# Proof for mean of non-negative discrete random variable

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• March 17th 2009, 09:43 PM
blues360
Proof for mean of non-negative discrete random variable
For a nonnegative integer-valued discrete random variable X, show that
E(X)=http://upload.wikimedia.org/math/3/d...c702e182f0.pngx=0 (to infinity) P(X>x). Can anyone help me solve this please?
• March 18th 2009, 08:03 AM
meymathis
So

$\sum_{x=0}^\infty P(X>x) = \sum_{x=0}^\infty \sum_{y=x+1}^\infty P(X=y)$

This is a triangular region. So we can switch the order of the sum if we do it carefully.

$= \sum_{y=1}^\infty \sum_{x=0}^{y-1} P(X=y)$

$= \sum_{y=1}^\infty y P(X=y)$

Can you finish it?
• March 18th 2009, 05:03 PM
blues360
Thank so much for your reply!
So to finish it then:

http://www.mathhelpforum.com/math-he...b874ace9-1.gif=http://upload.wikimedia.org/math/3/d...c702e182f0.pngx=0 (to infinity)xP(x)=E(x)?
Because x=0 --> y=1?
• March 18th 2009, 08:12 PM
meymathis
Well, there are no x's anymore. The x and y are just the summing variables.

The main observation at this point is that the sum:
$\sum_{y=1}^\infty yP(X=y)$
almost is $E[X]$ (whether it is x or y doesn't matter, it is just a summing variable). The only thing missing is the 0 term. The sum should go from 0 to $\infty$, whereas this one starts at 1. But the 0 term is
$0P(X=0)=0$.

So $\sum_{y=1}^\infty yP(X=y) = 0P(X=0)+\sum_{y=1}^\infty yP(X=y) = \sum_{y=0}^\infty yP(X=y) =E[X]$