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My wayProve that for random variable with natural numbers from 1 to infinity the expected value $E(X)$ is equal to $\sum_{n = 1}^\infty P(X \geq n)$. Is this the mathematically correct way to prove it? And are there any steps I missed or mistakes made?

Zhanxiong
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Ste0l
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  • Yes i will upload a picture of my current way, im not sure if its the mathematically correct way to prove it. – Ste0l Nov 12 '23 at 13:33
  • This is called summation by parts. – whuber Nov 12 '23 at 14:55
  • See [https://math.stackexchange.com/questions/1795529/why-is-mathbbex-1-sum-infty-k-1-mathbbpx-k-true/1798539#1798539] in math StackExchange. I am pretty sure this famous identity has been asked here before as well so it is 99.9% a duplicate. – Zhanxiong Nov 12 '23 at 14:56
  • It's a duplicate almost surely @Zhanxiong :-) – User1865345 Nov 12 '23 at 15:50
  • @whuber, off topic comment. But it seems that the upvote in my post didn't add to my rep. I might be missing something for it shows +15 but not the additional +10. Nothing important but got curious. – User1865345 Nov 13 '23 at 03:26
  • @User1865345 I see that you didn't get any rep for six upvotes, not just one. Did you perhaps receive a notice yesterday stating that "serial voting was reversed"? I am just guessing, but this sort of thing happens to me all the time: the powers that be disapprove of users reading through one person's post and upvoting bunches of them in a short period. It's a mechanism to inhibit voting rings where groups of people vote each others' posts up to increase their reputations. – whuber Nov 13 '23 at 13:12
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    @whuber, yes. You are right. Someone spontaneously voted some of my old posts (it is evident they didn't read those; the pace at which the reps were received was high) but within half an hour or so, it was reversed. This upvote came way after that. Is this related to that too? But at least it is showing the +1 along the post (unlike the ones that were reversed). Thanks for the reply. I don't want to continue further this off-topic comment here. I really don't have any problem whether it gets counted or not. – User1865345 Nov 13 '23 at 13:21

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Yes. This is looking legit. For the sake of clarity, the second equality occurs due to this:

\begin{align}\sum_{i=1}^\infty i\Pr(X=i) &=\sum_{i=1}^\infty i\Pr(i\leq X<i+1)\\&=\sum_{k=1}^\infty [\Pr(k\leq X<k+1)+\Pr(k+1\leq X<k+2)+\ldots]\\&=\sum_{k=1}^\infty\Pr(X\geq k).\end{align}

User1865345
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