Consider the Weibull distribution with parameter $\theta$, fixed $\lambda$ and p.m.f : $$ f_Y(y)=\frac{\lambda y^{\lambda -1}}{\theta^{\lambda}}exp(-(\frac{y}{\theta})^{\lambda}) $$$$ f_Y(y)=\frac{\lambda y^{\lambda -1}}{\theta^{\lambda}}\exp(-(\frac{y}{\theta})^{\lambda}) $$ It can be shown that this distribution is from exponential family because: $$ f_Y(y)={\lambda y^{\lambda -1}}exp(-y^{\lambda}\theta^{-\lambda}-\lambda\log\theta) $$$$ f_Y(y)={\lambda y^{\lambda -1}}\exp(-y^{\lambda}\theta^{-\lambda}-\lambda\log\theta) $$ 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.Ng's notes here: http://cs229.stanford.edu/notes/cs229-notes1.pdf , It can be seen naturally that : $$ \theta=exp{({\frac{\log\eta}{-\lambda}})} $$$$ \theta=\exp{({\frac{\log\eta}{-\lambda}})} $$ or $$ \theta=exp{({\frac{\log(\beta X)}{-\lambda}})} $$$$ \theta=\exp{({\frac{\log(\beta X)}{-\lambda}})} $$
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.
