Is Weibull distribution a exponential family? I'm wondering is Weibull distribution a exponential family?
 A: The answer is NO. The logarithm of the weibull density is given by
$$
   \log f(x) = \log(k/\lambda) + (k-1)\log(x/\lambda)  - (x/\lambda)^k
$$
where $x>0$, $k>0$ (shape parameter), $\lambda>0$ (scale parameter).  The problem is the last term. IF $k$ were known (prespecified), this would be a one-parameter exponential family. So one could say (maybe) that the two-parameter weibull family is an (infinite) union of one-parameter exponential families, but if that is of any help I do not know. 
To see this, the general form of the multi-parameter exponential family is 
$$
   f(x;\theta) = h(x) \exp( \sum_1^s \eta_i(\theta) T_i(x) - A(\theta)
$$
we can se that the parameter function $\eta_i(\theta)$ and the data function $T_i(x)$ always combines multiplicatively, which do not happen for the last term in the weibull formula above. 
A: The two parameter Weibull distribution (with $k$ and $\lambda$ as described on wikipedia) is not an exponential family.  However, if you fix $k$ to anything, then it is an exponential family having sufficient statistics $x^k$ on support $[0, \infty)$.
