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I already found the topic which shows that $$\sum_{i=0}^{\infty} P(X>i) = E(X)$$ Find the Mean for Non-Negative Integer-Valued Random Variable.

In my case I need to show that $$\sum_{i=0}^{\infty} i P(X>i) = 1/2 (E(X^2)+E(X))$$

Hence this assumption can I say that it's not or am I missing something.

tinlyx
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vicR
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  • Are you sure that you have $\sum_{i=0}^{\infty} i P(X>1)$? I am just asking because $\sum_{i=0}^{\infty} i P(X>1)=\infty.$ – zoli Mar 18 '17 at 17:07
  • its X>i i guess thats the same isnt it ? – vicR Mar 18 '17 at 19:28
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    Yes, but do you pay attention at all to the problem you are to solve? – zoli Mar 18 '17 at 20:02
  • that was a typo, back to the actual question, so its not cos the actual equation is E(X) – vicR Mar 19 '17 at 09:12
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    @vicR Are there any more typos? Look to your problem carefully. Say, for $X$ with Bernoulli distribution, $$0\cdot P(X>0)+1\cdot P(X>1) =0 \neq \frac12(E(X^2)+E(X))=\frac12(p+p)=p.$$ – NCh Mar 19 '17 at 14:28

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This question is a bit old,I hope my answer helps you or someone else.

First: $\frac{1}{2}(E[X^2] + E[X]) = \frac{1}{2}(\sum x^2P[X=x] + \sum xP[X=x]) = \frac{1}{2}(\sum x(xP[X=x] + P[X=x])) = \frac{1}{2}(\sum x(x+1)P[X=x]) = \sum \frac{x(x+1)}{2}P[X=x]$

And since $$P[X \geq 1] = P[X=1] + P[X=2] + P[X=3] + ...$$ $$2P[X \geq 2] = 2P[X = 2] + 2P[X=3] + 2P[X = 4]+...$$ In general $$xP[X \geq x] = xP[X = x] + xP[X=x+1] + ...$$ So we have $\sum xP[X\geq x] = \sum \frac{x(x+1)}{2}P[X=x]$

Hence the Equality holds.

Cochemuacos
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