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The problem is:

Suppose $X\ge 0$ is a random variable, $p(x)$ is its probability density function. If $\operatorname EX$ exists, $r>0$, prove:

$$\operatorname EX^r = \int_0^\infty rx^{r-1}P(X>x) \, dx.$$

I see $\operatorname E X^r$ as $\int_{-\infty}^\infty x^r p(x) \, dx$, while $$\text{RHS}=\int_0^\infty \left(\int_x^\infty p(t)\,dt \right) dx^r = \left.\left(x^r\int_x^\infty p(t) \, dt\right)\right|_0^\infty-\int_0^\infty x^r p(x) \, dx.$$

However, when calculating $x^r \int_x^\infty p(t) \, dt$ as $x\to \infty$, it comes to a $0 \times \infty$ problem. How can I solve it?

Thanks.

StubbornAtom
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James Yu
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2 Answers2

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\begin{align} & \int_0^\infty rx^{r-1}P(X>x) \, dx \\[8pt] = {} & \int_0^\infty rx^{r-1} \left( \int_x^\infty p(w)\,dw \right) \, dx \\[8pt] = {} & \iint\limits_{(x,w)\,:\, 0\,<\,x\,<\,w} rx^{r-1} p(w) \, d(x,w) \\[8pt] = {} & \int_0^\infty \left( p(w) \int_0^w rx^{r-1} \, dx \right) \, dw \\[8pt] = {} & \int_0^\infty w^r p(w) \, dw \end{align} You said $X\ge0,$ so this last integral is only from $0$ to $+\infty,$ not from $-\infty$ to $+\infty.$

Michael Hardy
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$$E(X^r)=\int_0^{\infty} r x^{r-1} [1-F_X(x)]dx$$

by parts you get

$$E(X^r)=\Big[x^r(1-F_X(x))\Big]_0^{\infty}+\int_0^{\infty} x^r f_X(x)\,dx$$

$$x^r\left[1-F_X(x)\right]_{x=\infty}=\frac{x^r}{\frac{1}{1-F}}=\frac{\infty}{\infty}\rightarrow 0$$

Using the l'hopital $r$ times

Thus you are all set

Michael Hardy
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tommik
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  • Is it always true that $x^rP(X>x)\to 0 $ as $x\to \infty$? What if we take $p(x)=\frac{2}{x^3}\cdot \mathbb{1}_{[1,+\infty)}$ and $r=2$. Then, $x^2P(X>x)\to 2$ as $x\to \infty$. I think we have to assume that $EX^r$ is finite in order to get $x^rP(X>x)\to 0$. Am i somewhere wrong? – ChrisNick92 Nov 12 '20 at 17:04