ST301 Actuarial Mathematics Summer 2017 examination
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Summer 2017 examination
ST301
Actuarial Mathematics (Life)
1. Select life table. A special survival model has a select period of three years. Functions for this model are denoted by an asterisk, * . Functions without an asterisk are taken from the Canada Life Tables 2000-02, Males. You are given that, for all values of x,
p [x(*)] = 4px-5 , p [x(*)]+1 = 3px-1 , p [x(*)]+2 = 2px+2, px(*) = px+1 .
A life table, tabulated at integer ages, is constructed on the basis of the special survival model and the value of l2(*)5 is taken as 98363 (i.e. l26 for Canada Life Tables 2000-02, Males). Some lx values for this table are shown in the table below.
(a) Construct the l [x(*)] , l [x(*)]+1 , l [x(*)]+2 , and lx(*)+3 columns for x = 20, 21. [10 marks]
(b) Calculate 2|38q 
, 40p 
, 40p 
, and 40p2(*)2 . [6 marks]
[total=16 marks]
|
Age, x |
lx |
|
15 16 17 18 19 20 21 22 23 24 25 26
62 63 64 65 |
99180 99 135 99 079 99 014 98 942 98 866 98 785 98 700 98 615 98 529 98 444 98 363
87 503 86 455 85 313 84 074 |
2. Expenses and premiums. Consider a 20-year annual premium endowment insurance with sum insured 100000 issued to a select life aged 35. Assume initial expenses of 3% of the basic sum insured and 20% of the first premium, and renewal expenses of 3% of the second and subsequent premiums. Assume that the death benefit is payable at the end of the year of death. The interest rate is 5% per year. The following values will be useful for answering the questions below:
[35]:20
= 13.02489, A[35]:20
= 0.37977, 2 A[35]:20
= 0.14511, 19p[35] = 0.98466.
(a) Write down an expression for the gross future loss random variable. [5 marks]
(b) Calculate the gross annual premium. [5 marks]
(c) Calculate the standard deviation of the gross future loss random variable. [7 marks]
(d) Calculate the probability that the contract makes a profit. [7 marks]
[total=24 marks]
3. Multi-life insurance. The following values might be useful to you for this question. They all refer to a force of mortality µx . At a force of interest 0.03
30:40 = 21.8, 
30:30:40 = 20.3, 
30:40:40 = 20.5,
30:30:40:40 = 19.8, 
30:40:30:40 = 19.7.
(a) Show that
xy x(1)y + 
xyxy(1) = 
xyxy .
We recall that the notation above corresponds to a joint life status for four indepen- dent lives with respective ages x, y, x, y . [8 marks]
(b) Two sisters aged 30 and 40 are subject to the mortality forces µ30+t and µ40+t at time t. They are taking care of a disabled relative who is subject to a mortality force
0.01 + µ30+t + µ40+t
at time t. The deaths of the three lives are independent events. An insurance contact provides a benefit of 200000 when one of the two sisters dies while the disabled relative is still alive. If the second sister dies as well while the disabled relative remains alive, then the relative will receive a life annuity of 50000 per annum payable continuously. Calculate the single premium to be paid at the outset for these benefits using a force of interest of 0.02 per annum. [10 marks]
[total=18 marks]
4. Diversifiable vs non-diversifiable risks.
(a) The coefficient of variation for a random variable X is defined as the ratio of the standard deviation of X to the mean of X . Show that for a random variable X =
j = 1N Xj, if the risk is diversifiable, then the limiting value of the coefficient of
variation, as N → ∞, is zero. [6 marks]
(b) An insurer issues a portfolio of identical 15-year term insurance policies to indepen- dent lives aged 65. The sum insured for each policy is 100000, payable at the end of the year of death.
The mortality for the portfolio is assumed to follow Makeham’s law with A = 0.00022 and B = 2.7 ×10-6 . The insurer is uncertain whether the parameter c for Makeham’s mortality law is 1.124, as in the Standard Ultimate Survival Model, or 1 .114. The insurer models this uncertainty assuming that there is a 75% probability that c = 1.124 and a 25% probability that c = 1.114. Assume the same mortality applies to each life in the portfolio. The effective rate of interest is assumed to be 6% per year. The following values will be useful to answer some of the questions below:
A6(1)5:15
= 0.11642, A
= 0.06410,
where the unstarred function is calculated using the Standard Ultimate Survival Model and the starred function is calculated using Makeham’s law with c = 1.114.
i. Calculate the coefficient of variation of the present value of the benefit for an individual policy. [8 marks]
ii. Demonstrate that the mortality risk is not fully diversifiable, and find the limiting value of the coefficient of variation. [8 marks]
[total=22 marks]
5. Surplus. An office issued with profits 30 year endowment assurance policies to 30 year old lives with a sum assured of 200000. The premium is 3340 per annum payable continuously and was calculated using a force of mortality µx, where x a life’s age, a force of interest of 0.05 per annum and no expenses. The real force of interest is stochastic and can take the values 0.04 and 0.07. The transition rates between the two values are 0.8. The real
mortality force is also µx at age x. The force of interest is 0.07 initially. The policyholders will receive all surplus in the form of a terminal bonus.
(a) Set out a system of equations and explain how you would use it to predict the terminal bonus. [15 marks]
(b) Suppose that we are at time t = 5 and the force of interest has been 0.07 for the first year and 0.04 for the next four years and it still is 0.04. Explain how you would predict the terminal bonus
(Hint: You have to calculate the accumulated surplus at t = 5) [5 marks]
[total=20 marks]
2023-01-13