# Proof of Expected Value

• Nov 17th 2011, 07:00 AM
I-Think
Proof of Expected Value
RTP: $E[X]=\sum_{i=1}^{\infty} P[X\geq{i}]$

If I expand this algebraically, I can see it
$\sum_{i=1}^{\infty} P[X\geq{i}]= (P[X=1]+P[X=2]+...)+(P[X=2]+P[X=3]+...)+....$
$=P[X=1]+2P[X=2]+.....$

But I'm having trouble understanding the proof as given by the text and wikipedia, in particular the interchange of the order of summation

Given proof
$\sum_{i=1}^{\infty} P[X\geq{i}]= \sum_{i=1}^{\infty}\sum_{j=i}^{\infty} P[X=j]$
Then we interchange the order of summation
$\sum_{i=1}^{\infty} P[X\geq{i}]= \sum_{j=1}^{\infty}\sum_{i=1}^{j} P[X=j]$

This step, I do not understand. Given that $i$ does not appear in the summation of $\sum_{i=1}^{j} P[X=j]$, how am I supposed to expand this?
• Nov 17th 2011, 07:27 AM
Plato
Re: Proof of Expected Value
Quote:

Originally Posted by I-Think
Given proof
$\sum_{i=1}^{\infty} P[X\geq{i}]= \sum_{i=1}^{\infty}\sum_{j=i}^{\infty} P[X=j]$
Then we interchange the order of summation
$\sum_{i=1}^{\infty} P[X\geq{i}]= \sum_{j=1}^{\infty}\color{blue}\sum_{i=1}^{j} P[X=j]$

Look at the bit on blue.
For each $i=1\text{ to }j$ the term $P[X=j]$ is constant.
Thus $\sum_{i=1}^{j} P[X=j]=j\cdot P[X=j]$