3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Your approach works with an appropriate modification: A Weibull ($W$) for $X$ is an Exponential ($E$) distribution on $U=X^\beta$. Therefore $E_W[X^r]$ = $E_E[U^{r/\beta}]$ = $\Gamma(r/\beta+1)$. $\endgroup$ – whuber Mar 26 '12 at 18:37
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – P Sellaz Mar 26 '12 at 11:28
  • $\begingroup$ Nope, you've got it in one. I won't bother updating my answer as it is already in your comment. :-) $\endgroup$ – probabilityislogic Mar 26 '12 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.