Mathematics of Life Contingencies MATH 3281 - Life ...efurman/MATH3281_2011/Lec3.pdf · Mathematics of Life Contingencies MATH 3281 Life annuities contracts Edward Furman Department

Mathematics of Life Contingencies MATH3281

Life annuities contracts

Edward Furman

Department of Mathematics and StatisticsYork University

February 13, 2012

Edward Furman Mathematics of Life Contingencies MATH 3281 1 / 23

Definition 0.1 (Life annuity.)

Life annuity is a series of payments made continuously or atequal intervals while a given life survives.

Unlike in the context of life insurances, we have a series ofpayments.

Discrete payments can be made not only at the end of theyear, but also at the beginning of the year.

Question.

Given a life status (u) what is the expected future lifetime? Ifone dollar is payed at the beginning of every year while (u) isalive, what would the random present value of the contract be?What if the payment was made at the end of the year?

Recall that, e.g., K (u) : Ω → Nu ⊆ [0, ∞].


Solution.

The expected future lifetime is E[K (u)] full years.

The random present value if the payments are made at thebeginning of each year is

..

aK (u)+1 = 1+v + · · ·+vK (u)+1−1.

Also, the random present value of the payments if they aremade at the end of each year is aK (u) = v + · · ·+ vK (u).

Once again, recall that we find the price by taking expectationoperator (under the identity utility principle).

Net premium for an annuity contract

Fix a life status (u) (be careful with joint life statuses). The netpremium for a continuously payable annuity is

E[aT (u) ] =

∫

Ru

at tpuµ(u + t)dt = −

∫

Ru

at d tpu := au.


Net premium for an annuity contract. cont.

The net premium for an annuity due payable annually is

E[..

aK (u)+1 ] =∑

k∈Nu

..

ak+1 kpuqu+k = −∑

k∈Nu

..

ak+1∆kpu :=..

au.

Last but not least, the net premium for an annuity certainpayable annually is

E[aK (u) ] =∑

k∈Nu

ak kpuqu+k = −∑

k∈Nu

ak ∆kpu := au.

Example 0.1

Whole life annuity. Let (u) = (x). Then Ru = [0,∞) andNu = 0, 1, . . .. Thus


Cont.

..

ax = E[..

aK (x)+1 ] =

∞∑

k=0

..

ak+1 k |qx .

Further

ax = E[aK (x) ] =∞∑

k=0

ak k |qx .

And

ax = E[aT (x) ] =

∫ ∞

0at tpxµ(x + t)dt .

Recall that.

..

an =1 − vn

d, an =

1 − V n

i, an =

1 − vn

δ, n ≥ 0.


Proposition 0.1

Fix a life status (u), and let i , d , δ be the effective and discountinterests and the force of interest, respectively. Then,

au =1 − Au

δ, au =

1 − (1 + i)Au

i,

..

au =1 − Au

d.

Proof.

We clearly have that

au = E[

aT (u)

]

= E

[

1 − vT (u)

δ

]

=1 − Au

δ.

The same can be done for..

au. For the last one

au = E

[

1 − vK (u)

i

]

= E

[

1 − (1 + i)vK (u)+1

i

]

,

as required.Edward Furman Mathematics of Life Contingencies MATH 3281 6 / 23

Proposition 0.2

Fix (u). The variance of the random present value can be givenby

Var[aT (u) ] =2Au − (Au)

2

δ2 ,

and

Var[..

aK (u)+1 ] =2Au − (Au)

2

d2 ,

as well as

Var[aK (u) ] =(1 + i)2

(

2Au − (Au)2)

i2.

Proof.

The proof is straightforward and similar to the proof of theprevious proposition. It is thus omitted.


Proposition 0.3

Consider a whole life annuity due. Then

..

ax =

∞∑

k=0

vkkpx =

∞∑

k=0

kEx .

Proof.

We have that

..

ax =∞∑

k=0

..

ak+1 kpxqx+k =∞∑

k=0

k∑

i=0

v iP[K (x) = k ]

=

∞∑

i=0

(

∞∑

k=i

P[K (x) = k ]

)

v i =

∞∑

i=0

P[K (x) ≥ i]v i

=

∞∑

i=0

P[K (x) > i − 1]v i =

∞∑

i=0

ipxv i ,


Proposition 0.4

Let again (u) = (x). Then

..

ax = 1 + Ex..

ax+1.

Here..

a∞ = 0.

Remark.

Note that..

a∞ 6=..

a∞ = 1/d .

Proof.

We have that

..

ax = 1 +

∞∑

k=0

k+1pxvk+1 = 1 + vpx

∞∑

k=0

vkkpx+1,

as required.


Example 0.2 (n year term life annuity.)

Let Ru = [0, ∞) and Nu = 0, 1, . . . ,∞. And let (u) = (x : n ).In the discrete case, K (x : n ) is K (x) if K (x) < n andK (x : n) = n if K (x) ≥ n. The above is in fact the number ofpayments for the discretely payable annuity due.

E[..

aK (x:n )+1 ] =

n−1∑

k=0

..

ak+1 k |qx +

∞∑

k=n

..

an k |qx :=..

ax:n .

(Recall the endowment insurance case.) Recall, however thatthe p.m.f. is

kpx:n · q(x:n )+k =

kpx · qx+k , k ≤ n − 2n−1px , k = n − 10, k ≥ n

Thus

E[..

aK (x:n )+1 ] =n−2∑

k=0

..

ak+1 k |qx +..

an n−1pxEdward Furman Mathematics of Life Contingencies MATH 3281 10 / 23

Example 0.2 (cont.)

Therefore

E[..

aK (x:n )+1 ] =

n−2∑

k=0

..

ak+1 k |qx +..

an

∞∑

k=n−1k |qx

=n−1∑

k=0

..

ak+1 k |qx +..

an

∞∑

k=nk |qx

=n−1∑

k=0

..

ak+1 k |qx +..

an npx

Also

E[aT (x:n ) ] =

∫ n

0at tpxµ(x + t)dt + an · npx .


Proposition 0.5

For (u) = (x : n), we have that

..

ax:n =n−1∑

k=0

kEx .

Proof.

..

ax:n =

n−1∑

k=0

k∑

i=0

v ik |qx +

n−1∑

i=0

v inpx

=

n−1∑

i=0

n−1∑

k=i

v ik |qx +

n−1∑

i=0

v inpx

=

n−1∑

i=0

v i

(

n−1∑

k=ik |qx + npx

)

=

n−1∑

i=0

ipxv i ,


Proposition 0.6

For a general (u) (be careful with the joint life statuses), wehave that

..

au = −..

ab+1 b+1pu +..

aa apu +

b∑

k=a

kpuvk .

Proof.

Write the annuity as, for a and b being any in 0, 1, . . . ,∞.

..

au = −b∑

k=a

..

ak+1∆kpu.

..

au = −

(

..

ak+1 kpu∣

∣

b+1a −

b∑

k=a

k+1pu∆..

ak+1

)

= −..

ab+2 b+1pu +..

aa+1 apu +b∑

k=a

k+1puvk+1.


Proof.

Further

..

au = −..

ab+2 b+1pu +..

aa+1 apu +b+1∑

k=a+1

kpuvk

= −..

ab+1 b+1pu +..

aa apu +b∑

k=a

kpuvk


Example 0.3 (Deferred annuities.)

We are sometimes interested in an annuity that starts to payone dollar after n time units only. We thus have, e.g.,

..

au −..

au:n := n∣

∣

..

au =∞∑

k=n

kpuvk .

Also,

n∣

∣

..

ax =∞∑

k=n

..

ak+1 k∣

∣qx −..

an

∞∑

k=n

k∣

∣qx =∞∑

k=n

(

..

ak+1 −..

an

)

k∣

∣qx

=

∞∑

k=0

(

..

ak+n+1 −..

an

)

k+n∣

∣qx =

∞∑

k=0

(

vn(1 − vk+1)

d

)

k+n∣

∣qx

= vn∞∑

k=0

..

ak+1 n+kpx · qx+n+k = nEx ·..

ax+n.


Example (cont.)

For the continuous case, we have that (make sure you canprove that),

n|ax = nExax+n,

as well as

n|ax =

∫ ∞

ntExdt .

Note that we can obtain, say the discretely payable deferredwhole life annuity as an expectation of the r.v.

K ∗(x) :=

..

aK (x)+1 −..

aK (x)+1 ≡ 0, K (x) < n..

aK (x)+1 −..

an , K (x) ≥ n.


Example 0.4 (n year certain and then life annuity)

Consider an r.v.

K ∗(x) :=

..

an , K (x) < n..

aK (x)+1 , K (x) ≥ n.

Then taking expectation, we have that

..

ax:n =..

an nqx +

∞∑

k=n

..

ak+1 k |qx =..

an nqx −∞∑

k=n

..

ak+1∆kpx

=..

an nqx −

(

..

ak+1 kpx∣

∣

∞

n −∞∑

k=n

k+1pxvk+1

)

=..

an + vnnpx +

∞∑

k=n

k+1pxvk+1 ..

an + n|..

ax .

In a similar fashion ax:n = an + n|ax = an + nExax+n.


Proposition 0.7

The variance of the n year certain and whole life annuity isequal to the variance of the n years deferred annuity.

Proof.

Note that if we denote by K ∗(x) the random present value ofthe deferred annuity and by k∗∗(x) the random present value ofthe n year certain whole life annuity, then

k∗∗(x) =..

an + K ∗(x).

Thus

Var[k∗∗(x)] = Var[..

an ] + Var[K ∗(x)] + Cov[..

an ,K ∗(x)],

as required.


Recall.

We know thatsn := (1 + i)nan ,

for n ≥ 0. The above is an accumulated present value of anannuity at time n instead of time 0. Similarly, we have that

sx:n := (nEx )−1 ax:n ,

that represents the actuarial accumulated value at the end ofthe term of an n-year temporary life annuity of 1 per yearpayable continuously while (x) survives. The benefit isavailable at age x + n if x survives till then


Summary of discrete life annuities of 1 per annum payable at thebeginning of each year (due) or at the end of each year (immediate)


Useful relations between specific discrete annuities and insurances.


Summary of continuous life annuities of 1 per annum payablecontinuously.


Typical distributions for the present value rv T .


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Mathematics of Life Contingencies MATH 3281 - Life ...efurman/MATH3281_2011/Lec3.pdf · Mathematics of Life Contingencies MATH 3281 Life annuities contracts Edward Furman Department