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I want to prove that for non-negative random variables with distribution F: $$E(X^{n}) = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx$$

Is the following proof correct?

$$R.H.S = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx = \int_0^\infty n x^{n-1} (1-F(x)) dx$$

using integration by parts: $$R.H.S = [x^{n}(1-F(x))]_0^\infty + \int_0^\infty x^{n} f(x) dx = 0 + \int_0^\infty x^{n} f(x) dx = E(X^{n})$$

If not correct, then how to prove it?

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up vote 4 down vote accepted

Here's another way. (As the others point out, the statement is true if $E[X^n]$ actually exists.)

Let $Y = X^n$. $Y$ is non-negative if $X$ is.

We know $$E[Y] = \int_0^{\infty} P(Y \geq t) dt,$$ so $$E[X^n] = \int_0^{\infty} P(X^n \geq t) dt.$$ Then, perform the change of variables $t = x^n$. This immediately yields $$E[X^n] = \int_0^{\infty} n x^{n-1} P(X^n \geq x^n) dx = \int_0^{\infty} n x^{n-1} P(X \geq x) dx.$$

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A better way to prove it would be to use Fubini's theorem to change the order of integration. This also gives you a condition when the result you have is true.

Consider $I = \displaystyle \int_0^{\infty}n x^{n-1} (1-F(x))dx$.

Using the fact that $\displaystyle \int_x^{\infty} f(y)dy = 1 - F(x)$, we get $I = \displaystyle \int_0^{\infty}n x^{n-1} \int_x^{\infty} f(y) dy dx$.

Now we first integrate with respect to $y$ (the inner integral) and $y$ goes from $x$ to $\infty$ and then integrate with respect to $x$ (the outer integral), $x$ goes from $0$ to $\infty$.

Change the order of integration. i.e. integrate with respect to $x$ first and then with respect to $y$.

Note that this can be done provided the integral $I < \infty$ (See Fubini's theorem). This is the condition svenkatr and trutheality get as well.

Changing the order of integration, we get

$I = \displaystyle \int_0^{\infty} \displaystyle \int_{0}^{y} nx^{n-1}f(y)dxdy$.

Note that now $x$ in the inner integral goes from $0$ to $y$ and $y$ goes from $0$ to $\infty$.

Now the inner integral with respect to $x$ can performed easily and now we get

$I = \displaystyle \int_0^{\infty} y^{n}f(y)dy = E[X^n]$.

Hence, we have $\displaystyle \int_0^{\infty}n x^{n-1} (1-F(x))dx = E[X^n]$.

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If ${\rm E}(X^n) < \infty$, then $\int_0^\infty {x^n f(x){\rm d}x} < \infty $, and in turn, $\int_M^\infty {x^n f(x){\rm d}x} \to 0$ as $M \to \infty$. Since $\int_M^\infty {x^n f(x){\rm d}x} \geq \int_M^\infty {M^n f(x){\rm d}x} = M^n [1 - F(M)]$, we have that $M^n [1 - F(M)] \to 0$ as $M \to \infty$. Hence, your solution is correct (assuming that ${\rm E}(X^n) < \infty$).

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Actually, the result is true also if ${\rm E}(X^n) = \infty$. – Shai Covo Nov 18 '10 at 1:45

Only if $\lim_{x\rightarrow \infty} x^n(1-F(x)) = 0$.

It doesn't look like that's true in general.

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As a matter of notation, you should write the expectation as $E(X^n)$ instead of $E(x^n)$. The expectation is not a function of $x$.

Your proof would work if $\lim_{x \to \infty} x^n (1-F(x)) = 0$. I'm not sure this is true for any distribution.

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I corrected it! – Osama Gamal Nov 18 '10 at 0:21

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