# Proof for mean of non-negative discrete random variable

• Mar 17th 2009, 10:43 PM
blues360
Proof for mean of non-negative discrete random variable
For a nonnegative integer-valued discrete random variable X, show that
• Mar 18th 2009, 09:03 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?
• Mar 18th 2009, 06:03 PM
blues360
So to finish it then:

Because x=0 --> y=1?
• Mar 18th 2009, 09:12 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 $\sum_{y=1}^\infty yP(X=y) = 0P(X=0)+\sum_{y=1}^\infty yP(X=y) = \sum_{y=0}^\infty yP(X=y) =E[X]$