I am currently tasked with proving an alternative definition of the expected value function.

Considering X to be a random variable that takes all positive integers, I have to prove that $E[X]=\sum\limits_{i=1}^{\infty} P[X\ge i]$.

So far I've gotten to the fact that $P[X\ge i] = \sum\limits_{j=i}^{\infty} P[X=j]$ however I just have no idea where to go from here. I have no idea how that relates to expected value in any way.

I'm most certainly not looking for the answer, but a friendly nudge in the right direction would be extremely helpful.

  • 1
    $\begingroup$ Is this homework? $\endgroup$ – robjohn Oct 20 '11 at 8:41
  • $\begingroup$ For even more details than @BrianScott has provided, see this answer by Mike Spivey to a similar question. $\endgroup$ – Dilip Sarwate Oct 20 '11 at 11:41
  • $\begingroup$ This post itself could have served as a "mother post" to duplicates. The current choice of mother post has the same number of duplicate-links and and more other links. Please see the meta post on (abstract) duplicates. $\endgroup$ – Lee David Chung Lin Nov 13 '18 at 13:21

Let $p_i = P[X=i]$. You know that $$E[X]=\sum_{i=1}^\infty ip_i$$ and that $$P[X\ge i] = \sum_{j=i}^\infty p_j,$$ so you want to show that $$\sum_{i=1}^\infty ip_i = \sum_{i=1}^\infty \sum_{j=i}^\infty p_j\;.$$ Try reversing the order of summation in the double summation.


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