Expected value of random variable

It is known that for any non-negative Y random variable:

$\displaystyle E(Y)= \int_0^\infty P\{Y>t\}\,\mathrm{d}t$

Show that for any non-negative X random variable:

$\displaystyle E(X^n)= \int_0^\infty nx^{n-1}P\{X>x\}\,\mathrm{d}x$

Thank you very much in advance!

Re: Expected value of random variable

As it is written, I'm not sure it exists.

Anyway, it looks like a simple job for an Integration by Parts. Let's see what you get.

Re: Expected value of random variable

Quote:

Originally Posted by

**TKHunny** As it is written, I'm not sure it exists.

Anyway, it looks like a simple job for an Integration by Parts. Let's see what you get.

You are right there was typo in the text, I corrected the variable t to x in the second integral.

I suppose that $\displaystyle E(X^n)=\int_0^\infty P(X^n>t)\,\mathrm{d}t$ and then changing the variable: $\displaystyle t:=x^n$. But to tell the truth I don't know how to continue.

Re: Expected value of random variable

Hello,

Since X is almost surely positive, you can say that $\displaystyle \{X^n>t\}\Leftrightarrow \{X>t^{1/n}\}$

and then change $\displaystyle x=t^{1/n}$