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I'm wondering is Weibull distribution a exponential family?

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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.

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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 $[0, \infty)$.

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  • $\begingroup$ Here is a similar question: stats.stackexchange.com/questions/87501/… $\endgroup$ May 4, 2017 at 8:12
  • $\begingroup$ OK, planetmath agree with you: planetmath.org/exponentialfamily but they don't argue ... $\endgroup$ May 4, 2017 at 8:24
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    $\begingroup$ @kjetilbhalvorsen I updated my answer. I only remember having read that it was an exponential family, but I never tried to put it into canonical form. You're right: unless you fix $k$, it's not an exponential family. $\endgroup$
    – Neil G
    May 4, 2017 at 8:29

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