# Show that Weibull distribution belongs to a one dimensional exponential family

It is given that

$f_\eta(y) = h(y)exp(\eta T(y)-A^*(\eta))$

$P_Y(y)= \frac{k}{\lambda} (\frac{y}{\lambda})^{k-1}exp(-(\frac{y}{\lambda})^k)$

What i did was by arranging $P_Y(y)$ to get $\frac{k}{\lambda ^k} y^{k-1}exp(-(\frac{y}{\lambda})^k)$

so that $lnP_Y(y) = ln(\frac{k}{\lambda^k})+(k-1)ln(y)-(\frac{y}{\lambda})^k$

Therefore

$T(y) = y^{k-1}$

$\eta = -(\frac{1}{\lambda})^k$

$h(y)=\frac{k}{\lambda^k}$

$A^* = -ln(\frac{k}{\lambda^k})= -ln(-k\eta)$

Am i correct?

and how do you work out the mean and variance of Y in terms of $A^*$？

## 1 Answer

For part one of your question:

No. In case you want to use this particular representation of exponential family [over a more relevant form $h(x)g(\eta) \exp(\eta T(x))$]

\begin{align*} f_\eta(y) &= h(y)\exp(\eta T(y)-A^*(\eta))\\ \ln f_{\eta}(y) &= \ln h(y) + \eta T(y) - A^*(\eta)\\ P_Y(y) &= \frac{k}{\lambda} (\frac{y}{\lambda})^{k-1}\exp(-(\frac{y}{\lambda})^k)\\ \ln P_Y(y) &= \ln y^{k-1} -(\frac{y}{\lambda})^k - \ln k - \ln \lambda^k\\ \ln P_Y(y) &= \ln ky^{k-1} + \eta (-y^k) - \ln \eta \text{ where } \eta = \lambda^k\\ \end{align*}