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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.

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

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