MATH 403A, fall 98, CHAPTER 4 Probabilities involving the present value random variable

Step 1: Express Z as a function of T. And then express F_{Z}(z) = P [Z ≤ z]. Step 2: Algebraically rearrange the stuff in brackets so that F_{Z}(z) = P [T ≥ something], where “something” is a function of z. Step 3: the probability on the right side now is in the form 1 − F_{T} (something). Step 4: Now you have F_{Z}(z). To find the p.d.f. take derivative.

1. You are givent that q_{x} = .2x for x = 0, 1, 2, …, 5. Death are uniformly distributed within each year of age. The force of interest is .06. A 4–year continuous endowment insurance of 1 is issued to a life aage 1. The single premium is determined in such a way that the probability of a loss on the policy is .25. There are no expenses other than the death benefit and endowment. Find the premium.

2. With the mortality and interest assumptions of problem 1, find the p.d.f. of Z, the present value random variable for a 4–year term insurance of 100 issued to a life aged 0.

3. A special continuous whole life insurance pays a death benefit b_{t} = (1.04)^{t}. Mortality follows deMoivre’s Law with w = 85. Using a force of interest of .08, find the probability that the present value of the death benefit is less than .5.