# 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.

• This sounds like a routine textbook type question. Is this for some subject? – Glen_b -Reinstate Monica Feb 22 '14 at 23:03
• While I wait for a response on that, a hint: (0) forget the general functional form of the density for the the exponential family for the moment, but write down a list of the well known distributions in the family; (1) write down the cdf for a Weibull - the correct power should be immediately obvious from that; (2) Then the cdf for $z$ can be written down immediately, and confirmed to be a member of the exponential family – Glen_b -Reinstate Monica Feb 22 '14 at 23:08
• thanks - i've just tried this and im still confused though – user40124 Feb 23 '14 at 18:58
• (1) Could you address the question in the first comment please? How does this question arise? (2) What is the cdf for a Weibull, and what probability statement does it correspond to? If you won't show any working, it's hard to explain to you where your problems lie? – Glen_b -Reinstate Monica Feb 23 '14 at 21:19

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

$f(x;\lambda,k) =\frac{k}{\lambda}\left(\frac{z^{1/k}}{\lambda}\right)^{k-1}e^{-(z^{1/k}/\lambda)^{k}}\cdot \frac{1}{k}\,z^{\frac{1}{k}-1}=...$