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Let $X$ be a discrete random variable over $\mathbb{N}$, $E$ its expected value.

I want to show that $\displaystyle E(X)=\sum_{n\in\mathbb{N}}P(X\ge n)$

I thought of an "Euler sum" approach :

$\displaystyle E(X)=\sum_{n\in\mathbb{N}}nP(X=n)=P(1)+2P(2)+3P(3)+\cdots=\\ [P(1)+P(2)+P(3)+\cdots]+[P(2)+2P(3)+\cdots]+\dots=[P(1)+P(2)+P(3)+\cdots]\\ +[P(2)+P(3)+\cdots]+[P(3)+\cdots]+\cdots=\sum_{n\in\mathbb{N}}P(X\ge n)$

but that approach seems "too easy" and not rigorous enough to me (I do know that changing the order of the terms in the expected value's sum does not affect its final value, though).

What would be a rigorous way to prove it ?

Gabriel Romon
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Hippalectryon
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    This can also be seen as an instance of summation by parts. In any case, the only sensible way we could make this more rigorous is to begin by noting that the equality holds for all finite $n$, then taking the limit as $n \to \infty$ "of both sides". – Ben Grossmann Dec 08 '14 at 17:14
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    Integrate the pointwise equality $X=\sum_{n\in\mathbb{N}} \mathbf{1}_{X\geqslant n}$ and use Tonelli to interchange sum and integration. – Stefan Hansen Dec 08 '14 at 17:17

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$\displaystyle \sum_{n\geq 1}P(X\geq n)=\sum_{n\geq 1}\sum_{k\geq n}P(X=k)=\sum_{k\geq 1}\sum_{n\leq k}P(X=k)=\sum_{k\geq 1}kP(X=k)=E(X)$

Gabriel Romon
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