Actuarial mathematics 2

Actuarial mathematics 2

Life insurance contracts

Edward Furman

Department of Mathematics and Statistics York University

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 the insurance 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)?

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 is E[vK(u)+1_]. I.e., E[vT(u)] = Z Ru vttpuµ(u+t)dt =− Z Ru vtdtpu :=Au.

Also E[vK(u)+1] = X k∈Ru vk+1_kpuqu+k =− X k∈Ru vk+1∆kpu :=Au.

Example 0.1 (Whole life insurance.)

Let(u) = (x). ThenRu = [0, ∞), and Ax :=E[vT(u)] = Z ∞ 0 vttpxµ(x+t)dt, as well as Ax :=E[vK(u)+1] = ∞ X k=0 vk+1kpxqx+k.

Example 0.2 (n-year term insurance.)

Let(u) = (x1 :n). ThusRu= [0, n)∪ {∞}, and A1 x:n :=E[v T(u)_{] =}Z n 0 vttpxµ(x+t)dt+0, as well as A1 x:n :=E[v K(u)+1_{] =} n−1 X k=0 vk+1kpxqx+k. Proposition 0.1

We have that, for A1

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

A1

Proof. A1 x:n = n−1 X k=0 vk+1kpxqx+k =vqx + n−1 X k=1 vk+1kpxqx+k = vqx +vpx n−1 X k=1 vkk−1px+1qx+k = vqx +vpx n−2 X k=0 vk+1kpx+1qx+k+1 = vqx +vpxA 1 x+1:n−1, as required.

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

Au UDD = i δAu. Proof. We have that tpuµ(u+t) UDD = qu, 0≤t ≤1, u=0,1, . . . . Then Au = = Z 1 0 vttpuµ(u+t)dt+ Z ∞ 1 vttpuµ(u+t)dt

Proof (cont.) Au = Z 1 0 vttpuµ(u+t)dt+ Z ∞ 0 vt+1t+1puµ(u+1+t)dt = Z 1 0 vttpuµ(u+t)dt+ Z ∞ 0 vt+1pu·tpu+1µ(u+1+t)dt = Z 1 0 vttpuµ(u+t)dt+vpu Z ∞ 0 vttpu+1µ(u+1+t)dt UDD = Z 1 0 vtqudt+vpuAu+1=qu Z 1 0 e−δtdt+vpuAu+1 = 1−e −δ δ qu+vpuAu+1= 1−v δ qu+vpuAu+1.

Proof (cont.)

Thus we have that:

Au UDD

= i

δvqu+vpuAu+1.

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

A∞=0. Further, Au UDD = i δvqu+vpu i δvqu+1+vpu+1Au+2 = i δvqu+ i δv 2_p uqu+1+v2pupu+1Au+2 = i δ

v0puqu+0+v2puqu+1+v3pupu+1qu+2+· · ·+0 = i δ ∞ X k=0 vk+1kpuqu+k = i δAu,

Example 0.3 (Pure endowment insurance.)

Let(u) = (x :n)1 . Then it only makes sense to speak of the discrete case. The r.v. representing the future payment is

vK(x: 1 n)_{, Thus} A x:n1 := vn ∞ X k=n kpxqx+k =vnP[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 := Z n 0 vttpxµ(x+t)dt+vnnpx =A1 x:n +Ax:n1 . Note that vT(x:n)=vT( 1 x:n)_+vT(x:n1)_.

Example 0.5 (Example 0.4 (cont.)) Do we have that T(x :n) =T( 1 x :n) +T(x :n)?1 Also, Ax:n :=E[vK( 1 x:n)+1_+vK(x:n1)_],i.e., Ax:n = n−1 X k=0 vk+1kpxqx+k +vnnpx =A1 x:n +Ax: 1 n . Note:

Of course, we have that, lim

Proposition 0.3

The variance of the r.v. representing the future payment due to death of(u)is

Var[vT(u)] =2Au−(Au)2, where2Auis an insurance payable usingδ∗ =2δ.

Proof. We have that E[(vT(u))2] = Z Ru e−δ2ttpuµ(u+t)dt = Z Ru e−δ∗ttpuµ(u+t)dt,

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

Corollary 0.2

The variance of the r.v. representing the present value of due to the pure endowment insurance is

Var[vT(x: 1

n)_{] =v}2n_npx(nqx).

Proof.

Note that for the pure endowment insurance we have that the second moment is

E[v2T(x: 1

n)_{] =v}2n

npx.

Thus the variance is

v2nnpx −(vnnpx)2, as required.

Proposition 0.4

The variance of the general endowment insurance is

Var[vT(x:n] = 2A1 x:n− A1 x:n 2 +v2nnpx nqx−2A1 x:nv n npx. Proof. Note that Var[vT(x:n] = Var[vT( 1 x:n _+vT(x:n1_] = Var[vT( 1 x:n_{] +Var[v}T(x:n1_{] +2Cov[v}T(x1:n_,vT(x:n1)_] = Var[vT( 1 x:n_{] +Var[v}T(x: 1 n_]−2E[vT(x1:n_]]E[vT(x: 1 n_],

Insurances mentioned above can be used as building blocks to create other forms of insurances.

Example 0.6 (Deferred insurance.)

An m year deferred n years term insurance provides for a benefit following the death of the insured only if the insured dies at least m years following policy issue and before the end of the policy. Thus we have that

m|nAx :=A1 u:m+n −A1u:m = m+n−1 X k=m vk+1kpuqu+k.

The continuous counterpart is then:

m|nAu:=A1

u:m+n −A1u:m =

Z m+n

m

Example 0.7 (Deferred whole life insurance.)

Take n→ ∞in the previous example, and get

m|Au=Au−A1 u:m = ∞ X k=m vk+1_kpuqu+k,

with the continuous counterpart given then by:

m|Au=Au−A1 u:m =

Z ∞

m

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] = ∞

X

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)] =

Z ∞ 0

Example 0.10 (Annually increasing continuous whole life insurance.)

Let the r.v. of interest be⌊T(u) +1⌋VT(u)_{. Then}

(IA)u :=E[⌊T(u) +1⌋VT(u)] =

Z ∞ 0

⌊t+1⌋vttpuµ(u+t)dt.

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

Let the r.v. of interest be ⌊T(u)_mm+1⌋vT(u)_{. Then:}

(I(m)A)u:=E ⌊T(u)m+1⌋ m v T(u) = Z ∞ 0 ⌊tm+1⌋ m v t tpuµ(u+t). Remark.

Of course, for m→ ∞, the just mentioned insurance reduces to the continuously increasing one.

Proposition 0.5 We have that (IA)u= Z ∞ 0 s| Auds. Proof. (IA)u = Z ∞ 0 Z t 0 ds vttpuµ(u+t)dt = Z ∞ 0 Z ∞ s vttpuµ(u+t)dtds = Z ∞ 0 s| Auds, as required.

Figure:Auxiliary plot.

Note:

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

(IA)u= ∞

X

j=0

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(u1:n+1_{. Then}

(DA)1 u:n :=E[(n−K(u))v K(u1:n+1_{] =} n−1 X k=0 (n−k)vk+1kpuqu+k. Proposition 0.6 We have that (DA)1 u:n = n−1 X j=0 A1 u:n−j.

Proof. Note that n−k = n−k−1 X j=0 (1). Then (DA)1 u:n = n−1 X k=0 (n−k)vk+1kpuqu+k = n−1 X k=0 n−k−1 X j=0 (1)vk+1kpuqu+k = n−1 X j=0 n−j−1 X k=0 (1)vk+1kpuqu+k = n−1 X j=0 A1 u:n−j, as needed.

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

δE[b(K(u) +1)v

K(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

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)1−J(u)], where E[(1+i)1−J(u)] = Z 1 0 (1+i)1−sds=− Z 1 0 d(1+i)1−s ln(1+i) , which is E[(1+i)1−J(u)] =i/δ.

Corollary 0.3

Thus, we easily have that

Ax UDD

= i

δAx, with b(K +1)≡1. And also

(IA)x UDD= i

δ(IA)x, with b(K +1) =K+1.

Question:

Proposition 0.8

Under the UDD, we have that

(IA)x UDD= i δ (IA)x− 1+i i − 1 δ Ax . 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+1vJ−1

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

Proof (cont.) E[Z] UDD = E[(K +1)vK+1(1+i)1−J]−E[vK+1]E[(1−J)(1+i)1−J] = i δ(IA)x −AxE[(1−J)(1+i) 1−J_] = i δ(IA)x −AxE[(J)(1+i) J_] = i δ(IA)x −Ax Z 1 0 jd(1+i) j ln(1+i) = i δ(IA)x − Ax ln(1+i) Z 1 0 jd(1+i)j ! = i δ(IA)x − Ax ln(1+i) j(1+i) j_|1 0− Z 1 0 (1+i)jdj ! = i δ(IA)x − Ax ln(1+i) (1+i)− i ln(1+i) .

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(xm):= ∞ X k=0 (v1/m)k+1k mpx· 1 mqx+ 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 the number of total m-thly periods(x)was alive. Thus

A(xm) :=E[(v1/m)Km+J+1] =E[(v)K+(J+1)/m] = ∞ X k=0 m−1 X j=0 vk+(j+1)/mP[K =k,J =j].

Proposition 0.9

We have that under the UDD for integer ages,

A(xm) UDD = i i(m)Ax, where i(m)_=m((1+i)1/m_−1). Proof. We have that E[vK+(J+1)/m] = ∞ X k=0 m−1 X j=0 vk+(j+1)/m_kpx·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+k − j mqx+k = 1 mqx+k.

Proof (cont.) Thus A(xm) UDD = ∞ X k=0 vk+1 m−1 X j=0 v(j+1)/m−1kpx· 1 mqx+k = ∞ X k=0 vk+1kpxqx+k m−1 X j=0 (1+i)1−(j+1)/m1 m = ∞ X k=0 vk+1kpxqx+k(1+i) m−1 X j=0 (v(1/m))j+11 m = ∞ X k=0 vk+1kpxqx+k(1+i) 1 mv 1/m1−v(m−1+1)/m 1−v1/m = ∞ X k=0 vk+1kpxqx+k(1+i) 1 m 1−v v−1/m_(1−v1/m₎. Moreover

Proof. A(_xm) UDD= ∞ X k=0 vk+1kpxqx+k(1+i) 1 m 1−v (1+i)1/m₋₁ = ∞ X k=0 vk+1kpxqx+k (1−v)(1+i) i(m) = ∞ X k=0 vk+1kpxqx+k i i(m) = i i(m)Ax,

Proposition 0.10 We have that Ax:y +Ax:y =Ax +Ay, as well as Ax:y +Ax:y =Ax +Ay. Proof.

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

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

Taking expectations throughout then completes the proof.

Example 0.14

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

(x)and(y) is

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

Z ∞

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)] =

Z ∞ 0

vttpx:yµ((x :y) +t)dt.

Example 0.16

An insurance that pays one dollar upon the death of(x)if he/she dies first is A1

x:y. Of course it is an expectation of v T(x1:y)_. The latter is vT(x) _{if T(x)<T(y)and v}∞_{=0 if T(x)≥T(y).}

Thus A1 x:y :=E[v T(x1:y)_{] =}Z ∞ 0 vt Z ∞ t f_T(x),T(y)(t, s)dsdt

Ex. (cont.)

The latter expression is rewritten as

A1 x:y :=E[v T(x1:y ] = Z ∞ 0 vt Z ∞ t f_T(x),T(y)(t, s)dsdt = Z ∞ 0 vt Z ∞ t fT(y)|T(x)(s|t)ds·fT(x)(t)dt ind = Z ∞ 0 vt Z ∞ t f_T(y)(s)ds·f_T(x)(t)dt = Z ∞ 0 vtF_T(y)(t)·f_T(x)(t)dt = Z ∞ 0 vttpy·tpxµ(x+t)dt,

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

Example 0.17

An insurance payable upon death of(y)if it precedes the death of(x)is A x:y2 :=E[v T(x:y2)_{] =}Z ∞ 0 vt Z t 0 f_T(x),T(y)(s,t)dsdt. Proposition 0.11

Under independence of the future lifetimes of(x)and(y), we have that A x:y2 ind = Ay −A x:y1. Proof. By definition A x:y2 = Z ∞ 0 vt Z t 0 f_T(x)|T(y)(s|t)ds·f_T(y)(t)dt.

Proof.

By independence and in actuarial notation

A x:y2 = Z ∞ 0 vttqx ·tpyµ(y +t)dt = Z ∞ 0 vt(1−tpx)·tpyµ(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), we have that A x:y2 ind = Z ∞ s A_y ·_sp_xµ(x +s)ds.

Proof.

From the previous proposition, changing the order of integration and by substitution u =t−s, A x:y2 = Z ∞ 0 vttqx·tpyµ(y+t)dt ind = Z ∞ 0 vt Z t 0 spxµ(x+s)ds·tpyµ(y +t)dt = Z ∞ 0 Z ∞ s vtspxµ(x +s)·tpyµ(y +t)dtds = Z ∞ 0 Z ∞ 0 vu+sspxµ(x+s)·u+spyµ(y +u+s)duds = Z ∞ 0 vsspy spxµ(x+s) Z ∞ 0 vuupy+sµ(y+u+s)duds = Z ∞ 0 vsspy spxµ(x+s)Ay+sds.

Proof. cont. s A_y =A_y −A1 y:s = Z ∞ 0 vu+su+spyµ(y+u+s)du = vsspy Z ∞ 0 vuupy+sµ(y +u+s)du = vsspyAy+s.

This completes the proof.

Remark.

We will often use the notationsEx :=vsspx. This is refereed to

Example 0.18

Assume m sources of decrement and a whole life insurance contract due to each one. Also, let b(x +t)(j)_{, j =1, . . . ,m be} the payment due to the decrement j. Then the overall price is

A= m X j=1 Z ∞ 0 b(x+t)(j)vttpτxµ(j)(x +t)dt. Example 0.19

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

A= Z ∞ 0 tvttpxτµ(1)(x+t)dt = ∞ X k=0 Z k+1 k tvttpτxµ(1)(x +t)dt = ∞ X k=0 Z 1 0 (k+s)vk+sk+spτxµ(1)(x +k +s)ds

Example 0.20 (Example. cont.)

Then by the UDD

tpxµ(x+t)UDD= qx, and A= ∞ X k=0 vkkpxτ Z 1 0 (k +s)vsspτx+kµ(1)(x+k+s)ds UDD = ∞ X k=0 vk+1kpxτq (1) x+k Z 1 0 (k +s)vs−1ds = ∞ X k=0 vk+1kpxτq (1) x+k Z 1 0 (k +s)(1+i)1−sds = ∞ X k=0 vk+1_kp_xτq(1)_x+ki δ k+1 δ − 1 i.

Remark

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

Z 1

0