Question

Use the result that for a nonnegative random variable$Y$,

$E\left[Y\right]={\int }_0^{\infty }P\left\{Y>t\right\}dt$

to show that for a nonnegative random variable$X$,

$E\left[X^n\right]={\int }_0^{\infty }n{x}^{n-1}P\left\{X>x\right\}dx$

Hint: Start with

$E\left[X^n\right]={\int }_0^{\infty }P\left\{X^n>t\right\}dt$

and make the change of variables$t=x^n$.

Short answer

Therefore, we’ve shown$E\left[X^n\right]={\int }_0^{\infty }n{x}^{n-1}P\left\{X>x\right\}dx={\int }_0^{\infty }n{x}^{n-1}P\left\{X>t\right\}dt$as required.

Step-by-step solution

Given information:

Use the result that for a non-negative random variable$Y$,

$E\left[Y\right]={\int }_0^{\infty }P\left\{Y>t\right\}dt$

Hint: Start with

$E\left[X^n\right]={\int }_0^{\infty }P\left\{X^n>t\right\}dt$

and make the change of variables$t=x^n$

Explanation:

By letting $Y=X^n$we know that

$E\left[X^n\right]={\int }_0^{\infty }P\left\{X^n>t\right\}dt$

Making the change of variables$t=x^n$and$dt=n{x}^{n-1}dx$, we get that

$E\left[X^n\right]={\int }_0^{\infty }n{x}^{n-1}P\left\{X^n>x^n\right\}dx$

Explanation:

Since $u\mapsto u^n$is an increasing function of $u\ge 0$the events $\left\{X^n>x^n\right\}$and$\left\{X>x\right\}$are identical. So we’ve shown

$E\left[X^n\right]={\int }_0^{\infty }n{x}^{n-1}P\left\{X>x\right\}dx={\int }_0^{\infty }n{x}^{n-1}P\left\{X>t\right\}dt$as required.