# Is a maximum likelihood estimator in an exponential family always sufficient?

An exponential family (under natural parameterization) is such that $$p(X|\eta)=h(X)\exp\{\eta^\top T(X)-A(\eta)\}$$, where $$X$$ is the data, $$\eta$$ is the natural parameter, and $$h,T,A$$ are some functions (that are interrelated).

The statistic $$T(X)$$ is sufficient, and the maximum likelihood estimator is $$\hat\eta(X)$$ is such that $$\nabla A(\hat\eta(X))=T(X)$$.

It therefore seems to me that the density can be written as $$p(X|\eta)=h(X)\exp\{\eta^\top\nabla A(\hat\eta(X))-A(\eta)\}=h(X)\cdot f(\hat\eta(X),\eta)$$, which entails that $$\hat\eta(X)$$ is a sufficient statistic.

Am I mistaken or is it sound to conclude that a maximum likelihood estimator in an exponential family is always sufficient?

• Thank you. Which part of the "proof" fails if the parameterization is not minimal? Aug 10, 2020 at 8:43
• Thanks! Is it possible to point me to a textbook or journal reference? Aug 10, 2020 at 9:24
• I deem your proof is enough: Provided$$\nabla A(\hat\eta(t))=t\tag{1}$$has one and only one solution$$\hat\eta(t)$$for almost every realisation $t$ of $T(X)$, the factorisation theorem applies. Cases where this condition fails may be for discrete exponential families, on the boundary of the support of $T(X)$ and curved exponential families when the constraints on $\eta$ may be incompatible with (1). (But it is debatable this is a "natural" exponential family.) Aug 10, 2020 at 9:39
• Michael Jordan mentions the sufficiency of the mean parameter MLE in his notes.. Aug 10, 2020 at 9:40

You can also turn it around. If the MLE is a (one-to-one) function of the sufficient statistic $$\sum T(X_i)$$, then it is itself a sufficient statistic as well.