a) The moment generating function of X if t < 4/3. Otherwise, MX(t) is infinite.
b) The mean of X is 3/7. The variance of X is 9/7.
(a) The moment generating function of X is given by
MX(t) = E[etX] = 2/5 * e^0t + (3/4) * e^t / 5 + (3/4)^2 * e^2t / 5 + ...
= (2/5 + 3/20 * (3/4)t + 9/80 * (3/4)^2 t^2 + ...)
= (2/5 + 3t/20 + 9/40 t^2 + ...) * 1/(1 - 3t/4)
This is the moment generating function of X if t < 4/3.
Otherwise, MX(t) is infinite.
(b) The mean of X is given by
E[X] = M'X(0) = (d/dt) [MX(t)]t=0
= (3/20 + 9/40 * 2 + ...) * 1/(1 - 3t/4)
= 3/4 * 1/(1 - 3t/4)
= 3/4 * 4/7
= 3/7
The variance of X is given by
Var [X] = E[X^2] - E[X]^2
= M''X(0) - M'X(0)^2
= (d^2/dt^2) [MX(t)]t=0 - M'X(0)^2
= (9/40 + 9/20 * 2 + ...) * 1/(1 - 3t/4) - (3/7)^2
= 9/4 * 1/(1 - 3t/4) - (3/7)^2
= 9/4 * 4/7 - (3/7)^2
= 9/7.
Question
Suppose that the random variable X has probability mass function given by
P(X=0)=2/5, P(X=k)=((3/4)^(k))(1/5), k=1,2,...
(a) Compute the moment generating function MX(t) of X. Be careful about the possibility that MX(t) might be infinite.
(b) Use the moment generating function to compute the mean and the variance of X.