Question

Find the CDF of the Weibull distribution, which has pdf: \[ f(x) =\left(\frac{\beta}{\alpha} \right) \left(\frac{x}{\alpha} \right)^{\beta-1}e^{-\left(\frac{X}{\alpha} \right)^{\beta}} \] with \(\alpha,\beta>0,x\ge 0\)

The answer is supposed to be
\[F(x)=1-e^{\left(\frac{X}{\alpha} \right)^{\beta}}\]

Answer 1

(Just as a note, I wrote "x" as "X" in some places just to make it look big and visible)
\[F(x) = \int_{-\infty}^{X} \left(\frac{\beta}{\alpha} \right) \left(\frac{x}{\alpha} \right)^{\beta-1}e^{-\left(\frac{X}{\alpha} \right)^{\beta}}dx \]
\[F(x) = \frac{\beta}{\alpha^{\beta}}\int_{-\infty}^{X} x^{\beta-1}e^{-\left(\frac{X}{\alpha} \right)^{\beta}}dx \]
So, now let \(u=x^{\beta}\), then \(\frac{du}{\beta-1}=x^{\beta-1}dx\)
\[F(x) = \frac{\beta}{\alpha^{\beta}(\beta-1)}\int_{-\infty}^{X^{\beta}}e^{-u\left(\frac{1}{\alpha} \right)^{\beta}}du \]
\[F(x) = \frac{\beta(-\alpha^{\beta})}{\alpha^{\beta}(\beta-1)}e^{-u\left(\frac{1}{\alpha} \right)^{\beta}}\big |_{u=-\infty}^{u=X^{\beta}}\]
\[F(x) = \frac{-\beta}{\beta-1}(e^{-X^{\beta}\left(\frac{1}{\alpha} \right)^{\beta}}-\lim_{u \rightarrow -\infty}e^{-u\left(\frac{1}{\alpha} \right)^{\beta}})\]

I get stuck at the limit part because won't it diverge? I'm assuming I did something wrong but I am not sure where :S. Any help would be appreciated!!

Answer 2

Well one quick thing is that du is incorrect but that doesn't deal with the divergence problem it seems

Answer 3

oh right it should be \(du/\beta\)

Answer 4

Does x>=0 mean we can start integrating at x=0 intstead of negative infinity? That seems to do the trick

Answer 5

omg yes it does!! wow I can't believe I missed that. Wow thank you so much.

Answer 6

Sure thing

Answer 7

It totally worked now :D