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To show: $X$ discrete random variable with values in $\mathbb{N}$. Show: $\mathbb{E}[X] = \sum_{n=1}^{+\infty}\mathbb{P}[X \geq n]$.

My attempt: Since $X$ is a discrete-integer-valued random variable, we have that:

$\mathbb{E}[X] = \sum_{k=1}^{+\infty}k \mathbb{P}[X=k]$.

Intuitevely, I can see why the claim should hold.

$\sum_{k=1}^{+\infty}k \mathbb{P}[X=k] = 1\mathbb{P}[X=1]+2\mathbb{P}[X=2]+ \dots$

and

$\sum_{n=1}^{+\infty}\mathbb{P}[X \geq n] = \mathbb{P}[X\geq1] + \mathbb{P}[X\geq 2]+\dots$

and for example:

$\mathbb{P}[X\geq1] = \mathbb{P}[X=1]+\mathbb{P}[X=2]+\dots$

$\mathbb{P}[X\geq2] = \mathbb{P}[X=2]+\mathbb{P}[X=3]+\dots$

But I don't see how to prove this formally. Any hint?

MyGanton
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    Do you know that if $a_{ij}\geq 0$ then $\sum_i \sum_j a_{ij}=\sum_j\sum_i a_{ij}$? – Ciarán Ó Raghaillaigh Jun 27 '21 at 17:40
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    In my opinion, you’re almost there unless something different is expected for “formal” proof here. Just add the equations you have written at the end, and show that how each term appears exactly as many times. – bigbang Jun 27 '21 at 17:41

1 Answers1

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The idea is $$E[X]=\sum_{k \in \mathbb{N}}k\,P(X=k)=\sum_{k \in \mathbb{N}}\sum_{n \in \mathbb{N}}\mathbb{I}_{(0,k]}(n)\,P(X=k)=\\ =\sum_{n \in \mathbb{N}}\sum_{k \in \mathbb{N}}\mathbb{I}_{(0,k]}(n)\,P(X=k)=\sum_{n \in \mathbb{N}}P(X\geq n)$$ In fact a much more general statement holds for arbitrary $\sigma$-finite measure spaces and positive measurable functions, and this is a particular case.

Snoop
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