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I am currently tasked with proving an alternative definition of the expected value function.

Considering X to be a random variable that takes all positive integers, I have to prove that $E[X]=\sum\limits_{i=1}^{\infty} P[X\ge i]$.

So far I've gotten to the fact that $P[X\ge i] = \sum\limits_{j=i}^{\infty} P[X=j]$ however I just have no idea where to go from here. I have no idea how that relates to expected value in any way.

I'm most certainly not looking for the answer, but a friendly nudge in the right direction would be extremely helpful.

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Is this homework? –  robjohn Oct 20 '11 at 8:41
    
For even more details than @BrianScott has provided, see this answer by Mike Spivey to a similar question. –  Dilip Sarwate Oct 20 '11 at 11:41
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1 Answer

up vote 5 down vote accepted

Let $p_i = P[X=i]$. You know that $$E[X]=\sum_{i=1}^\infty ip_i$$ and that $$P[X\ge i] = \sum_{j=i}^\infty p_j,$$ so you want to show that $$\sum_{i=1}^\infty ip_i = \sum_{i=1}^\infty \sum_{j=i}^\infty p_j\;.$$ Try reversing the order of summation in the double summation.

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