Differentiate with respect to t.
Then E(Y) = m'(0), E(Y^2) = m''(0), and V(Y) = E(Y^2)-[E(Y)]^2 = m''(0) - [m'(0)]^2.
Differentiate the moment-generating function in Exercise 3.147 to find E(Y) and E(Y^2). Then find V(Y).
Exercise 3.147
If Y has a geometric distribution with probability of success p, show that the moment generating function for Y is
m(t) = pe^t/(1-qe^t) , where q= 1-p
Im confused... Differentiate with respect to what? After I take the derivative, how do I find E(Y), E(Y^2) and V(Y)?
Thanks