+k

= vqx + vpx

n2k=0

vk+1kpx+1qx+k+1

= vqx + vpxA 1x+1:n1

,

as required.



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Corollary 0.1

We have that

Ax = vqx + vpxAx+1.

Proposition 0.2

Under the UDD assumption for each year of age, we have that

AuUDD

=i

Au.

Proof.

We have that

tpu(u+ t)

UDD

= qu, 0 t 1, u = 0,1, . . . .

Then

Au = = 1

0

vttpu(u+ t)dt +

1

vttpu(u+ t)dt



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Proof (cont.)

Au =

10

vttpu(u+ t)dt +

0

vt+1t+1pu(u+ 1 + t)dt

= 1

0

vttpu(u+ t)dt +

0

vt+1pu tpu+1(u+ 1 + t)dt

=

1

0

vttpu(u+ t)dt + vpu

0

vttpu+1(u+ 1 + t)dt

UDD=

1

0

vtqudt + vpuAu+1 = qu1

0

etdt + vpuAu+1

=1 e

qu + vpuAu+1 =

1 v

qu + vpuAu+1.



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Proof (cont.)

Thus we have that:

Au

UDD

=

i

vqu + vpuAu+1.

The domain for the latter relationship is: u = 0,1, . . . andA = 0. Further,

Au UDD=i

vqu + vpu

ivq

u+1 + vpu+1Au+2

=i

vqu +

i

v2puqu+1 + v

2pupu+1Au+2

=i

v

0p

uq

u+0+ v2p

uq

u+1+ v3p

up

u+1q

u+2+ + 0

=i

k=0

vk+1kpuqu+k =i

Au,

which completes the proof.Edward Furman Actuarial mathematics MATH 3280 10/45


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Example 0.3 (Pure endowment insurance.)

Let (u) = (x :1n). Then it only makes sense to speak of the

discrete case. The r.v. representing the future payment is

vK(x:1n), Thus

Ax:

1n

:= vn

k=nkpxqx+k = v

nP[T(x) n] = vnnpx

Example 0.4 (General endowment insurance.)

Let (u) = (x : n). Then by definition of T(x : n) we have that

Ax:n :=n

0vttpx(x + t)dt + vnnpx = A1

x:n+ A

x:1n.

Note that

vT(x:n) = vT(1x:n) + vT(x:

1n).

Edward Furman Actuarial mathematics MATH 3280 11/45


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Example 0.5 (Example 0.4 (cont.))

Do we have that

T(x : n) = T( 1x : n) + T(x : 1n)?

Also,

Ax:n := E[vK(

1x:n)+1 + vK(x:

1n)], i.e.,

Ax:n =n1k=0

vk+1kpxqx+k + vn

npx = A1x:n

+ Ax:

1n.

Note:

Of course, we have that,

limn

A1x:n

= Ax.



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Proposition 0.3

The variance of the r.v. representing the future payment due to

death of(u) isVar[vT(u)] = 2Au (Au)

2,

where2Au is an insurance payable using = 2.

Proof.We have that

E[(vT(u))2] =

Ru

e2ttpu(u+ t)dt =

Ru

et

tpu(u+ t)dt,

that is an insurance payable with a new force of interest. This

completes the proof.



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Proposition 0.4

The variance of the general endowment insurance is

Var[vT(x:n]

= 2A1x:n

A1x:n

2+ v2nnpx nqx 2A1

x:nvnnpx.

Proof.

Note that

Var[vT(x:n]

= Var[vT(1x:n + vT(x:

1n]

= Var[vT(1

x:n] + Var[vT(x:1

n] + 2Cov[vT(1

x:n, vT(x:1

n)]

= Var[vT(1x:n] + Var[vT(x:

1n] 2E[vT(

1x:n]]E[vT(x:

1n],

which completes the proof.



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Insurances mentioned above can be used as building blocks to

create other forms of insurances.

Example 0.6 (Deferred insurance.)

An myear deferred nyears term insurance provides for a

benefit following the death of the insured only if the insured dies

at least myears following policy issue and before the end of the

policy. Thus we have that

m|nAx := A1u:m+n

A1u:m

=m+n1

k=m

vk+1kpuqu+k.

The continuous counterpart is then:

m|nAu := A1u:m+n

A1u:m

=

m+nm

vttpu(u+ t)dt.



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Example 0.7 (Deferred whole life insurance.)

Take n in the previous example, and get

m|Au = Au A1u:m

=

k=mvk+1kpuqu+k,

with the continuous counterpart given then by:

m|Au = Au A1u:m

=

m

vttpu(u+ t)dt.



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Note, that we have been interested in calculating expectations

of transformed future life time r.v.s. These r.v.s can be more

general.

Example 0.8 (Annually increasing whole life insurance.)

Consider the r.v. (K(u) + 1)vK(u)+1. The present value is

(IA)u := E[K(u) + 1)vK(u)+1] =

k=0

(k + 1)vk+1kpuqu+k.

Example 0.9 (Continuously increasing whole life insurance.)

Consider the r.v. T(u)vT(u), then

(IA)u := E[T(u)vT(u)] =

0

tvttpu(u+ t)dt.



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Example 0.10 (Annually increasing continuous whole life

insurance.)

Let the r.v. of interest be T(u) + 1VT(u). Then

(IA)u := E[T(u) + 1VT(u)] =

0

t + 1 vttpu(u+ t)dt.

Example 0.11 (m-thly per year increasing continuous whole life

insurance.)Let the r.v. of interest be

T(u)m+1m

vT(u). Then:

(I(m)A)u := E T(u)m+ 1

mvT(u) =

0

tm+ 1

mvttpu(u+t).

Remark.

Of course, for m, the just mentioned insurance reduces tothe continuously increasing one.



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Proposition 0.5

We have that

(IA)u =

0 s|Auds.

Proof.

(IA)u =

0

t0

ds

vttpu(u+ t)dt

=

0

s

vttpu(u+ t)dtds

=

0s|Auds,

as required.



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Figure: Auxiliary plot.

Note:

For the totally discrete counterpart, i.e., for (IA)u, we should

again have (why?)

(IA)u =

j=0

j|Au = Au + 1|Au + 2|Au + ...



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Benefits must not be increasing.

Example 0.12 (Annually decreasing n-year term insurance.)

Let the r.v. of interest be (n K(u))vK(1u:n+1. Then

(DA

)1u:n :=E

[(n

K(u

))vK(

1u:n+1

] =

n1

k=0(

n

k)vk+1

kp

uq

u+k.

Proposition 0.6

We have that

(DA)1u:n

=n1j=0

A1u:nj

.



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Proof.

Note that

n k =nk1

j=0

(1).

Then

(DA)1u:n

=n1

k=0

(n k)vk+1kpuqu+k

=

n1k=0

nk1j=0

(1)vk+1kpuqu+k

=n1j=0

n

j

1k=0

(1)vk+1kpuqu+k =n1j=0

A1u:nj

,

as needed.



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Figure: Annually decreasing 8-year term insurance.



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Proposition 0.7

Let the actuarial present value r.v. be generally given by

Z = b(K(u) + 1)vT(u). And assume the UDD approximation for

each integer u. Then

E[b(K(u) + 1)vT(u)]UDD

=i

E[b(K(u) + 1)vK(u)+1]

for any life status (u).

Proof (cont.)

Recall that T = K + J with J U(0,1) because of the UDD.We have also proven that under the UDD K and J are

independent. Namely

P[K = k, J j] = kpu qu+k j = P[K = k]P[J j].



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Proof (cont.)

Also

E[b(K(u) + 1)vT(u)

]= E[b(K(u) + 1)vK(u)+1+J(u)1]

= E[b(K(u) + 1)vK(u)+1vJ(u)1]UDD

= E[b(K(u) + 1)vK(u)+1]E[vJ(u)1]

= E[b(K(u) + 1)vK(u)+1]E[(1 + i)1J(u)],

where

E

[(1

+i)

1J(u)

] =1

0 (1

+i)

1sds=

10

d(1 + i)1s

ln(1 + i) ,

which is

E[(1 + i)1J(u)] = i/.



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Corollary 0.3

Thus, we easily have that

AxUDD

=i

Ax,

with b(K + 1) 1. And also

(IA)xUDD

= i

(IA)x,

with b(K + 1) = K + 1.

Question:

What about (IA)x?


Proposition 0.8


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Proposition 0.8

Under the UDD, we have that

(IA)x

UDD=

i

(IA)

x1 + i

i

1

A

x .

Proof.

The r.v. corresponding to the price above is

Z = TvT = (K + J)vK+J = (K + 1)vK+J + (J 1)vK+J

= (K + 1)vK+J (1 J)vK+1vJ1

= (K + 1)vK+1(1 + i)1J (1 J)vK+1(1 + i)1J.

Also, note that 1 J

U(0,1). Indeed

P[1 J j] = P[J 1 j] = j.

Recall that J and K are independent because of the UDD

assumption and take expectationsEdward Furman Actuarial mathematics MATH 3280 28/45

Proof (cont.)


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Proof (cont.)

E[Z]UDD

= E[(K + 1)vK+1(1 + i)1J] E[vK+1]E[(1 J)(1 + i)1J]

=i

(IA)x AxE[(1 J)(1 + i)

1J]

=i

(IA)x AxE[(J)(1 + i)

J]

=i

(IA)x Ax

10

jd(1 + i)j

ln(1 + i)

=i

(IA)x

Ax

ln(1 + i) 1

0

jd(1 + i)j=

i

(IA)x

Ax

ln(1 + i)

j(1 + i)j|10

10

(1 + i)jdj

=i

(IA)xAx

ln(1 + i)(1 + i) i

ln(1 + i) .


An insurance can be payable m thly


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An insurance can be payable m-thly.

Example 0.13 (m-thly payable whole life insurance.)

The price of an m-thly payable whole life insurance is

A(m)x :=

k=0

(v1/m)k+1 km

px 1m

qx+ k

m.

Note that this is not the expectation of the transformed K(x),but rather an expectation of a transformation of

K(m)(x) = K(x)m+ J(x), where this time J(x) counts thenumber of total m-thly periods (x) was alive. Thus

A(m)x := E[(v1/m)Km+J+1] = E[(v)K+(J+1)/m]

=

k=0

mj=0

vk+(j+1)/mP[K = k, J = j].


Proposition 0.9


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p

We have that under the UDD for integer ages,

A(m)

x

UDD=

i

i(m)A

x,

where i(m) = m((1 + i)1/m 1).

Proof.

We have that

E[vK+(J+1)/m] =

k=0

m1j=0

vk+(j+1)/mkpx j/m|1/mqx+k = A(m)x .

In addition, under the UDD

j/m|1/mqx+k = (j+1)/mqx+k (j)/mqx+k

UDD=

j + 1

m

qx+kj

m

qx+k =1

m

qx+k.


Proof (cont.)


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( )

Thus

A

(m)

x

UDD

=

k=0

v

k+1m1j=0

v

(j+1)/m1

kpx

1

mqx+k

=

k=0

vk+1kpxqx+k

m1j=0

(1 + i)1(j+1)/m1

m

=

k=0

vk+1kpxqx+k(1 + i)m1j=0

(v(1/m))j+11

m

=

k=0

vk+1kpxqx+k(1 + i)1

m

v1/m1 v(m1+1)/m

1 v1/m

=

k=0

vk+1kpxqx+k(1 + i)1

m

1 v

v1/m(1 v1/m).

Moreover Edward Furman Actuarial mathematics MATH 3280 32/45


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Proof.

A(m)x

UDD=

k=0

vk+1kpxqx+k(1 + i)1

m

1 v

(1 + i)1/m 1

=

k=0

vk+1kpxqx+k(1 v)(1 + i)

i(m)

=

k=0

vk+1kpxqx+ki

i(m)=

i

i(m)Ax,

which completes the proof.


Proposition 0.10


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We have that

Ax:y + Ax:y = Ax + Ay,

as well asAx:y + Ax:y = Ax + Ay.

Proof.

Because of the definition of say T(x : y) and T(x : y), we have

thatvT(x:y) + vT(x:y) = vT(x) + vT(y).

Taking expectations throughout then completes the proof.

Example 0.14An insurance that pays one dollar upon the death of the first of

(x) and (y) is

Ax:y := E[vT(x:y)] =

0

vttpx:y((x : y) + t)dt.



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Example 0.15

An insurance that pays one dollar upon the death of the last

one of (x) and (y) is

Ax:y := E[vT(x:y)] =

0

vttpx:y((x : y) + t)dt.

Example 0.16

An insurance that pays one dollar upon the death of (x) if

he/she dies first is A1x:y

. Of course it is an expectation of vT(1x:y).

The latter is vT(x) if T(x) < T(y) and v = 0 if T(x) T(y).

Thus

A1x:y

:= E[vT(1x:y)] =

0

vt

t

fT(x),T(y)(t, s)dsdt


Ex (cont )


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Ex. (cont.)

The latter expression is rewritten as

A1x:y := E[vT(

1x:y

] =

0v

tt

fT(x),T(y)(t, s)dsdt

=

0

vt

t

fT(y)|T(x)(s| t)ds fT(x)(t)dt

ind=

0

vtt

fT(y)

(s)ds fT(x)

(t)dt

=

0

vtFT(y)(t) fT(x)(t)dt

=

0

vttpy tpx(x + t)dt,

i.e., if (x) dies at any time t when (y) is alive, then vt dollars arepayed. Check at home that if = 0, then A1

x:y= q1

x:y.


Example 0.17


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An insurance payable upon death of (y) if it precedes the deathof (x) is

Ax:

2y

:= E[vT(x:2y)] =

0

vt

t

0

fT(x),T(y)(s, t)dsdt.

Proposition 0.11

Under independence of the future lifetimes of(x) and(y), wehave thatA

x:2y

ind= Ay A

x:1y.

Proof.

By definition

Ax:

2y

=

0

vtt

0

fT(x)| T(y)(s| t)ds fT(y)(t)dt.


Proof.


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By independence and in actuarial notation

Ax:

2

y

=

0

vttqx tpy(y + t)dt

=

0

vt(1 tpx) tpy(y + t)dt,

as required.

Remark.

The result holds for dependent future lifetimes too. Prove at

home by looking at the corresponding r.v.s.

Proposition 0.12

Under independence of the future lifetimes of(x) and(y), wehave that

Ax:

2y

ind=

0

sAy spx(x + s)ds.Edward Furman Actuarial mathematics MATH 3280 38/45

Proof.


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From the previous proposition, changing the order of integration

and by substitution u = t s,

Ax:

2y

=

0

vttqx tpy(y + t)dt

ind=

0

vtt

0spx(x + s)ds tpy(y + t)dt

=

0

s

vtspx(x + s) tpy(y + t)dtds

=

0

0

vu+sspx(x + s) u+spy(y + u+ s)duds

=

0 v

s

spy spx(x + s)

0 v

u

upy+s(y + u+ s)duds

=

0

vsspy spx(x + s)Ay+sds.

Also, we can write the deferred insurance as


P f t


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Proof. cont.

sAy = Ay A1y:s

=

0

vu+su+spy(y + u+ s)du

= vsspy

0

vuupy+s(y + u+ s)du

= vsspyAy+s.

This completes the proof.

Remark.We will often use the notation sEx := v

sspx. This is refereed to

as the stochastic discount factor.


Example 0.18


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Assume m sources of decrement and a whole life insurance

contract due to each one. Also, let b(x + t)(j), j = 1, . . . ,m bethe payment due to the decrement j. Then the overall price is

A =m

j=1

0

b(x + t)(j)vttpx

(j)(x + t)dt.

Example 0.19Let b(x + t)(1) = t and b(x + t)(2) = 0 for all t> 0. AssumeUDD for each year of death. Then

A =

0tvttpx(1)(x + t)dt =

k=0

k+1k

tvttpx(1)(x + t)dt

=

k=0

10

(k + s)vk+sk+spx

(1)(x + k + s)ds


Example 0.20 (Example. cont.)


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Example 0.20 (Example. cont.)

Then by the UDD

tpx(x + t)UDD

= qx,

and

A =

k=0vkkp

x

10

(k + s)vsspx+k

(1)(x + k + s)ds

UDD=

k=0

vk+1kpxq

(1)x+k

1

0

(k + s)vs1ds

=

k=0

vk+1kpxq

(1)x+k

1

0

(k + s)(1 + i)1sds

=

k=0

vk+1kpxq

(1)x+k

i

k +

1

1

i.



43/45

Remark

Note that if b(x + k + s)(j) is more cumbersome than (k + s)than using, e.g., the midpoint rule

1

0

b(x + k + s)(j)(1 + i)1sds b(x + k + 1/2)(j)(1 + i)11/2.



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Figure: Insurances payed at the end of the year of death.



45/45

Figure: Immediately payed insurances.


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