# Exponential Family: Observed vs. Expected Sufficient Statistics

My question arises from reading reading Minka's "Estimating a Dirichlet Distribution", which states the following without proof in the context of deriving a maximum-likelihood estimator for a Dirichlet distribution based on observations of random vectors:

As always with the exponential family, when the gradient is zero, the expected sufficient statistics are equal to the observed sufficient statistics.

I haven't seen maximum likelihood estimation in the exponential family presented in this way, nor have I found any suitable explanations in my search. Can someone offer insight into the relationship between observed and expected sufficient statistics, and perhaps help to understand maximum likelihood estimation as minimizing their difference?

This is a usual assertion about the exponential family, but in my opinion, most of the times it is stated in a way that may confuse the less experienced reader. Because, taken at face value, it could be interpreted as saying "if our random variable follows a distribution in the exponential family, then if we take a sample and insert it into the sufficient statistic, we will obtain the true expected value of the statistic". If only it were so... More over it does not take into account the size of the sample, which may cause further confusion.

The exponential density function is

$$f_X(x) = h(x)e^{\eta(\theta) T(x)}e^{-A(\theta)} \tag{1}$$

where $T(x)$ is the sufficient statistic.

Since this is a density, it has to integrate to unity, so ($S_x$ is the support of $X$)

$$\int_{S_x} h(x)e^{\eta(\theta) T(x)}e^{-A(\theta)}dx =1 \tag{2}$$

Eq. $(2)$ holds for all $\theta$ so we can differentiate both sides with respect to it:

$$\frac {\partial}{\partial \theta} \int_{S_x} h(x)e^{\eta(\theta) T(x)}e^{-A(\theta)}dx =\frac {\partial (1)}{\partial \theta} =0 \tag{3}$$

Interchanging the order of differentiation and integration, we obtain

$$\int_{S_x} \frac {\partial}{\partial \theta} \left(h(x)e^{\eta(\theta) T(x)}e^{-A(\theta)}\right)dx =0 \tag{4}$$

Carrying out the differentiation we have

$$\frac {\partial}{\partial \theta} \left(h(x)e^{\eta(\theta) T(x)}e^{-A(\theta)}\right) = f_X(x)\big[T(x)\eta'(\theta) - A'(\theta)\big] \tag{5}$$

Inserting $(5)$ into $(4)$ we get

$$\int_{S_x} f_X(x)\big[T(x)\eta'(\theta) - A'(\theta)\big]dx =0$$

$$\Rightarrow \eta'(\theta)E[T(X)] - A'(\theta) = 0 \Rightarrow E[T(X)] = \frac {A'(\theta)}{\eta'(\theta)} \tag{6}$$

Now we ask: the left-hand-side of $(6)$ is a real number. So, the right-hand-side must also be a real number, and not a function. Therefore it must be evaluated at a specific $\theta$, and it should be the "true" $\theta$, otherwise in the left-hand-side we would not have the true expected value of $T(X)$. To emphasize this we denote the true value by $\theta_0$, and we re-write $(6)$ as

$$E_{\theta_0}[T(X)] = \frac {A'(\theta)}{\eta'(\theta)}\Big |_{\theta =\theta_0} \tag{6a}$$

We turn now to maximum likelihood estimation. The log-likelihood for a sample of size $n$ is

$$L(\theta \mid \mathbf x) = \sum_{i=1}^n\ln h(x_i) +\eta(\theta)\sum_{i=1}^nT(x_i) -nA(\theta)$$

Setting its derivative with respect to $\theta$ equal to $0$ we obtain the MLE

$$\hat \theta(x) : \frac 1n\sum_{i=1}^nT(x_i) = \frac {A'(\theta)}{\eta'(\theta)}\Big |_{\theta =\hat \theta(x)} \tag {7}$$

Compare $(7)$ with $(6a)$. The right-hand sides are not equal, since we cannot argue that the MLE estimator hit upon the true value. So neither are the left hand-sides. But remember that eq. $2$ holds for all $\theta$ and so for $\hat \theta$ also. So the steps in eq. $3,4,5,6$ can be taken with respect to $\hat \theta$ and so we can write eq. $6a$ for $\hat \theta$:

$$E_{\hat\theta(x)}[T(X)] = \frac {A'(\theta)}{\eta'(\theta)}\Big |_{\theta =\hat\theta(x)} \tag{6b}$$

which, combined with $(7)$, leads us to the valid relation

$$E_{\hat\theta(x)}[T(X)] = \frac 1n\sum_{i=1}^nT(x_i)$$

which is what the assertion under examination really says: the expected value of the sufficient statistic under the MLE for the unknown parameters (in other words, the value of the first raw moment of the distribution that we will obtain if we use $\hat \theta(x)$ in place of $\theta$), equals (and it is not just approximated by) the average of the sufficient statistic as calculated from the sample $\mathbf x$.

Moreover, only if the sample size is $n=1$ then we could accurately say, "the expected value of the sufficient statistic under the MLE equals the sufficient statistic".

• Could you further elaborate why the transition from 6a to 6b is valid please? Sep 16, 2016 at 17:03
• @Theoden In between eq. $(2)$ and $(3)$ I write "eq. $(2)$ holds for all $\theta$" - and therefore for $\hat \theta$ also. So all the steps in eq. $3,4,5,6$ can be taken with respect to $\hat \theta$. I repeated this remark in the text for clarity. Sep 16, 2016 at 18:58
• @AlecosPapadopoulos your proof below seems to suggest that what you say at the start - "if our random variable follows a distribution in the exponential family, then if we take a sample and insert it into the sufficient statistic, we will obtain the true expected value of the statistic" is true. I mean I can just always do that for (2), replacing it with observed sufficient stat and get the result. What am I missing here? I don't quite get it. Dec 14, 2016 at 4:15
• @user136266 The true expected value of the statistic is $6a$, and in order to be calculated, one needs to know the, by design unknown, parameter $\theta$. So what we can actually calculate is $6b$ which is the expected value of the statistic under the assumption that our point estimate has hit the true value. Dec 14, 2016 at 16:05
• Could you explain why we can interchange the order of differentiation and integration in eq. (3) please? Aug 2, 2017 at 7:47
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