Chapter 3

Chapter 3

Life Annuities

3.1 Continuous Life Annuities

A life annuity is a series of payments made to an individual (x) while he is still living.

3.1.1 n-year Pure Endowment

Simplest form of an annuity. Pays $1 at the end of n years if (x) is still living.

Let T = future lifetime of an individual (x) Z = present value of benefit.

Then Z=

{

0if T ≤ n

vn_{if T} >n

E(Z)=A1_x:n|=vnP{T>N}=vnnpx=nEx

¿ net single premium (n.s.p.) at age (x)

¿ actuarial present value of $1 due at the end of n years if (x) is living.

Alternatively,

1 Ex n

=a mount payable at the end of n years if(x)is living such that n. s . p . at age(x)=$1.

Interpretation:

Consider $1 payable by each of l_x lives age (x). Accumulated value of fund at the end of n

years ¿l_x(1+i)n.

Number of survivors at end of n years = l_x+n

i .e . share per survivor=lx(1+i) n

l_x+n = 1

[

(1+i1 )n

]

l_x+n l_x

= 1

¿ 1

Ex n

Notation: 1

Ex n

=actuarial accumulated value of $1at the end of n years.

¿ accumulated value of $1 at the end of n years by interest and survivorship.

¿ accumulated value of $1 at the end of n years by interest only ¿(1+i)n

3.1.2 Continuous Life Annuities

Recall in Theory of Interest:

a_t|=present value of $1per year payable continuously for t years .

¿

∫

0 1

vkdk= v

k

logv

|

¿t0=

−vk

δ

|

¿t0=

(1+vt) δ

Let Z = present value of $1 per year payable continuously to (x) while (x) is living.

Then Z=a_T|∧E(Z)=a_x=

∫

0

∞

a_t|tp_xμ_x+tdt

¿

∫

0

∞

a_t| ∂

∂t

(

−tpx

)

dt

¿

[

a_t|

(

−_tp_x

)

]

|

∞ 0−

∫

0

∞

(

−_tp_x

)

∂ ∂ t at|dt

¿

∫

0

∞

p_x

t

∂ ∂ tat|dt

But a_t|=

∫

0

t

vkdk⟹ ∂ ∂ tat|=v

t

i .e . , ax=

∫

0

∞

a_t|tpxμx+tdt=

∫

0

∞

3.1.3 n-year Temporary Life Annuity

Z = present value of $1 per annum payable continuously while (x) is living during the next n

years.

E(Z)=a_x:n|=

∫

0

n

a_t|tp_xμ_x+tdt+a_n|np_x

¿

[

a_t|

(

−_tp_x

)

]

|

n 0−

∫

0

n

(

−_tp_x

)

∂

∂t at|dt+an|npx

¿

∫

0

n

vt _p x

t dt

3.1.4 n-year Deferred Life Annuity

Z = present value of $1 per annum payable continuously after n years if (x) is living.

E(Z)=_n|a_x=vn

∫

n ∞

a_t−n|tp_xμ_x+tdt

¿vn

[

a_t−n|tp_x

]

|

∞

n−v

n

∫

n ∞

(

−_tp_x

)

∂

∂ tat−n|dt

¿vn

∫

n ∞

(

_tp_x

₎

(

∂

∂ t at−n|

)

dt

But a_t−n|=1−v

t−n

δ

i .e . ∂

∂ tat−n|=

−v−nvtlogv

δ =v

t−n

i .e . a_{n| x}=vn

_∫

n ∞

p_x

t v

t−n

dt=

∫

n ∞

vt_tp_xdt

¿a_x−a_x:n|

3.1.5 m-year Deferred, n-year Temporary Life Annuity

E(Z)=_m|na_x=vm

∫

m m+n

a_t−m|tp_xμ_x+tdt+vma_n|m+np_x

¿vm

[

a_t

−m|

(

−tpx

)

]

|

m+n

¿m +v

m

∫

m m+n

p_x

t v

t−m_dt

+vma_n|m+np_x

¿

∫

m m+n

vt _p x

t dt

¿m|ax−m+n|ax

¿a_x:m+n|−a_x:m|

¿

∫

m m+n

vmvt−m

p_x

m t−mpx+mdt

¿vm_mp_x

∫

0

n

vk_kp_x+mdk

¿vmmpxax+m:n|

¿mExax+m:n|

3.1.6 Relationship to Insurance

Life annuity:

Let Y = present value of continuous life annuity of $1 to (x).

Let Z = present value of continuous whole life insurance of $1 to (x).

Then E(Z)=a_x=

∫

0

∞

a_t|tp_xμ_x+tdt=

∫

0

∞

1−vt

δ tpxμx+tdt=

1

δ(1−Ax)

i .e . A_x+δ a_x=1

Var(Y)= 1

δ2Var(Z)= 1

δ2

[

Ax−

(

Ax

)

2 2

]

n-year temporary life annuity

Z = present value of continuous n year endowment

E(Y)=a_x:n|=

∫

0

n

(

1−vt

)

δ tpxμx+tdt+an|npx

¿1

δ nqx−

A_x1_:n| δ +

(1−vn) δ npx

¿1

δ− A1_x:n|

δ − A1_x:n|

δ

¿1−Ax:n|

δ

Var(Y)= 1

δ2Var(Z)= 1

δ2

[

Ax:n|−

(

Ax:n|

)

2 2

]

ALternatively ,

Var(Y)=E

(

Y2

)

−

[

E(Y)

]

2∧

[

E(Y)

]

2=

₍

a_x:n|

₎

2

E

(

Y2

₎

=

∫

0

n

(

at|

)

2 _p

x

t μx+tdt+

(

a_n|

)

2 _p

x n

¿

∫

0

n

(

at|

)

2 ∂

∂ t

(

−tpx

)

dt+

(

a_n|

)

2 _p

x n

¿−_tp_x

₍

a_n|

₎

2

|

n

¿0−

∫

₀

n

(

−_tp_x

)

∂ ∂ t

(

an|

)

2_dt

+

₍

a_n|

₎

2_np_x

¿

∫

0

n

p_x

t

2_a

t|dt since ∂ ∂t

(

at|

)

2

=2a_t|vt

¿2

δ

∫

₀

n

vt

₍

₁

−vt

)

_tp_xdt

¿2

δ

∫

₀

n

(

vt−v2t

)

_tp_xdt

¿2

δ

(

ax:n|− ax:n| 2

where a2 _x:n|=a_x:n|calculated at force of interest2δ

i .e .Var(Y)=2

δ

(

ax:n|− ax:n| 2

)

−

₍

a_x:n|

₎

2

n-year deferred life annuity:

a_x

n| =ax−ax:n|=

1−Ax

δ −

1−A_x:n|

δ =

A_x:n|−A_x δ

If Y = present value of n-year deferred life,

Var(Y)=2 δv

2n _p x

n

(

ax+n− ax+n

2

)

−

₍

_n|a_x

₎

2

m year deferred, n year temporary life annuity:

a_x

m|n =ax:m+n|−ax:n|=

1−A_x:m

+n|

δ −

1−A_x:n|

δ =

A_x:m|−A_x:m

+n| δ Notation:

Recall 1 Ex n

= 1

vnnpx

=actuarial accumulated value of $1at the end of n years .

s_x:n|=a_x:n| 1 Ex n

=actuarial accumulated value at the end of n years of an n−year

life annuity of $1per annum .

Example 3.1.1 You are given:

(i) A_x and a_x are based on force of interest δ and force of mortality μ_x+t

(ii) A'x and ax' are based on force of interest k+δ and force of mortality μx+t (iii) A' 'x is based on force of interest δ and force of mortality k+μx+t

Which of the following equals A' 'x−Ax ? A. k a_x

B. A'x−Ax C. Ax

'

D. (k−δ)ax '

+δ ax

E. δ

(

ax−ax'

)

Solution:

A' '_x=

∫

0

n

(

v''

)

t_tp' '_xμ_x''_+tdt , where

(

v' '

)

t=vt=e−δt, μ' '_x+t=k+μ_x+t

px ' '

t =e

−∫

x x+t

μy ' '

dy

=exp

{

−

∫

x x+t

(

k+μ_y

)

dy

}

=exp{−kt}tpx

i .e . A_x''=

∫

0

∞

e−δt

e−kt

p_x

t

(

k+μx+t

)

dt

¿k

∫

0

∞

e−(δ+k)t

p_x

t dt+

∫

0

∞

e−(δ+k)t

p_x

t μx+tdt

¿k ax'+A'x

¿k ax'+1−(δ+k)ax' since Ax=1−δ ax

¿1−δ a'x

i .e . Ax ''

−Ax=

[

1−δ ax

'

]

−

_[

1−δ ax

]

¿δ

(

ax−ax'

)

Answer: E.

Example 3.1.2 T is random variable for the future lifetime of x. Determine

C ov

₍

a_T|, vT

)

A. 1_δ

_[

(

Ax

)

2 −2A_x

]

B.

(

Ax

)

2 −2A_x

C. 0

E. 1_δ

_[

2A_x−

₍

A_x

₎

2

]

Solution:

For two random variables X, Y:

C ov(X , Y)=E

[

X−E(X)

]

E

_[

Y−E(Y)

]

=E(XY)−E(X)E(Y)

E

₍

a_T|vT

₎

=E

[

v

T

₍

_1−vT

₎

δ

]

= 1 δ E

[

v

T

−v2T

]

¿1

δ

[

(

Ax

)

2 −2A_x

]

E

₍

a_T|

₎

=E

[

1−v

T

δ

]

= 1−A_x

δ ∧E

(

v

T

₎

=Ax

C ov

₍

a_T|, vT

₎

=1 δ

[

(

Ax

)

2

−2A_x

]

−1−Ax δ Ax

¿1

δ

[

Ax− Ax

2

−A_x+

₍

A_x

₎

2

]

¿1

δ

[

(

Ax

)

2 −2A_x

]

Answer: A.

3.2 Discrete Life Annuities

3.2.1 Whole Life Annuity Due

If K = curtate future lifetime of (x), then when K = k, Y = a¨_k+1|=1+v+v2+…+vk,

i .e .:Y= ¨a_K+1|∧E(Y)= ¨ax=

∑

k=0

∞

¨ a_k+1|k|q_x

Recall Theorem 1.3.1. For given function Z(K)

E

[

Z(K)

]

=

∑

t=0

ω−x−1

z(t)_t|q_x=z(0)+

∑

t=0

ω−x−2

∆ z(t)_t+1p_x

For Z(K)= ¨a_K+1|, z(0)= ¨a_1|=1

∆ z(k)= ¨a_k+2|− ¨a_k+1|=vk+1

¨ a_x=

∑

k=0

∞

¨

a_k+1|k|q_x=1+

∑

k=0

∞

vk+1 _p

x

k+1 =

∑

k=0

∞

vk _p x k

Relationship to insurance:

¨

a_k+1|=1+v+v2+…+vk=1−v

k+1 1−v =

1−vk+1 d

E(Y)_{= ¨}ax=E

(

a¨K+1|

)

=E

(

1−vk+1 d

)

=

1−Ax

d i .e . A_x=1−da¨x

Var(Y)=Var

₍

a¨_K+1|

₎

= 1 d2Var

(

v

K+1

)

¿ 1

d2

[

Ax 2

−

₍

A_x

₎

2

]

3.2.2 Whole Life Annuity Due

Y = present value of $1 payable at the beginning of each year for n years while (x) survives.

Y=

{

a¨k+1|when K<n ¨

i .e . E(Y)=

∑

k=0

n−1

¨

a_k+1|k|q_x+ ¨an|npx= ¨ax:n|

Alternatively ,Y=Y1+Y2+…+Yn,where

Yk=present value of $1payment made at the beginning ot the kt hyear ,k=1,2, … , n

P

{

Y1=1

}

=1, P

{

Yk+1=v

k

}

=kpx, P

{

Yk+1=0

}

=kqx

i .e . , E(Y)= ¨a_x:n|=E

₍

Y₁

₎

+E

₍

Y₂

₎

+…+E

₍

Y_n

₎

¿1+v px+v22px+…+vn−1n−1px

¿

∑

k=0

n−1 vk _p

x k

Relationship to insurance:

E(Y)= ¨ax:n|=

∑

k=0

n−1

1−vk+1

d k|qx+

1−vn d npx

¿1

d− 1 dAx:n|

1

−1

d Ax:n| 1

¿1

d

(

1−Ax:n|

)

i .e . A_x:n|+da¨_x:n|=1

Var(Y)= 1

d2

[

Ax:n| 2

−

₍

A_x:n|

₎

2

]

Notation:

¨

s_x:n|= a¨x:n| vn_kp_x=

∑

k=0

n−1

vk _p x k

E_x

n

=

∑

k=0

n−1

E_x

k

E_x

n

=

∑

k=0

n−1 1 E_x+k

n−k

Interpretation:

¨

s_{x:n| = single payment at the end of n year if (x) is living which is equivalent to $1 payable}

3.2.3 n-year Deferred Life Annuity Due

Y = present value of $1 payable at the beginning of each year while (x) survives from age x + n onward.

Y=

{

v

n

¨

a_k+1−n|= ¨_n|a_k+1−n|when K=k ≥n 0when K<n

Alternatively ,Y=Y_n+1+Y_n+2+… ,

where Y_n+k=present value of $1payment made at the beginning of the(n+k)t hyear

if(x)is living , k=1,2, … , n .

E

(

Yn+k

)

=v n+k−1

px n+k−1

and

E(Y)= ¨_n|a_x=

∑

k=n ∞

¨ a_k+1−n|

n| k|qx=

∑

k=n ∞

E

(

Yn+k

)

¿

∑

k=n ∞

vk_kp_x

¿

∑

k=n ∞

vk _p x

k −

∑

k=0

n−1 vk _p

x k

¿a¨x− ¨a_x:n|

¿vnnpxa¨x+n

¿nExa¨x+n

Relationship to insurance:

¨ a_x

n| =

∑

k=n

∞

¨ a_k+1−n|

n| k|qx=

∑

k=n

∞

vn1−vk+1−n

¿1

dv

n _p

x

n −

∑

k=n ∞

vnvk+1

d k|qx

¿1

d Ax:n| 1

−1

dn|Ax

¿1

d Ax:n| 1

−1

d

(

Ax−Ax:n| 1

)

¿1

d

[

Ax:n|−Ax

]

Alternatively ,n|a¨x= ¨ax− ¨ax:n|= 1

d

(

1−Ax

)

−

1

d

(

1−Ax:n|

)

¿1

d

[

Ax:n|−Ax

]

3.3 Life Annuities Immediate

Here payments are made at the end of each year while (x) survives.

3.3.1 Whole Life Annuity Immediate

Y = present value of $1 payable at the end of each year while (x) survives.

E(Y)=a_x=

∑

k=0

∞

a_k|k|q_x=

∑

k=0

∞

vk_kp_x

Relationships:

(i) a_x= ¨ax−1

(ii) Since A_x+da¨x=1, ax= ¨ax−Ax+da¨x= ¨ax(1−d)−Ax=va¨x−Ax

i .e . A_x=va¨x−ax

(iii) ¿(i), Ax+da¨x=Ax+d

(

ax+1

)

=1

3.3.2 n-year Temporary Life Annuity Immediate

Y = present value of $1 payable at the end of each year for n years while (x) survives.

E(Y)=a_x:n|=

∑

k=0

n−1

a_k|k|q_x+a_n|np_x=

∑

k=1

n

vk _p x k

Relationships:

(i)a_x:n|= ¨a_x:n|−1+vnnpx= ¨ax:n|−1+nEx

(ii)a¨_x:n|=1+a_x:n

−1|

(iii)Since A_x:n|+da¨_x:n|=1,

¿(i), a_x:n|= ¨a_x:n|−

₍

A_x:n|+da¨_x:n|

₎

+vnnpx

¿a¨_x:n|(1−d)−A1_x:n|

i .e . A_x1_:n|=va¨_x:n|−a_x:n|

(iv)¿(i), A_x:n|+da¨_x:n|=A_x:n|+d

₍

a_x:n|+1−vn_np_x

₎

=1 i .e . ,(1+i)A_x:n|+i

₍

a_x:n|+1−vn_np_x

₎

=1+i

i .e . ,i a_x:n|+A_x:n|+i A1_x:n|=1

(v)¿(iii), A1_x:n|=va¨_x:n|−a_x:n|=va¨_x:n|−

₍

a¨_x:n|−1+vn_np_x

₎

i .e . A_x:n|=va¨_x:n|−a_x:n−1|

3.3.3 n-year Deferred Life Annuity Immediate

Y = present value of $1 payable at the end of each year while (x) survives after age x + n.

E(Y)=_n|a_x=

∑

k=n ∞

a_k−n|

n| k|qx=

∑

k=n+1

∞

vk_kp_x

Relationships:

(ii)n|a¨x=n|ax+v n

px n

Example 3.3.1 You are given s_x:n|=ax:n| E_x

n Which of the following is true?

I . An| x−n+1|Ax=v qx+nnEx

II . A_x:n|−A_x:n+1|=d En x

III .s¨_x:n|−s_x:n|= 1 Ex n+1

Answer:

I . A_{n| x}–_n+1|A_x=

∑

k=n ∞

vk+1 _q

x

k| –

∑

k=n+1

∞

vk+1 _q

x

k| =v

n+1 _q

x n|

¿v vnnpxqx+n

¿v q_x+nnE_x

I . is true

II . A_x:n|−A_x:n

+1|=

∑

k=0

n−1 vk+1

q_x

k| +v

n

p_x

n −

(

∑

k=0

n

vk+1 q_x

k| +v

n+1 p_x

n+1

)

¿vnnpx−vn+1n+1px−v n+1

px

n qx+n

¿vn_np_x−vn+1_np_x

(

px+n+qx+n

)

¿vnnpx(1−v)

¿d En x

II .is true

III .s¨_x:n|−s_x:n|= a¨x:n| vn+1

p_x

n+1

− ax:n| vn_np_x=

¨ a_x:n

+1|– v px vn+1

p_x

≠ 1 vn+1

px n+1

¿ 1

Ex n+1

III . is false

Example 3.3.2 You are given: (i)1000A_x:n|=563

(ii)1000A_x=129

(iii)d=0.057 (iv)1000nEx=543

Calculate an| x

Answer:

a_x

n| =

∑

k=n

∞

a_k−n|

n| k|qx=v

n

∑

k=n

∞

1−vk−n

i k|qx

¿v

n

i npx−

1 iv

∑

k=n

∞

vk+1 q_x

k|

¿1

i nEx−1_dn|Ax

¿1

i nEx−

1

d

(

Ax−

[

Ax:n|−nEx

])

i= d 1−d=

0.057 0.943=0.06

i .e . , a_{n| x}= 1

0.06(0.543)− 1

0.057

(

0.129−

[

0.563−0.543

])

=7.07

(ii)a¨x=10.0 (iii)s¨_{x: 10|}=15.0

(iv)v=0.94

Calculate A1_{x: 10|} .

Answer:

¨

a_x:n|=

∑

k=0

n−1 ¨

a_k+1|k|q_x+ ¨a_n|

∑

k=n ∞

q_x

k|

¿

∑

k=0

n−1

1−vk+1 d k|qx+

1−vn d

∑

k=n

∞

q_x

k|

¿1

d− 1 dAx:n|

1

−1

dv

n

p_x

n

i .e . , A1_x:n|=1−vnnpx−da¨x:n|

For n=10,

¨

a_{x: 10|}= ¨a_x− ¨_10|a_x=10−4=6

¨

s_{x: 10|}= a¨x: 10| vn_np_x

i .e . , v10₁₀p_x=a¨x: 10| ¨ s_{x: 10|}=

6 15=0.4

d=1−v=1−0.94=0.06

i .e . A1_{x: 10|}=1−v1010px−da¨x: 10|=1−0.4−0.06(6)

¿0.24

Consider a life annuity which pays $1/m per month at the beginning of each month of a year while (x) is living.

Y = present value of payments made

1/m 1/m 1/m 1/m 1/m 1/m 1/m …….

x x+(1/m) x+(2/m) x+(3/m) x+(4/m) x+(t/m) x+((t+1)/m)

If (x) dies between x+(t/m) and x+(t+1/m) , i.e.: with probability t qx m

|

1

m .

Y= ¨a_t+1

m

|

(m)

=1

m

(

1+v 1

m

+v

2

m_+…+v

t m

)

¿ 1−v

t+1

m

m

(

1−v 1

m

)

¿1−v

t+1

m

d(m)

where d(m) _{= monthly nominal rate of discount satisfies}

(

1−d (m) m

)

m

=1−d=v

i .e .1−d

(m)

m =v 1

m_∧d(m)

=m

(

1−v

1

m

)

E(Y)= ¨a(xm)=

∑

t=0

∞

¨ a_t

+1

m

|

(m) _q

x t m

|

1

m

=

∑

t=0

∞

1−v

t+1

m

d(m) t qx

m

|

1

m

¿1−Ax

(m)

d(m)

Altenatively ,a¨x

(m)

=1

m

∑

t=0

∞

v

t

m _p

Relationship to a¨x :

¨

a_x=1−Ax d ,a¨(xm)=

1−A_x(m) d(m) ,

i .e . ,1=A_x+da¨x=Ax

(m)_+d(m)

¨ a(_xm)_⟹

¨ a(_xm)

=

[

d

d(m)

]

a¨x−

Ax

(m)

−A_x

d(m)

Note

¨ a(_1|m)

=1

m

(

1+v 1

m_+v

2

m

+…+v

m−1

m

)

=1

m 1−v

(

1−v 1

m

)

= d

d(m)

¨ a(_∞m_|)

=1

m

(

1+v 1

m_+v

2

m

+…

)

=1

m 1

(

1−v 1

m

)

= 1

d(m)

i .e .a¨x

(m)

= ¨a1|

(m)

¨ a_x− ¨a∞|

(m)

[

A(_xm)

−A_x

]

Consider A(_xm)

=v 1 m _q x 1 m

|

1 m

+…+v

m

m _q

x

(m−1)

m

|

1

m

+v p_x

(

v 1

m _q

x+1 1

m

+v

2

m _q

x+1 1

m

|

1

m

+…+v

m

m _q

x+1

(m−1)

m

|

m1

)

+v2₂p_x

(

v 1

m _q

x+2 1

m

+v

2

m _q

x+2 1 m

|

1 m +…+v m m _q

x+2 (m−1)

m

|

1

m

)

+…

Consider St=v

1

m _q

x+t

1

m

+v

2

m _q

x+t

1 m

|

1 m +…+v m m _q

x+t (m−1)

m

|

1

m

Under the assumption of Uniform Distribution of Deaths,

q_x+t

t1|t2 =

l_x+t+t1−l_x+t+t1+t2 l_x+t =

t2dx+t

i .e . , S_t=qx+t

(

1+v 1

m_+v

2

m_+…+v

m m

)

m =

q_x+t(1−v)

i(m) =

iv i(m)qx+t

i .e . , Ax

(m)

= iv

i(m)qx+v px

iv

i(m)qx+1+v

2 px

2 iv

i(m)qx+2+…

¿ i

i(m)

(

v qx+v 2

p_xq_x+1+v3₂p_xq_x+2+…

₎

¿ i

i(m)Ax

Thus ,a¨(_xm)= d d(m)a¨x−

1 d(m)

[

Ax

(m)

−A_x

]

¿ d

d(m)a¨x−

Ax

(

i i(m)−1

)

d(m) under UDD … …..…(1)

Sincea¨_x=1−Ax

d ⟹Ax=1−da¨x

Substituting ,a¨x

(m)

= d

d(m)a¨x−

(

1−da¨x

)

(

i i(m)−1

)

d(m)

¿ id

i(m)_d(m)a¨x−

(

i i(m)_d(m)−

1 d(m)

)

¿ id

i(m)

d(m)a¨x− i−i(m) i(m)

d(m)

¿α(m)a¨x−β(m)… … … . …(2)

where α(m)= id

i(m)_d(m), β(m)=

i−i(m)

Then α(m)= d

d(m)a¨x−β(m)Ax¿(1)

¿ 1

d(m)−

(

β(m)+

1

d(m)

)

Ax, substitutinga¨x=

1−A_x d

Example 3.4.1 Find the net single premium for a monthly life annuity due of $100 per month for a person age 50. Assume De Moivre’s Law for mortality with ω = 100.

i = 0.05, i(12) _{= 0.0489, d = 0.0476, d}(12) _{= 0.0487 ,} a

50| = 18.2559

Solution:

Note: De Moivre’s Law implies UDD within each year of death

Check: Under De Moivre’s Law, tpxμx+t = 1/(ω−x)

i.e. q_x =

_∫

0 1

p_xμ_x+sds

s = 1/(ω−x)

For0≤ t ≤1, q_{t x}=

∫

0

t

p_xμ_x+sds

s =

t

ω−x=(t)qxunder UDD

Thusa¨₅₀(12)= 1

d(12)−

[

β(12)+

1 d(12)

]

A50

A_x=

∑

t=0

ω−x−1 vt+1

q_x

t| i . e . , A50=

∑

t=0 49

vt+1 q₅₀

t|

But qt| 50=

∫

t t+1

p50

k μ50+kdk=

1

(100−50)=

1 50

i .e . A₅₀= 1 50

[

v+v

2

+…+v50

]

= 1

50a50|= 1

5018.2559=0.3651

i d(12)=

1

0.487=20.5339

β(12)= i−i (12)

i(12) d(12)=

0.05−0.489

i .e .a¨50(12)= 1

d(12)−

[

β(12)+ 1 d(12)

]

A50

¿20.5339−(0.4622+20.5339)0.3651

¿$12.87

Monthly lifeannuity due of $100per month=1200a¨50(12)=1200(12.87)=$15,444

3.4.1 Temporary Life Annuity with Monthly Payments

¨

a(_xm_:n)_{| = expected value of $1/m per m}th _{of a year payable at the beginning of each m}th _{of a}

year for n years while (x) is living. Under UDD and using the relationship

¨ a_x(m_:n)_|

= ¨ax(m)−nExa¨(xm+)n∧ ¨a(xm)=

d

d(m)a¨x−β(m)Ax,

where β(m)=i−i

(m)

i(m)_d(m)

We geta¨(_xm_:n)_|= d

d(m)a¨x−β(m)Ax−nEx

[

d

d(m)a¨x+n−β(m)Ax+n

]

i .e .a¨_x(m_:n)_|= d

d(m)a¨x:n|−β(m)Ax:n| 1

In terms of life annuities only, we use the relationship

A1_x:n|

=A_x:n|−nEx and a¨x:n|=(1−Ax:n|)/d

Substitutinga¨(_xm_:n)_|= d

d(m)a¨x:n|−β(m)

(

1−da¨x:n|−nEx

)

¿α(m) ¨a_x:n|−β(m)

₍

1−nEx

)

where α(m)= id

i(m)_d(m), β(m)=

i−i(m)

i(m)_d(m)

¨ a

n| x(

m) _{= expected value of $1/m per m}th _{of a year payable at the beginning of each m}th _{of a}

year for n years while (x) is living, with the first payment starting n years from now.

Under UDD and using the relationship

¨ a

n| x (m)

= ¨ax(m)− ¨ax:n| (m)

We get_n|a¨x(m)=

d

d(m)a¨x−β(m)Ax−

(

d

d(m)a¨x:n|−β(m)Ax:n| 1

)

¿ d

d(m)n|a¨x−β(m)n|Ax

In terms of life annuities only, we use the relationships

¨ ax

(m)

=α(m)a¨x−β(m)

¨ a(_xm_:n)_|=α

(m)a¨_x:n|−β(m)

₍

1−nEx

)

and

¨ a

n| x

(m)

= ¨a(xm)− ¨ax:n|

(m) _{to get}

¨ a

n| x

(m)

=

_[

α(m)a¨x−β(m)

]

−

[

α(m)a¨x:n|−β(m)

(

1−nEx

)

]

¿α(m)_n|a¨x−β(m)nEx

3.4.3 Life Annuities Immediate with Monthly Payments

Use the relationships:

a_x(m) = ¨ax

(m)

−1

m

a(_xm_:n)_|

= ¨a(_xm_:n)_|−1 m+

1 mnEx

¿a¨(_xm_:n)_|− 1

m

(

1−nEx

)

a

n| x

(m)

=ax

(m)

−a(_xm_:n)_|

Example 3.4.2 Find the net single premium of an annuity providing $550 at the end of every 3 months for life to a person now age 50, assuming 1958 CSO mortality and 3% interest. Use the assumption of Uniform Distribution of Deaths for fractional ages and the following:

¨

a_x=$16.66at i=3 %interest1958CSO

i(4)

=0.0297,d(4)=0.0294, d=0.0291at i=3 % Solution:

We want2000a50(4)

a50(4)= ¨a50(4)− 1 4

¨

a50(4)=α(4)a¨50−β(4)

¨

a₅₀=16.66

α(4)= id i(4)_d(4)=

0.03(0.0291)

0.0297(0.0294)=0.9998

β(4)=i−i (4) i(4)

d(4)=

0.03−0.0297

0.0297(0.0294)=0.3436

i .e .a50¨(4)=0.9998(16.66)−0.3436=16.3130

i .e . a₅₀(4) = ¨a50(4)−

1

4=16.3130−0.25=16.0630

n . s . p .=2000a50 (4)

¿2000(16.0630)

¿$32,125.99

¨

a(xm)≈a¨x−m−1

2m − m2−1 12m2 μx+δ

or

¨ a(_xm)

≈a¨_x−m−1 2m

Formulas for other life annuity functions can be derived based on the above approximations.

3.5 Complete Annuities Immediate and

Apportionable Annuities Due

3.5.1Complete Annuities Immediate

Benefit payment is made at the end of each mth _{of a year while (x) is living.}

Upon (x) death, an adjustment payment is made to recognize the fraction of the mth _{of the}

year lives by (x).

An mth_{ly annuity immediate with an appropriate adjustment payment at death is called a}

complete annuity immediate.

e.g., monthly pension benefits paid at the end of the month while the retiree is living.

3.5.2 Apportionable Annuities Due

Here, insured (x) is making payments at the beginning of each mth _{of a year while (x) is}

living.

Upon (x)’s death, a refund of the unearned portion of the payment is made to the insured.

An mth_{ly annuity due which has this refund feature is called an apportionable annuity due.}

3.5.3 Basic Technique in Calculating Adjustment/Refund

Payment

Calculate an equivalent annual continuous payment sequence equal to either mth_ly

payments at the end or beginning of each mth _{of a year.}

Then calculate exact amount of adjustment/refund assuming equivalent annual continuous payments made.

3.5.4 Complete Annuity Immediate

1/m payable at the end of each mth _{of a year} ⟺1/

(

m s1

m

|

)

per payable continuously, where

s₁

m

|

=

∫

0 1

m

(1+i)tdt=i

(m) mδ

Then the exact adjustment upon death occurring, at x + k/m + t, 0≤ t ≤ 1

m , assuming 1/

(

ms1

m

|

)

payable continuously, is

s_t| m s₁

m

|

=(1+i)

t

−1

δ 1 m

mδ i(m)=

(1+i)t−1 i(m)

If exact adjustment payment is made, then n.s.p. for an mth_{ly annuity immediate with}

adjustment payment at death.

⟺n . s. p . for a continuous lifeannuity of 1 m s₁

m

|

= ax

m s₁

m

|

= δ

i(m)ax

i .e . ,a˙x

(m)

=n . s . p . for a complete annuity immediate= δ

i(m)ax

Note

1.¿ax

(m) < ˙ax

(m) <ax

Formulas:

1.¿a˙(xm)=

δ i(m)ax

2.¿a˙(_xm_:)_n|=complete temporary lifeannuity immediate

¿a˙_x(m)−_nE_xa˙(_xm₊)_n= δ

i(m)ax−nEx δ i(m)ax+n

¿ δ

i(m)ax:n|

3.¿n|a˙x(m)=complete deferred life annuity immediate

¿ δ

i(m)n|ax

3.5.5 Apportionable Annuity Due

1/m payable at the beginning of each mth _{of a year}

⟺ 1

ma₁

m

|

per year payable continuously ,

where a₁

m

|

=

∫

0 1

m

vtdt=d (m) mδ

Then exact refund upon death occurring at x + k/m +t, 0≤ t ≤1/m , assuming 1/(ma1

m

|

) _per

year payable continuously, is

a ₁

m−t

|

ma₁

m

|

=1−v

1

m−t

δ 1 m

mδ d(m)=

1−v 1

m−t

d(m)

If an exact refund payment is made, then n.s.p. for an mth_{ly annuity due with unearned}

⟺n . s. p . for a continuous lifeannuity of 1 m a₁

m

|

per year= ax ma₁

m

|

= δ

d(m)ax

Note

1.¿a¨(xm)> ¨ax{m}>ax

2.¿∈practice ,the refund payment at deathis approximated by(1/m−t).

(

mth_{ly payment}

₎

Formulas:

1.¿a¨{xm}=

δ d(m)ax

2.¿a¨{_xm_:}_n|=apportionable temporarylife annuity due= δ

d(m)ax:n|

3.¿_n|a¨{_nm}=apportionable deferredlife annuity due= δ

d(m)n|ax

3.5.6 Relationship between Complete and Apportionable Life

Annuities

¨ a_x{m}₌ δ

d(m)ax∧ ˙ax (m)

= δ

i(m)ax

⟹a¨x

{m}

=i

(m) d(m)a˙x

(m) > ˙ax

(m)

Since i d=

i iv=1+

i∧i(m) d(m) =

i(m) m d(m)

m

=(1+i)

1

m

It follows that

¨

ax{m}=(1+i)

1

m

˙ ax(m)

¨ a_x{m_:n}_|

=(1+i)

1

m

¨ an{m}

n| =(1+i)

1

m

˙ a(xm) n|

For m=1,a¨_x{1}=δ d ax,a˙x

(1)

= ˙a_x=δ

i ax∧ ¨ax

{1}

=(1+i)a˙_x

In Particular, for m=1

˙ a_x(1)

= ˙ax=

δ

i ax; a¨x{1}=

δ d ax

˙ a_x(1)_:n|

=δ

i ax:n|; a˙x:n| (1)

=δ

i ax:n|

˙ a_x(1)

n| =

δ

i n|ax; a¨n

{1}

n| =

δ d∗n|ax

Example 3.5.1 Which of the following is equal to a˙x

(m)_?

1.¿ δ

i(m)ax 2.¿v 1

m

¨

a{_xm} 3.¿E

[

1−v

T

d(m)

]

Solution:

1.¿a˙(xm)=

δ

i(m)ax1.¿istrue

2.¿a¨{x

m} ₌ δ

d(m)ax = i(m) d(m)a˙x

(m)

=(1+i)

1

m

˙ ax

(m)

i .e . ,a˙x

(m)

=(1+i)

−1

m ¨

ax{ m}

=v

1

m ¨

ax{

m} 2.¿istrue

3.¿E

[

1−v

T

d(m)

]

= 1−A_x

d(m) = δ a_x

d(m)≠a˙x (m)

= δ

i(m)ax3.¿is false

Example 3.5.2 Calculate a˙x

(2) _{given A}

Solution:

˙ a_x(2)

= δ

i(2)ax

δ=log(1+i)=log1.21=0.19

i(2)

satisfies

(

1+i (2) 2

)

2

=1+i

i .e .1+i (2)

2 =

√

1+i=

√

1.21=1.1⟺i (2)

=0.2

a_x=1−Ax

δ =

1−

(

i δ

)

Ax

δ under UDD

¿

1−

(

0.21 0.19

)

0.1 0.19

¿4.68

i .e .a˙x(2)=

(

0.19 0.2

)

4.68

¿4.45

Example 3.5.3 Y is the present value random variable for a life annuity that pays 1 at the end of each year that (x) survives plus a final adjustment payment at the moment of death.

The adjustment payment is t, where t is the portion of the year between the date of the last regular payment and the date at death. Assume that deaths are uniformly distributed over each year of age.

Calculate E(Y).

Solution:

Let Y1 = present value of annuity benefit. Let Y2 = present value of adjustment payment at

E(Y)=E

₍

Y₁

₎

+E

₍

Y₂

₎

=a_x+E(Y)

If death occurs at x+k+t ,0<t<1, present value of death benefit=t vk+t

Since1=

∫

0

∞

px

t μx+tdt=

∑

k=0

∞

∫

0 1

px

t+k μx+t+kdt=

∑

k=0

∞

px

k

∫

0 1

px+k

t μx+t+kdt ,

E

₍

Y₂

₎

=

∑

k=0

∞

p_x

k

∫

0 1

t vk+t _p x+k

t μx+t+kdt=

∑

k=0

∞

vk _p x

k

∫

0 1

t vt _p x+k

t μx+t+kdt

Under UDD pt xμx+t=qx,0<t<1

i .e . E

(

Y2

)

=

∑

k=0

∞

q_x+kvk_kp_x

_∫

0 1

t vtd t

But

_∫

0 1

t vtdt= t v

t

logv

|

¿10−

∫

₀

1 vt logvdt=

−v δ –

vt δ2

|

1

¿0

¿1−v

δ2 – v δ

¿iv

δ2− v δ

¿v

(

i−δ

δ2

)

Substituting , E

₍

Y₂

₎

=

∑

k=0

∞

vk _p x

k qx+kv

(

i−δ δ2

)

¿

(

i−δ

δ2

)

∑

k=0

∞

vk+1 q

k| x

¿

(

i−δ

δ2

)

Ax

3.6 Commutation Functions

Consider the following sequence of annuity payments to (x): d₀, d₁, d₂, … , d_n−1 , where d_t = benefit payment at the beginning of the (t+1)st year if (x) is living.

d0 d1 d2 ………. dn-1

--- --- --- -- x x+1 x+2 ……… x+n-1 x+n

Then n . s . p .:=d0+d1px+d2v22px+…+dn−1v

n−1 px n−1

¿d₀+d₁v

(

lx+1

l_x

)

+d2v 2

(

lx+2

l_x

)

+…+dn−1v

n−1

(

lx+n−1 l_x

)

¿d0

(

vxl_x vxlx

)

+d1

(

vx+1

l_x+1 vxlx

)

+d2

(

vx+2

l_x+2 vxlx

)

+…

+dn−1

(

vx+n−1_l

x+n−1 vx_l

x

)

¿ 1

D_x

(

d0Dx+d1Dx+1+d2Dx+2+dn−1Dx+n−1

)

where Dx+n=v

x+n

lx+n

Define N_x=D_x+D_x+1+D_x+2+… Then D_x=N_x−N_x+1

¿n . s . p .= 1

D_x

(

n0Nx+n1Nx+1+n2Nx+2+nn−1Nx−1+nnNx

)

where n₀=d₀,

n_t=d_t−d_t−1, t=1,2, … , n−1 n_n=−d_n−1

Then N_x=S_x−S_x+1

¿n . s . p .= 1

D_x

(

s0Sx+s1Sx+1+s2Sx+2+sn−1Sx−1+snSx

)

where s₀=n₀,

s_t=n_t−n_t−1, t=1,2, … , n s_n+1=−n_n

Example 3.6.1 Consider an increasing life annuity immediate to (x) where the benefit paid is t at the end of the tth _{year, t = 1, 2, 3, …}

Age: x – 1 x x+1 x+2 x+3 x+4 …

D: 0 0 1 2 3 4 …

N: 0 0 1 1 1 1 …

S: 0 0 1 0 0 0 …

n . s . p .=(Ia)x=

∑

t=1

∞

t vt_tp_x= 1 Dx

∑

t=1

∞

t D_x+t

¿ 1

Dx

∑

t=1

∞

N_x+t

¿Sx+n

Dx

Example 3.6.2 n-year temporary increasing life annuity due

Age: x – 1 x x+1 x+2 … x+n–1 x+n x+n+1…

D: 0 1 2 3 … n 0 0…

N: 0 1 1 1 … 1 –n 0…

S: 0 1 0 0 … 0 –(n+1) n…

n . s . p .=(Ia¨)_x:n|=Sx−n Sx+n+1−(n+1)Sx+n Dx

¿Sx−Sx+n−n Nx+n

Example 3.6.3 n-year temporary decreasing life annuity immediate

Age: x – 1 x x+1 x+2 … x+n–1 x+n x+n+1 x+n+2

D: 0 0 n n – 1 … 2 1 0 0

N: 0 0 n – 1 … – 1 – 1 – 1 0

S: 0 0 n –(n+1) … 0 –(n+1) 0 1

n . s . p .=(Da)_x:n|=n Sx+1−(n+1)Sx+2+Sx+n+2 Dx

¿n Nx+1−Sx+2+Sx+n+2

Dx

3.6.1 Recursion Equations

1.¿a¨_x=1+₁E_xa¨_x+1

2.¿a¨_x:n|=1+1Exa¨x+1 :n|−v

n

px n

3.¿n|a¨x=1Exn|a¨x+1+v

n

px n

4.¿(Ia¨)_x=1+2v px+3v

2 px

2 +4v 3

px

3 +…

¿1+v px

(

2+3v px+1+4v22px+1+…

)

¿1+v p_x

[

(

2+3v p_x+1+4v2₂p_x+1+…

)

+

(

1+v p_x+1+v2₂p_x+1+…

)

]

¿1+v p_x

_[

(Ia¨)_x+1+ ¨ax+1

]

5.¿(Ia¨)_x:n|=1+2v px+3v22px+…+n vn−1n−1px

¿1+v px

(

2+3v px+1+4v

2 px+1

2 +…+n vn−2n−2px+1

)

¿1+v p_x

[

(

2+3v p_x+1+4v2₂p_x+1+…+(n−1)vn−2_n−2p_x+1

)

+

(

1+v px+1+v22px+1+…+vn−2n−2px+1

)

¿1+v p_x

_[

(Ia¨)_x+1:n−1|+ ¨a_{x+1 :n−1|}

_]

6.¿(Da¨)_x:n|=n+(n−1)v px+(n−2)v

2 px

2 +…+v

n−2 px

n−2 +v