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I stumbled upon an expression in an article of statistics for an $n$-th moment with $X$ being a random variable over $[0, \infty)$.

$\mathbb{E} X^{n} = \int^{\infty}_{0} nz^{n-1}\; \text{Pr}(X > z) \; \text{dz}$

Could someone enlighten me on why the above is true? It indeed works for the exponential distribution.

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    The functions P(X>z) and $P(X\geq z)$ can only differ at a countable set of $z$ values, and so it doesn't matter which of these you integrate.2012-06-14

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First, use Tonelli's theorem to conclude that (write the probability as an integral and interchange the two integrals) $ E[X^n]=\int_0^\infty P\left(X^n> z\right)\, \mathrm{d} z. $ Now write $P\left(X^n> z\right)=P\big(X> z^{1/n}\big)$ and use change of variables with $t=z^{1/n}$.

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    @DilipSarwate: Of course, I edited it now. Thanks for the heads up.2012-06-14