Let $X$ be a discrete random variable, then $\mathbb{E}[X]=\sum_{n=1}^{\infty}\mathbb{P}(X \geq n)$

Is this true or false ?

I think it's false , just by using the definition of $\mathbb{E}[X]=\sum_{n=1}^{\infty}x_n\mathbb{P}(X=x_n)$. But I'm not so sure, it seems to be so trivial. I'd like to find a counter-example

marked as duplicate by Did probability Apr 16 '17 at 15:12

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  • 1
    I suggest you try an example: a six-sided die, with equal probability of landing on each side. Evaluate both expressions for $E$, and see whether they match. If they do, the explicit combination may show you why. If not, it will convince you (and anyone else!) that the statement is false. – John Hughes Apr 16 '17 at 14:58
  • They match! Now it's claer. By using the first, I have: $1/6 + 2/6 +...+5/6 +1=3.5$. With the second I have the to sum $6$ times the probability to find a number greater or equal than $n$ and get the same result. Thank you very much – VoB Apr 16 '17 at 15:23
  • You might want to very carefully examine why drhab's extra condition is needed. – John Hughes Apr 16 '17 at 16:57

It is true under the extra condition that $\Pr(X\in\{0,1,2,\dots\})=1$.

If in that case $p_n:=\Pr(X=n)$ then: $$\mathbb EX=p_1+2p_2+3p_3+\cdots=(p_1+p_2+p_3+\cdots)+(p_2+p_3+\cdots)+(p_3+\cdots)+\cdots$$

  • Thanks, now it's clear. ;) – VoB Apr 16 '17 at 15:21

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