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I'm stuck with this proof:
$X:\Omega \to \mathbb{N}$ is random variable, prove that $\mathbb{E}[X]=\sum_{i=1,2,3...} \mathbb{P}(X\ge i)$?

How I'm proving it?
I'm starting with the definition: $$\mathbb{E}[X]=\sum_{x\in \Omega} \mathbb{P}(X = x) \cdot X(x)$$ How I'm continue from here??

Thank you!

CS1
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    This has been asked several times already. Try and search the site. (Side note: $\sum_{x\in\Omega} P(X=x)$ doesn't make sense, since $X$ takes values in $\mathbb{N}$, not $\Omega$). – Stefan Hansen Dec 13 '14 at 12:35
  • @StefanHansen, where I can find the the proof? I didn't found it :-(. About your note it's the definition of $E[X]$.... – CS1 Dec 13 '14 at 12:43
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    @Yoar No. Please think harder about Stefan's point before posting such remarks. The modified version $\mathbb{P}(X = x) \cdot X(x)$ is even more absurd. – Did Dec 13 '14 at 13:09
  • @Did, can you help me and tell me why it's more absurd? – CS1 Dec 13 '14 at 14:12
  • Because in the product $\mathbb{P}(X = x) \cdot X(x)$, the first $x$ can only belong to the target set $\mathbb N$ while the second $x$ can only belong to the source set $\Omega$ (as explained by @StefanHansen in the first comment). Anyway, this is not the formula for $\mathbb E(X)$ in your notes. – Did Dec 13 '14 at 19:55

2 Answers2

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Already posted $n$ times. The desired formula for $\mathbb E(X)$ is the integration of the pointwise identity $$X=\sum_{i=1}^\infty\mathbf 1_{X\geqslant i},$$ valid for every nonnegative integer valued random variable $X$.

Did
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$$\mathbb E(X)=\sum_{i=1}^{\infty}i\mathbb P(X=i)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\chi_{[1,i]}(j)\mathbb P(X=i)=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\chi_{[j,\infty)}(i)\mathbb P(X=i)=\sum_{j=1}^{\infty}\mathbb P(X\geq j)$$

where the next-to-last equality follows by fubini's theorem.

Marm
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  • Isn't Fubini's theorem for integrals? Or does it apply to sums too? – BCLC Dec 14 '14 at 10:34
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    @BCLC: Sums are integrals with respect to the counting measure. – Stefan Hansen Dec 14 '14 at 15:05
  • @stefan Cool. Like leb integral over {1, 2, ..., n} or $\mathbb{N}$ i guess? – BCLC Dec 14 '14 at 16:46
  • Ah sorry for confusion. I was wondering if you really needed something as strong as Fubini's to conclude such? Can't you switch from the fact that the summand is nonnegative? Never really got why that's true rigorously but my probability professor claimed with an argument of row and column sums (or is it that that was a mere intuitive explanation and the result actually follows from Fubini's?) – BCLC Dec 14 '14 at 16:48
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    @BCLC: Yes, Lebesgue integral over $\mathbb{N}$ with respect to the counting measure yields infinite sums. That you may interchange the order of double-sums when the summands are non-negative follows immediately by the non-negative version of Fubini (which is usually called Tonelli's theorem). – Stefan Hansen Dec 14 '14 at 18:10