Alternative formula for expected value of a discrete random variable

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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1 Answer

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 ? –  user26698 May 9 '12 at 15:39
ah, ok, thanks. –  user26698 May 9 '12 at 15:40
You are welcome. –  Did May 9 '12 at 15:43