# Generalized Linear Model for Weibull distribution

Consider the Weibull distribution with parameter $\theta$, fixed $\lambda$ and p.m.f : $$f_Y(y)=\frac{\lambda y^{\lambda -1}}{\theta^{\lambda}}\exp(-(\frac{y}{\theta})^{\lambda})$$ It can be shown that this distribution is from exponential family because: $$f_Y(y)={\lambda y^{\lambda -1}}\exp(-y^{\lambda}\theta^{-\lambda}-\lambda\log\theta)$$ where $$b(y)={\lambda y^{\lambda -1}}$$ $$\eta=\theta^{-\lambda}$$ $$T(y)=-y^{\lambda}$$ $$a(\eta)=\lambda\log\theta$$ As I saw in Andrew Ng's notes here: http://cs229.stanford.edu/notes/cs229-notes1.pdf , It can be seen naturally that : $$\theta=\exp{({\frac{\log\eta}{-\lambda}})}$$ or $$\theta=\exp{({\frac{\log(\beta X)}{-\lambda}})}$$

where $\beta$ is our line parameter we want to estimate.My problem is actually, how to calculate $E(T(y)|X,\beta)$ as response value the algorithm should return for a given $x$ because It seems algebraically hard for me to calculate it.

I should mention that in the examples that had been provided in Ng's notes, $T(y)$ was equal to $y$ and the calculation of $E(T(y)|X,\beta)$ was fairly easy and he gave tidings us that most of the time $T(y)=y$ is established but unfortunately in this distribution It didn't happen.

• Is your difficulty in computing $E(y^\lambda)$? – Glen_b Jul 10 '15 at 3:26
• yes.I don't know if my method is actually right. and If this approach is ok, then how should I calculate it? – CoderInNetwork Jul 10 '15 at 8:36

If the hurdle is $E(y^\lambda)$, a simple substitution serves:

$f_Y(y)=\frac{\lambda y^{\lambda -1}}{\theta^{\lambda}}\exp(-(\frac{y}{\theta})^{\lambda})$

$E(y^\lambda)=\int_0^\infty y^\lambda \frac{\lambda y^{\lambda -1}}{\theta^{\lambda}}\exp(-(\frac{y}{\theta})^{\lambda}) dy$

let $u=(\frac{y}{\theta})^{\lambda}$, $du=\frac{\lambda}{\theta}(\frac{y}{\theta})^{\lambda-1}dy$

$E(y^\lambda)=\theta^\lambda \int_0^\infty u\, \exp(-u)\,du =\theta^\lambda$

Alternatively, you could look at the Wikipedia page for the Weibull distribution, which gives (in your parameterization) $E(y^r)=\theta^r\,\Gamma(1+\frac{r}{\lambda})$, and substitute $r=\lambda$.

A little simple manipulation of information from your question will let you write $\theta^\lambda$ in terms of $X\beta$.