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Alternative Expected Value Proof

I am trying to study probability and this is one of the unsolved exercises I encountered. Please help.

Show that if $X$ takes only non-negative integer values, we have

$$ \mathrm{E}[X] = \sum_{n=0}^\infty \mathrm{P}(X>n) $$

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marked as duplicate by robjohn Nov 12 '12 at 1:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Did Nov 11 '12 at 23:05
Thanks did and Andre you are right. – user669083 Nov 11 '12 at 23:11
Additionnally this was already answered n times on the site. – Did Nov 11 '12 at 23:14
did would you please provide me the links and be a bit more contructive. I am a noob and trying to learn mathematics. I couldn't find exact question. I will be obliged if you help. – user669083 Nov 11 '12 at 23:28
Why don't you wander a little bit on the site and try to see if you don't find them yourself? Even your self-proclaimed noobness doesn't prevent you from doing that. – Did Nov 11 '12 at 23:33
up vote 2 down vote accepted

The claim in your question is known as the Tail Sum Formula. To prove it, we will employ the definition of expectation:

For $X$ a random variable that takes on only non-negative values, we have

$E(X)=\sum_{n=0}^\infty n P(X=n)$

Let's expand the above expression:

$E(X)=\underbrace{0 P(X=0)}_{=0} + 1P(X=1)+2 P(X=2)+3P(X=3)+\cdots$

and for each integer multiple $k P(X=k)$, we will expres it as $k$ summands $k P(X=k)= \underbrace{P(X=k)+ P(X=k)+\cdots + P(X=k)}_{k \text{ terms}}$.

Now, we will arrange in the following fashion:

E(X) = P(X=1)
      +P(X=2) + P(X=2)
      +P(X=3) + P(X=3) + P(X=3)
      +P(X=4) + P(X=4) + P(X=4) + P(X=4)
      + ...

Now, regroup the sum above column-wise, you will get what you seek. Hint, the first column sums to $P(X>0$), and the second column sums to $P(X>1)$.

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Thanks bonsoon. That was a very good explanation. – user669083 Nov 11 '12 at 23:32

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