问题及解答

Let $X$ have a Weibull distribution with the pdf from Expression (see alse Question2480)
\[
f(x;\alpha,\beta)=\begin{cases}
\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}e^{-(\frac{x}{\beta})^{\alpha}}, & x\geqslant 0,\\
0, & x < 0
\end{cases}
\]

Verify that

\[\mu=\beta\Gamma(1+\frac{1}{\alpha}).\] 

(Hint: In the integral for $E(X)$, make the change of variable $y=(\frac{x}{\beta})^{\alpha}$, so that $x=\beta y^{\frac{1}{\alpha}}$.)

(The proof can be found in the solution of Question2480.)