Since the working requires lots of symbols and notation, I am producing a hand written solution.
The symbols used in the solution below are either the ones contained in the question or others having usual meaning in the subject of insurance.
Please see the solution below:
Final Answer is Var(L0) = 0.124351

Consider a fully discrete whole life insurance of 1 to (r) with the following details: K...
Let Lı denote the loss-at-time-zero random variable for a unit benefit fully continuous whole life insurance on (x) where it is assumed that the premium rate, π, is calculated using the equivalence principle. Let L2 denote the loss-at-time-zero random variable for the same insurance on (x) assuming that the premium rate, π2, is (4/3) . Find E(L2) and Var(L2) given the following: (c) δ-08. (a) Var(L1)-.5652, (b.) āz-5,
Let Lı denote the loss-at-time-zero random variable for a unit benefit fully...
4. A fully discrete whole life insurance policy paying $50,000 at the end of the year of death is issued to an individual age 36. The net premium reserve at the end of 10 years is $8,000. The net premium for this policy is $900 and the net premium for an identical policy issued to an individual age 46 is P. The effective annual interest rate of interest is 6%. Determine P.
2. For a special fully discrete 10-payment whole life insurance on (30) with level annual net premium P: (i) The death benefit is equal to 1000 plus the refund, without interest, of the net premiums paid. (ii) i = 0.04 (iii) A30 = 0.30 (iv) 30. 10 = 0.014 (v) (14)zo:707 = 0.048 (vi) ä30:101 - 8.5 Calculate P
Let X be a discrete random variable with the following PMF. Px(k) = 1/4 for k = -2 1/8 for k = -1 1/8 for k = 0 1/4 for k = 1 1/4 for k = 2 0 otherwise Define a new random variable Y = (X + 1)2 a) Find E[X] and Var[X] b) Find the range of Y and write its PMF. c) Show that the PMF of Y is a valid PMF. d) Find P(Y ≤...
Let X be a discrete random variable with the following PMF 6 for k € {-10,-9, -, -1,0, 1, ... , 9, 10} Px(k) = otherwise The random variable Y = g(X) is defined as Y = g(x) = {x if X < 0 if 0 < X <5 otherwise Calculate E[X], E[Y], var(X), and var(Y) for the two variables X and Y
Problem 3 A discrete random variable Y takes values {k= 0, 1, 2, ...,} such that PLY Z k} = ()* for k 20. 1. Derive P[Y = k) for any k > 0. 2. Evaluate expectation, E[Y] = 3. Given E[Y(Y - 1)] = 15 , find variance of Y, Var[Y] =
Consider a discrete random variable X that can assume three values 1, 2, and k with respective probabilities 0.2, 0.5, and 0.3. If E(X) = 2.7, what is the value of k? Select one: a. 3 b. 1 c. 4 d. 5 e. 2
5. Consider a 10-year annual premium endowment insurance with sum insured $200,000 issued to a life aged 40, Assume initial expenses of 4% of the basic sum insured and 15% of the first premium,and renewal expenses of3% ofthe second and subsequent premiums. Assume that the death benefit is payable at the end of the year of death. (a) Write down an expression for the gross future loss random variable I4 (10 pts.) (b) Calculate the gross annual premium. (8 pts.)
Consider a discrete random variable X with pmf x)-(1-p1 p. defined for x - 1, 2, 3,..The moment generating function for this kind of random variable is M(t)Pe 1-(1-P)et. (a) What is E(X)? O p(1-P) 1-P (a) What is Var(x)? 1-p p2 p(1-P) O p(1-P) o -p
1. Consider a discrete random variable, X, where the outcome of this random variable is determined by throwing a 6-sided die. X takes on integer values 1,2,…,6. The die is fair. That is, P(X=1)= P(X=2)=…= P(X=6). i. Draw the probability distribution function for this random variable. Carefully label the graph. ii. Draw the cumulative distribution function for X. iii. Calculate the following: P(X=4) P(X≠5) P(X=1 or X=6) P(X4) E(X) Var(X) sd(X) iv. Consider the random variable Y where the outcome...