Question

Q1 (Joint probability). We know X|Y is a normal random variable with mean Y and variance 2. The probability distribution of Y is a binomial distribution with success probability 0.3 and the number of trials 5. What is the expected value of X?

Answer 1

We can find out the expected value of Y initially using the formula E(Y) = n*p = 5*0.3 = 1.5

Variance of Y = npq = 5*0.3*0.7 =1.05

E(X/Y) = 2

Variance of X = npq = 1.05

Expected value of X = E(X) = n*p = 1.5