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?

  • $\begingroup$ Thank you. Which part of the "proof" fails if the parameterization is not minimal? $\endgroup$
    – erezmb
    Commented Aug 10, 2020 at 8:43
  • $\begingroup$ Thanks! Is it possible to point me to a textbook or journal reference? $\endgroup$
    – erezmb
    Commented Aug 10, 2020 at 9:24
  • 3
    $\begingroup$ 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.) $\endgroup$
    – Xi'an
    Commented Aug 10, 2020 at 9:39
  • $\begingroup$ Michael Jordan mentions the sufficiency of the mean parameter MLE in his notes.. $\endgroup$
    – Xi'an
    Commented Aug 10, 2020 at 9:40

1 Answer 1


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.


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