Hello,
Get inspired by this : Geometric distribution - Wikipedia, the free encyclopedia
Or consider the equality and differentiate it as you wish to compute the series you're dealing with.
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
[/I]
Hello,
Get inspired by this : Geometric distribution - Wikipedia, the free encyclopedia
Or consider the equality and differentiate it as you wish to compute the series you're dealing with.