Question

Let $X$be a nonnegative random variable having a distribution function $F$. Show that if $\overline{F}\left(x\right)=1-F\left(x\right)$, then

$E\left[X^n\right]={\int }_0^{\infty }{x}^{n-1}\overline{F}\left(x\right)dx$

Hint: Start with the identity

$X^n=n{\int }_0^X{x}^{n-1}dx$

$=n{\int }_0^{\infty }{x}^{n-1}{I}_X\left(x\right)dx$

where
$I_x\left(x\right)=\left\{\begin{array}{l}1\\ 0\end{array}\right.$if$x<X$ otherwise

Short answer

$=\overline{\mathrm{F}}\left(x\right)$
Hence $\mathrm{E}\left({\mathrm{X}}^n\right)=n{\int }_0^{\infty }{x}^{n-1}\overline{F}\left(x\right)dx$

Step-by-step solution

Concept Introduction

Let $X$ be a Non-negative Random variable having distribution function $F$.
With $\overline{\mathrm{F}}\left(\mathrm{x}\right)=1-\mathrm{F}\left(x\right)$

:Explanation

Let $X$be a Non-negative Random variable having distribution function $F$.
With $\overline{\mathrm{F}}\left(\mathrm{x}\right)=1-\mathrm{F}\left(x\right)$

:Explanation

${\mathrm{X}}^{\mathrm{n}}=n{\int }_0^x{x}^{n-1}dx$

Step 4:Explanation

$=n{\int }_0^{\infty }{x}^{n-1}{I}_x\left(x\right)dx$$=n{\int }_0^{\infty }{x}^{n-1}{I}_x\left(x\right)dx$$=n{\int }_0^{\infty }{x}^{n-1}{I}_x\left(x\right)dx$$=n{\int }_0^{\infty }{x}^{n-1}{I}_x\left(x\right)dx$

Step 5:Explanation

Where ${\mathrm{I}}_{\mathrm{x}}\left(x\right)=\left\{\begin{array}{l}1\\ 0\end{array}\right.$if $x<X$$x<X$
otherwise

Step 6:Explanation

$\therefore \mathrm{E}\left[{\mathrm{I}}_{\mathrm{X}}\left(x\right)\right]=1.\mathrm{P}[x<X]$

Step 7:Explanation

$=\mathrm{P}[\mathrm{X}>x]$

Step 8:Explanation

$=1-\mathrm{P}[\mathrm{X}\le x]$$=1-\mathrm{P}[\mathrm{X}\le x]$$=1-\mathrm{P}[\mathrm{X}\le x]$
$=1-\mathrm{F}\left(x\right)$

Final Answer

$=\overline{\mathrm{F}}\left(x\right)$
Hence $\mathrm{E}\left({\mathrm{X}}^{\mathrm{n}}\right)=n{\int }_0^{\infty }{x}^{n-1}\overline{F}\left(x\right)dx$