# Weibull moment generating function and Gamma function

I'd be grateful for any hints or help with this question: Let $X$ follow the Weibull distribution with pdf

$f(x)=\beta x^{\beta-1}e^{-x^{\beta}}$

on $x>0$ with $\beta>0$. Show that

$E(X^r)=\Gamma(\frac{r}{\beta}+1)$

where $\Gamma(a)=\int_{0}^{\infty}x^{a-1}e^{-x} dx$

This is how far I have got.......

$E(X^r)=M_X^{(r)}(0)$

$M_X(t)=E(e^{tX}) = \int_{0}^{\infty} e^{tx}\beta x^{\beta-1}e^{-x^{\beta}}$

Let $u=x^\beta$

So

$E(e^{tx})=\int_{0}^{\infty} e^{tu^{1/\beta}}e^{-u}$

• Your approach works with an appropriate modification: A Weibull ($W$) for $X$ is an Exponential ($E$) distribution on $U=X^\beta$. Therefore $E_W[X^r]$ = $E_E[U^{r/\beta}]$ = $\Gamma(r/\beta+1)$.
– whuber
Mar 26, 2012 at 18:37

## 1 Answer

Resorting to mgf is not helpful here. It is easier to go for the expectation directly.

$$E[X^r]=\beta\int_0^{\infty}x^{\beta +r-1}\exp(-x^{\beta})dx$$

Make the same change of variables you did before. Ill post the rest of the answer later.

• Thank you. So, it's just: $E(X^r)=\int_{0}^{\infty }(u^{1\beta})^r e^{-u} du$ And then the required result follows immediately. It seems too easy ! ! It's great, but have I missed anything ? Mar 26, 2012 at 11:28
• Nope, you've got it in one. I won't bother updating my answer as it is already in your comment. :-) Mar 26, 2012 at 20:48