STAT 353: Sample Final
Important: All exercises/examples treated in class and all
assignments, the midterm and the sample midterm are part of the
sample final.
Problem 1: Please write in words what the following quantities mean?
Ax, a¨x:n|, ax, Px.
Problem 2: 1) Prove the following formula
Px:n| = nPx + (1− Ax+n)Px:n|1.
2) If 15P45 = 0.038, P45:15| = 0.056 and A60 = 0.625, then calculate P
1
45:15|.
Problem 3 An ordinary life contract for a unit amount on a fully discrete
basis is issued to a person age x with an annual premium of 0.048. Assume
d = 0.06, Ax = 0.4 and
2Ax = 0.2. Let L be the insurer’s loss function at
issue of this policy.
a) Calculate E(L).
b) Calculate V ar(L).
Put:
P = 0.048, Ax = 0.4,
2Ax = 0.2, d = 0.06.
Problem 4: If T (x) has an exponential distribution with parameter µ.
1) Determine the cumulative distribution function of the loss variable ,
L(P ), of a whole life insurance payable at the time of death.
2) Determine E[L(P )] in terms of P , µ and δ (the force of interest).
3) Use part 2) to confirm that E[L] = 0 when P = P (Ax).
4) Suppose now that µ = 0.03 and δ = 0.06.
a) Evaluate P (L ≤ 0) when P = P (Ax).
b) Determine P so that P (L > 0) = 0.5.
Problem 5: Consider the setting of Example 6.4.1. I.e. we consider the
case where the annual benefit level premium is payable in semi-annual
instalments for a 10,000, 20-year endowment insurance proceeds paid at the
end of the policy year of death (discrete) issued to (50), on the basis of
illustrative life table with i = 6%. Calculate P
(20)
50 . Assume that the UDD
assumption holds.
Problem 6: 1) Write P
(m)
x:n| as a function of P
1(m)
x:n| , P
1
x:n| and Px:n| ONLY.
2) Application: Suppose that
P
1(12)
x:20|
P 1
x:20|
= 1.032 and Px:20| = 0.040.
What is the value of P
(12)
x:20|?
Problem 7: Using the assumption of Uniform Distribution of Death
(UDD) in each age and the Illustrative Life Table with interest at the
effective annual rate i = 6%, calculate

(6)
50 , a¨
(6)
50:25|, and 25|a¨
(6)
50 . 