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Let $X$ be a random variable, where $X(s)\geq0$ for all $s \in S$, and let $A_k=\{s\in S: A(s)\geq k\}$ Show that $E(X)=\sum^\infty_{k=1} \mathbb{P}(A_k)$.

Pretty stumped on this one. My thought was just to start from the definition of $E(X)$ and try to somehow work $A_k$ into the definition through substitution, but not really seeing a way to do so. Help would be much appreciated!

Mittens
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goldenbudha
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    What have you tried? An obvious first step would be to work examples. Suppose the support of $X$ is just ${1,2,3}$ or such. Work it out explicitly in cases like that. Try to generalize. – lulu Aug 26 '21 at 12:46
  • Thanks that's helpful! I forgot that I could write $p(A_k)$ as a sum. When doing that I get $\sum\limits_{i = k}^\infty p(X = i)$. Then the original summation becomes $\sum\limits_{k = 1}^\infty \sum\limits_{i = k}^\infty p(X = i)$. At this point I can see how it is starting to look like the definition of expected value. I think I need to manipulate one of the summations somehow but not sure where to go from here. – goldenbudha Aug 26 '21 at 12:55
  • Exactly! $\quad $ – lulu Aug 26 '21 at 12:56
  • $$\begin{align}\sum^\infty_{k=1}P[X\geq k]&=\sum^\infty_{k=1}\sum^\infty_{j=k}P[X=j]=\sum^k_{k=1}\sum^\infty_{j=1}P[X=j]\mathbb{1}(j\geq k)\ &=\sum^\infty_{j=1}\sum^\infty_{k=1}P[X=j]\mathbb{1}(k\leq j)\ &=\sum^\infty_{j=1}P[X=j]\sum^\infty_{k=1}\mathbb{1}(k\leq j)=\sum^\infty_{j=1}P[X=j] j=E[X] \end{align}$$ – Mittens Aug 26 '21 at 13:36
  • Thank you! They both help. This might be a dumb question, but I'm having some trouble understanding how you/the other question swapped the summations to $\sum\limits_{k = 1}^k \sum\limits_{j = 1}^\infty$ – goldenbudha Aug 26 '21 at 13:52

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Let me have a simple explanation in continuous case: if $X$ is a non negative random variable

$$\mathbb{E}[X]=\int_0^{\infty}[1-F_X(x)]dx$$

in fact, integrating by parts you get

$$\int_0^{\infty}[1-F_X(x)]dx=\underbrace{\left[x(1-F) \right]_0^{\infty}}_{=0}+\int_0^{\infty}xf_X(x)dx$$

The expression $=0$ can be easy verified using de l'Hôpital on the resulting limit.

This prove what you are requested to do, as $1-F=P(X>k)$. For discrete rv, the proof is still valid substituting the integral with a sum

tommik
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