Mean of Weibull distribution

Question

I have the following CDF of Weibull distribution:

$$ F_X(t) = 1 - e^{-\lambda t^{\alpha}} $$

Where $\alpha$ is the shape parameter. PDF can be found by differentiation CDF:

$$ f_X(t) = \lambda \alpha t^{\alpha - 1} e^{-\lambda t^{\alpha}} $$

In order to find the expected value or mean, I proceeded as follows:

$$ E(X) = \int_0^\infty \lambda \alpha t^{\alpha - 1} e^{-\lambda t^{\alpha}} dt $$

Letting, $ u = t^\alpha $, I arrived at,

$$ E(X) = \lambda \int_0^\infty u^{\frac{1}{\alpha}} e^{-\lambda u} du $$

I know I am supposed to end up with a gamma notation, but I'm not sure how to deal with $ e^{-\lambda u} $ part. How can I go about integrating this? The end result should be: $ \frac{\Gamma(1+1/\alpha)}{\lambda^{1/\alpha}}$

Answer 1

You forgot to multiply by $t$ in the expectation. It should be $$\Bbb E[X] =\int_0^\infty tf(t)\,dt= \int_0^\infty t\cdot\lambda\alpha t^{\alpha-1} e^{-\lambda t^{\alpha}} dt$$ and the substitution $u=\lambda t^{\alpha}\implies t=(u/\lambda)^{1/\alpha}$ yields $$\Bbb E[X]=\int_0^\infty\frac{u^{1/\alpha}}{\lambda^{1/\alpha}}e^{-u}\,du=\frac1{\lambda^{1/\alpha}}\int_0^\infty u^{1+1/\alpha-1}e^{-u}\,du=\frac{\Gamma(1+1/\alpha)}{\lambda^{1/\alpha}}.$$

Answer 2

Just put $x=\lambda u$. You get $\lambda ^{-1 /\alpha}\int_0^{\infty} x^{(1+\frac 1 {\alpha}) -1} e^{-x}dx$ which is nothing but $\lambda ^{-1/ \alpha} \Gamma (1+\frac 1 {\alpha})$.