Actuarial Mathematics II Problems Class 2 Question. A term

Actuarial Mathematics II
Problems Class 2
Question. A term insurance policy for a 30-year old person is taken out and pays 20000 on
death if it occurs within the next 25 years. Assuming AER 5% and using commutation tables,
what is the fair price of this contract?
Solution.
• The point is to express everything in terms of value now. We work assuming that the
payment on death is one unit, and then multiply by 20000.
• x = 30 and n = 25, i = 0.05 and v = 1/1.05.
• If death is this year, the payout is 20000 with probability P (K = 0). The present value
of this payout is 20000v = 19047.62 in todays money.
• If death is next year, the payout is 20000 with probability P (K = 1). The present value
of this payout is 20000v 2 = 18140.59 in todays money.
• If death is in the final year, the payout is 20000 with probability P (K = 24). The present
value of this payout is 20000v 25 = 5906.06 in todays money.
• If death occurs later, there is no payout with probability P (K ≥ 25).
• The net single premium
is the sum of these 25 amounts multiplied by their probP24(NSP)
K+1
P (K = k).
abilities, i.e. E(Z) = k=0 v
• So this is the fair price, in today’s money, of the policy.
• In lectures we showed that for a unit payment the NSP for a term insurance is
1
=
E(Z) = Ax:n
Mx − Mx+n
M30 − M55
308013.589 − 213303.847
= 0.04308.
=
=
Dx
D30
2198405.5
• So the NSP if the payment is 20000 is 20000 × 0.04308 = 861.62.
• This is the fair price of the contract in today’s money. If the insurance company wants
to make a profit they will need to charge more than this.
• Note that this is a surprisingly small amount. It reflects that the probabilities of death
between 30 and 55 for someone aged 30 already are small, and most of the probability is
of death after 55 and with no payment.
3-5. Question. A whole life insurance issued to (x) pays £10,000 at the end of the year of
death if this person dies during the first 20 years of this policy, and £20,000 at the end of the
year of death if this person dies thereafter. In addition, if this person dies during the first 20
years of the policy, the net single premium is also returned, without interest. Express the net
single premium of this contract using commutation functions (assume constant AER i).
3-5. Solution.
• Let the net single premium be denoted by E(Z). The present value random variable Z
of this whole life insurance contract is Z = v K+1 [10, 000 + E(Z)] if K = 0, 1, . . . , 19 and
Z = v K+1 [20, 000] if K ≥ 20.
• Note that the first part is a life insurance with term duration, and with payment 10000 +
E(Z). The second part is a deferred whole life insurance with payment 20000. This leads
to
E(Z) = [10, 000 + E(Z)]
19
X
v
k+1
k px qx+k
+ 20, 000
k=0
1
E(Z)]Ax:n
∞
X
v k+1 k px qx+k
k=20
+ [20000]n| Ax
Mx+20
Mx − Mx+n
+ [20000]
,
= [10, 000 + E(Z)]
Dx
Dx
= [10, 000 +
using formulae from lectures.
• Solving this gives
Mx + Mx+20
E(Z) = 10, 000
.
Dx − Mx + Mx+20
• Note: this exercise shows how such whole life insurance contracts, with variable payments,
can often be written as the sum of contracts presented in the lectures, where the deferred
life insurances (the second term above) play a useful role.
3-17. Question. Assume that each of 100 independent lives is of age x at time 0, is subject
to a constant force of mortality µ = 0.04, and has a whole life insurance for a payment of
£100,000 payable at the moment of death. The payments involved are to be withdrawn from
an investment fund earning continuously compounded interest with force of interest δ = 0.06.
Calculate the minimum amount required in this fund, at time 0, so that the probability is
approximately 0.95 for the event that sufficient funds will be on hand to withdraw the benefit
payments at the deaths of all individuals. Briefly comment on this minimum required initial
amount.
(Hint: use the Central Limit Theorem (1H Probability), and the fact that, for a standard
normally distributed random quantity X, we have P (X ≤ 1.645) = 0.95.)
3-17. Solution.
• First note that constant force of mortality implies the exponential distribution. This is
because
Rt
P (T > t) = t px = e− 0 µx+s ds = e−0.04t ,
so an exponential distribution with parameter µ = 0.04.
• Next consider the individual policies. Let Zi be the present value of the contract to person
i, for i = 1, . . . , 100.
• For the exponential lifetime model, E(Zi ) =
µ
µ+δ
= 0.4, as shown in lectures.
• Also shown in lectures was that for whole life insurance, E(Zi2 ) is the same as E(Zi ), but
µ
with force of interest 2δ (i.e. 2δ replaces δ). Thus, E(Zi2 ) = µ+2δ
= 0.25.
• Hence Var(Zi ) = 0.25 − 0.16 = 0.09 and SD(Zi ) = 0.3.
P
• These lives are independent, so we have, for Z = 100
i=1 Zi , E(Z) = 40 and Var(Z) = 9.
• By the Central Limit Theorem, we have that Z ∼ N (40, 32 ) approximately.
• Let h be the minimum initial amount required in this fund to ensure sufficient funds for
all 100 death benefit payments, assuming 1-unit payments, with approximate probability
= 1.645, giving h = 44.935
0.95. Then we want, approximately, P (Z ≤ h) = 0.95, so h−40
3
per unit of payment
• As the payments per policy are £100,000, The total expected payout is 100 × 100, 000 ×
0.4 = £4, 000, 000, and an approximation for the required initial amount to meet this
criterion, is £4,493,500.
• The difference between this amount, and the expected total present value of all these
100 payments, which is £4,000,000, is called the ‘risk loading’ (for this criterion). In this
exercise, this risk loading is 12.34% of the net single premiums, which may perhaps be
surprisingly large.
• Of course, in this example, there is more than one would expect in real life, due to the
exponential distribution being used for the future lifetimes, giving high probabilities that
people die young. Also, because of this distribution, it is actually not relevant what the
current ages are of the people involved.
Supplementary question: The company wishes a risk loading of 10% of NSP. How many
policies should it sell?
√
• E(Z) = 0.4n √
and SD(Z) = 0.3 n. Per unit of payment per policy we calculate h =
(1.645) ∗ (0.3) n + 0.4n as before.
• The percentage increase relative to the total payment is
√
√
(1.645) ∗ (0.3) n + 0.4n − 0.4n
(1.645) ∗ (0.3) n
=
,
0.4n
0.4n
We want this to equal 0.1 to give 10% risk loading. It follows that we need n =
( (1.645)(0.3)
)2 , i.e. n = 152.21, or n = 153 rounding up. This gives a risk loading of
0.04
9.97%.