Math Is Fun Forum

SmileAli

Member

Registered: 2010-04-02

Posts: 10

Expected Value of Geometric random variable

Hi, so I am trying to understand this statement. There is a discrete random variable X that takes positive integers where E(X) = k = 1 to infinity P(X>= k). So, how would I add this to find the expected value of a geometric random variable?
For a geometric random variable the equation is P(X = k) = (1-p)^(k-1) * p. And, the expected equation is E(X) So, how would I make it go to infinity? Would I find the X<= k, and then 1 - of the answer? Thanks for you help.

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Expected Value of Geometric random variable

As far as I can tell you want to use the formula

(which is valid for a random variable X that takes only non-negative integer values) to compute the expected value of a Geometric random variable.

In order to compute

you could use that

but I think it's just as easy to use