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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.

This is a duplicate of this question – robjohn Nov 12 '12 at 1:10
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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