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Is there a formula like $$ EX=\sum_{i=1}^{\infty}P\left(X\geq i\right) $$

(which can be found on Wikipedia and holds for positive $X$) for $EX^{n}$ ?

And I don't mean this one,

$$ EX^{n}=\sum_{i=1}^{\infty}P\left(X\geq\sqrt[n]{i}\right), $$

which is immediate, if we take $Y=X^{n}$ and use the above formula for $Y$. I mean a "more elegant" one - if there is one.

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up vote 8 down vote accepted

One has $$ \mathrm E(X^n)=\sum_{i=1}^\infty(i^n-(i-1)^n)\,\mathrm P(X\geqslant i). $$ More generally, $$ \color{red}{\mathrm E(u(X))=u(0)+\sum_{i=1}^\infty(u(i)-u(i-1))\,\mathrm P(X\geqslant i)}. $$ To prove this, call $(\ast)$ the RHS and note that $\mathrm P(X\geqslant i)=\sum\limits_{k=i}^\infty\mathrm P(X=k)$, hence $$ (\ast)=u(0)+\sum_{i=1}^\infty(u(i)-u(i-1))\,\sum_{k=i}^\infty\mathrm P(X=k), $$ that is, $$ (\ast) = u(0)+\sum_{k=1}^\infty\mathrm P(X=k)\sum_{i=1}^ku(i)-u(i-1) =u(0)+\sum_{k=1}^\infty\mathrm P(X=k)(u(k)-u(0)), $$ and, finally, $$ (\ast) = u(0)\mathrm P(X=0)+\sum_{k=1}^\infty\mathrm P(X=k)u(k)=\mathrm E(u(X)). $$

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The same way you get the original, write $P(X \ge i) = \sum_{k=i}^\infty P(X=k)$ and interchange the sums. – Nate Eldredge May 9 '12 at 15:36
Could you give me an explanation or at least a hint, how one arrives at this formula ? – MyCatsHat May 9 '12 at 15:39
ah, ok, thanks. – MyCatsHat May 9 '12 at 15:40
You are welcome. – Did May 9 '12 at 15:43

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