A random variable X assumes only the values 0,1,2,3....., and the probabilities that X assumes these values are (2/3), (2/3)(1/3) , (2/3)(1/3)^2 , (2/3)(1/3)^3 , ................ respectively. Find the E(x) and Var(x).

Book Answer:

E(X) = 1/2

Var(x) = 3/4

I dont know where to start. Can someone show me how this problem should be done?

F(z) = P(X=0) +P(X=1)z+(p(x=2)z^2 + .......

F(z) = (2/3) + (2/3)(1/3)(z) +(2/3)(1/3)^2 z^2 + .........

F'(z) = (2/3)(1/3) + 2(2/3)(1/3)^2 z + ...........

F''(z) = .................

I know that

E(X) = P'(1)

and

Var(X) = f''(1) + f'(1) - [f'(1)]^2

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