Prove 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?
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$\begingroup$ Yes i will upload a picture of my current way, im not sure if its the mathematically correct way to prove it. $\endgroup$– Ste0lCommented Nov 12, 2023 at 13:33
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$\begingroup$ This is called summation by parts. $\endgroup$– whuber ♦Commented Nov 12, 2023 at 14:55
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$\begingroup$ See [math.stackexchange.com/questions/1795529/… in math StackExchange. I am pretty sure this famous identity has been asked here before as well so it is 99.9% a duplicate. $\endgroup$– ZhanxiongCommented Nov 12, 2023 at 14:56
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$\begingroup$ It's a duplicate almost surely @Zhanxiong :-) $\endgroup$– User1865345Commented Nov 12, 2023 at 15:50
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1$\begingroup$ @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. $\endgroup$– User1865345Commented Nov 13, 2023 at 13:21
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1 Answer
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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}
answered Nov 12, 2023 at 13:52
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