Will Fithian


August 29, 2023

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Sufficiency is a central concept in statistics that allows us to focus on the essential aspects of the data set while ignoring details that are irrelevant to the inference problem. If \(X\sim P_\theta\) represents the entire data set, drawn from a model \(\cP = \{P_\theta:\; \theta \in \Theta\}\), then this lecture will concern the idea of a sufficient statistic \(T(X)\) that carries all of the information in the data that can help us learn about \(\theta\).

A statistic \(T(X)\) is any random variable which is a function of the data \(X\), and which does not depend on the unknown parameter \(\theta\). We say the statistic \(T(X)\) is sufficient for the model \(\cP\) if \(P_\theta(X \mid T)\) does not depend on \(\theta\). This lecture will be devoted to interpreting this definition and giving examples.

Example (Independent Bernoulli sequence): We introduced the binomial example from Lecture 2 by telling a story about an investigator who flips a biased coin \(n\) times and records the total number of heads, which has a binomial distribution. All of the estimators we considered were functions only of the (binomially-distributed) count of heads.

But if the investigator had actually performed this experiment, they would have observed more than just the total number of heads: they would have observed the entire sequence of \(n\) heads and tails. If we let \(X_i\) denote a binary indicator of whether the \(i\)th throw is heads, for \(i=1,\ldots,n\), then we have assumed that these indicators are i.i.d. Bernoulli random variables:

\[ X_1,\ldots,X_n \simiid \text{Bern}(\theta). \]

Let \(T(X) = \sum_i X_i \sim \text{Binom}(n,\theta)\) denote the summary statistic that we previously used to represent the entire data set. It is undeniable that we have lost some information by only recording \(T(X)\) instead of the entire sequence \(X = (X_1,\ldots,X_n)\). As a result, we might wonder whether we could have improved the estimator by considering all functions of \(X\), not just functions of \(T(X)\).

The answer is that, no, we did not really lose anything by summarizing the data by \(T(X)\) because \(T(X)\) is sufficient. The joint pmf of the data set \(X \in \{0,1\}^n\) (i.e., the density wrt the counting measure on \(\{0,1\}^n\)) is

\[ p_\theta(x) = \prod_{i=1}^n \theta^{x_i}(1-\theta)^{1-x_i} = \theta^{\sum_i x_i}(1-\theta)^{n-\sum_i x_i}. \]

Note that this pmf depends only on \(T(x)\): it assigns probability \(\theta^t (1-\theta)^{n-t}\) to every sequence with \(T(X)=t\) total heads. As a result, the conditional distribution given \(T(X)=t\) should be uniform on all of the \(\binom{n}{t}\) sequences with \(t\) heads. We can confirm this by calculating the conditional pmf directly:

\[ \begin{aligned} \PP_\theta(X = x \mid T(X) = t) &= \frac{\PP_\theta(X=x, \sum_i X_i = t)}{\PP_\theta(T(X) = t)} \\[7pt] &= \frac{\theta^t (1-\theta)^{n-t}1\{\sum_i x_i = t\}}{\theta^t(1-\theta)^{n-t}\binom{n}{t}}\\[5pt] &= \binom{n}{t}^{-1}1\{T(x) = t\}. \end{aligned} \]

Since the conditional distribution does not depend on \(\theta\), \(T(X)\) is sufficient for the model \(\cP\).

As we will see next, we didn’t really need to go to the trouble of calculating the conditional distribution. Once we noticed that the density depends on \(x\) only through \(T(x)\), we could have concluded that \(T(X)\) was sufficient.

Factorization theorem

The easiest way to verify that a statistic is sufficient is to show that the density \(p_\theta\) factorizes into a part that involves only \(\theta\) and \(T(x)\), and a part that involves only \(h(x)\).

Factorization Theorem: Let \(\cP\) be a model having densities \(p_\theta(x)\) with respect to a common dominating measure \(\mu\). Then \(T(X)\) is sufficient for \(\cP\) if and only if there exist non-negative functions \(g_\theta\) and \(h\) for which

\[ p_\theta(x) = g_\theta(T(x)) h(x), \]

for almost every \(x\) under \(\mu\).

The “almost every \(x\)” qualification means that

\[ \mu\left(\{x:\; p_\theta(x) \neq g_\theta(T(x))h(x)\}\right) = 0. \]

It is needed to avoid counterexamples with continuous distributions where we could arbitrarily change the value of \(p_{\theta_0}(x_0)\) for a single \(\theta_0\) and \(x_0\) without actually changing any of the distributions.

Proof: The proof is easiest in the discrete case, so we don’t have to deal with conditioning on measure-zero events.

First, assume that there exists a factorization \(p_\theta(x) = g_\theta(T(x)) h(x)\). Then we

\[ \begin{aligned} \PP_\theta(X = x \mid T(X) = t) &= \frac{p_\theta(x)1\{T(x) = t\}}{\displaystyle\sum_{z:\;T(z) = t} p_\theta(z)}\\[7pt] &= \frac{g_\theta(t) h(x) 1\{T(x) = t\}}{g_\theta(t)\displaystyle\sum_{z:\;T(z) = t} h(z)}\\[7pt] &= \frac{h(x) 1\{T(x) = t\}}{\displaystyle\sum_{z:\;T(z) = t} h(z)}, \end{aligned} \]

which we see does not depend on \(\theta\).

Next consider the opposite direction. If \(T(X)\) is sufficient, then we can construct a factorization by writing

. First, define

\[ g_\theta(t) = \PP_\theta(T(X)=t) = \sum_{x:\;T(x) = t} p_\theta(x). \]

Then, let

\[ h(x) = \PP(X = x \mid T(X) = T(x)), \]

which does not depend on \(\theta\) by sufficiency. Then we have

\[ \PP_\theta(X = x) = \PP_\theta(T(X) = T(x)) \;\cdot\;\PP_\theta(X = x \mid T(X) = T(x)) = g_\theta(T(x)) h(x) \]

Sufficient statistics in exponential families