Formula similar to $EX=\sum\limits_{i=1}^{\infty}P\left(X\geq i\right)$ to compute $E(X^n)$?

Question

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.

Answer 1

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, $$ (\ast) = u(0)\mathrm P(X=0)+\sum_{k=1}^\infty\mathrm P(X=k)u(k)=\mathrm E(u(X)). $$