Show that Weibull distribution can be transformed to exponential family How do I show the Weibull distribution $f:(y; \lambda, \rho)$ can be transformed to the exponential family using the transformation $z=y^\lambda$?  
I know the form I need to express it in is 
$$\exp\lbrace(y.\theta - b(\theta)/a(\phi))+c(y,\phi)\rbrace$$ 
but am unsure how to get there.
 A: While this is far enough on in time that it might be reasonable to show the whole thing, I will leave some things still to do. Note that this is not the same parameterization in the question (again, this leaves at least something for the reader to do).
Really, there's nothing much to do here -- it's just the transformation it says to do, in any of the obvious ways to do it. (In fact if you're used to this kind of stuff you can do it by inspection, almost along this lines in "i)" - recognize that $(X/\lambda)^k$ is in the exp term, and its derivative is out the front... meaning that if you transform by the function inside the exp-term you get a standard exponential density -- then recognize that the $\lambda$ is a scaling factor that won't change anything but the exponential parameter, so the result of transforming by $X^k$ must be an exponential density)
There are a couple of ways to look at it.
i) I'll outline a closely related transformation that leads through the steps, but I'll leave the exact transformation still to be done.
$f(x;\lambda,k) =\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}e^{-(x/\lambda)^{k}} \: x\geq0$
(and 0 elsewhere, naturally)
So consider $Y=(X/\lambda)^k$, for which $dy = \frac{k}{\lambda} (x/\lambda)^{k-1} dx$
Hence $f(y) = e^{-y}\,,\quad y>0$, which is exponential family.
In similar fashion, one could tackle $Z=X^k$ as in the question
ii) First principles
$P(Z\leq z) = P(X^k\leq z)= P(X\leq z^{1/k})= F_X(z^{1/k})$
Hence $f_Z(z) = \frac{d}{dz} F_X(z^{1/k}) = f_X(z^{1/k}) . \frac{1}{k}\,z^{\frac{1}{k}-1}$
So
$f(x;\lambda,k) =\frac{k}{\lambda}\left(\frac{z^{1/k}}{\lambda}\right)^{k-1}e^{-(z^{1/k}/\lambda)^{k}}\cdot \frac{1}{k}\,z^{\frac{1}{k}-1}=...$ 
