Geometric distribution, E[X] and Var[X]

JeeBee · science forum beginner Joined: 18 Dec 2005 Posts: 9

Can somebody help me out with this?
(really, it's not homework :)

I need to know how the expectation and variance of the geometric
distribution is computed exactly. At many sites I can find the solution,
but not the exact computation. By myself I came this far:
(given failure probability p)

sorry if the notation is sloppy :(

E[X] = sum_{t=1..inf} t.p.(1-p)^(t-1) =
= p.sum_{t=1..inf} t.(1-p)^(t-1) =
= (using some calculus formula I don't understand here below)
= p.1/(p^2) = 1/p

That calculus formula I was referring to is:
sum_{t=1..inf} t.x^(t-1) = 1/((1-x)^2)

Now, I know that Var[X] = E[X^2] - E[X]^2.
I've read somewhere that E[X^2] should be (2-p)/(p^2)...

E[X]^2 = 1/(p^2)
E[X^2] = sum_{t=1..inf} ??????

Var[X] = 1/(p^2) - ??? = (1-p) / (p^2).

Thanks in advance,
JeeBee.

danielwells@gmail.com · science forum beginner Joined: 28 Feb 2006 Posts: 1

This formula:

sum_{t=1..inf} t.x^(t-1) = 1/((1-x)^2)

is the derivative of the geometric series.

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