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Alternative Expected Value Proof

If $X$ is a random variable that takes values in the range $\left \{ 1,2,3,4,5,6,\ldots \right \}$ how can I prove the following statement?

$$\mathbb{E}[X]=\sum_{i=1}^\infty \mathbb{P}[X \ge i]$$

I honestly have no idea where to even begin. Even just something to kick off the process would be great.

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marked as duplicate by Byron Schmuland, t.b., Mike Spivey, anon, Asaf Karagila Oct 25 '11 at 6:26

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.

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Haven't we seen this question or its continuous version about a half-dozen times in the past couple of weeks? –  Michael Hardy Oct 25 '11 at 4:03
    
If that's the case, do you have a link to one of the other half dozen questions? –  Steve S Holoman Oct 25 '11 at 4:09

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up vote 2 down vote accepted

$\rm\bf Note$: $$\sum_{k=1}^\infty\; \mathrm{P}(X\ge k)=\sum_{k=1}^\infty\; \sum_{\ell=k}^\infty\; \mathrm{P}(X=\ell)$$ $\rm\bf Rhetorical\text{ } question$: How many times does $\mathrm{P}(X=n)$ appear above for $n=1,2,3,\dots$?

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This is similar to the final proof I found by following this link to this link to this link. Thanks! –  Steve S Holoman Oct 25 '11 at 4:32

Consider the following doubly infinite array, where $p_k=P(X=k)$. We find the sum of all the elements of the array in two different ways.

$$\begin{matrix} p_1 &p_2 & p_3 &p_4 & p_5 &p_6 & p_7 &p_8 &p_9&p_{10}&p_{11} &p_{12} &\dots\\ &p_2 & p_3 &p_4 & p_5&p_6 & p_7 &p_8 &p_9&p_{10}&p_{11} &p_{12} &\dots\\ && p_3 &p_4 & p_ 5&p_6 & p_7 &p_8 &p_9&p_{10}&p_{11} &p_{12} &\dots\\ &&&p_4 & p_5 &p_6 & p_7 &p_8 &p_9&p_{10}&p_{11} &p_{12} &\dots\\ &&& & p_5 &p_6 & p_7 &p_8 &p_9&p_{10}&p_{11} &p_{12} &\dots\\ &&&& &p_6 & p_7 &p_8 &p_9&p_{10}&p_{11} &p_{12} &\dots\\ &&&& && p_7 &p_8 &p_9&p_{10}&p_{11} &p_{12} &\dots\\ &&&& &&&p_8 &p_9&p_{10}&p_{11} &p_{12} &\dots\\ &&&&&&&&\vdots&\vdots&\vdots&\vdots&\ddots \end{matrix} $$

The first way: The first row sum is $P(X \ge 1)$, the second row sum is $P(X\ge 2)$, the third row sum is $P(X \ge 3)$, and so on. So the sum of all the row sums is $\displaystyle\sum_{k=1}^\infty P(X\ge k)$.

The second way: The first column sum is $p_1$, the second column sum is $2p_2$, the third column sum is $3p_3$, and so on. So the sum of all the column sums is $p_1+2p_2+3p_3+\cdots$, that is, $E(X)$.

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