Sufficiency

This is one of the hardest topics in this course.

We have a single sample $Y \sim N(0, \sigma^2)$ where $\sigma^2$ is unknown. Suppose the sign of $Y$ is lost, so only $|Y|$ is retained. Does this affect the guess of $\sigma^2$?

Not really, because our distribution is centered at $0$, so the sign of the observations doesn’t really affect how far it’s away from the mean.

Definition

Suppose $Y_i \overset{i.i.d.}{\sim} Bern(p)$, $p \in [0, 1]$. If only $T = Y_1 + \cdots + Y_n$ is known, does it affect the estimation of $p$?

The answer is again no, because the estimator we use is

$$ \bar{Y}_n = \frac{1}{n}\sum_{i=1}^n Y_i = \frac{1}{n}T $$

When we say $T$ is known, we’re actually referring to the concept of conditioning in probability theory. Consider the conditional distribution

$$ \begin{gather*} P(Y_1 = y_1, Y_2 = y_2, \cdots, Y_n = y_n \mid T = t) = \frac{P(A \cap B)}{P(B)} \\
= \frac{P(Y_1 = y_1, \cdots, Y_n = y_n, T = t)}{P(T = t)} \end{gather*} $$

where each $y_i = 0$ or $1$. If we know the value of $t$, we actually know how many $1$'s there are in $y_1, \cdots, y_n$.

$$ \begin{aligned} &\cdots = \frac{P(Y_1 = y_1, \cdots, Y_n = y_n) I\left\{ \sum_{i=1}^n y_i = t \right\}}{P(T = t)} \\
&= \frac{P(Y_1 = y_1)^nI\left\{ \sum_{i=1}^n y_i = t \right\}}{P(T = t)} \\
&= \frac{p^{\sum_{i=1}^n Y_i}(1-p)^{n-\sum_{i=1}^n Y_i}I\left\{ \sum_{i=1}^n y_i = t \right\}}{P(T = t)} \\
&= \frac{p^{\sum_{i=1}^n Y_i}(1-p)^{n-\sum_{i=1}^n Y_i}I\left\{ \sum_{i=1}^n y_i = t \right\}}{\binom{n}{t}p^t(1-p)^{n-t}} \\
&= \begin{cases} \frac{p^t(1-p)^{n-t}}{\binom{n}{t}p^t(1-p)^{n-t}} = \frac{1}{\binom{n}{t}}, & \sum_{i=1}^n y_i = t, \\
0, & \text{otherwise} \end{cases} \\
&= \frac{1}{\binom{n}{t}} I\left\{\sum_{i=1}^n y_i = t\right\} \end{aligned} $$

Remark: If $T$ is given, then the joint distribution of $(Y_1, \cdots, Y_n)$ will no longer depend on $p$.

Def: Suppose $Y_1, \cdots, Y_n$ is a sample from a population distribution with unknown parameter $\theta$. A statistic $T$ is said to be sufficient for $\theta$ if the conditional distribution of $(Y_1, \cdots, Y_n)$ given $T$ does not depend on $\theta$. The interpretation is that knowing $T$ is sufficient for inferring $\theta$. After conditioning on $T$, the sample no longer reflect information about $\theta$.

Suppose statistic $T$ is sufficient for $\theta$. Let $f$ be an one-to-one known function, then $f(T)$ is still sufficient for $\theta$. For example, in the example above we know that $T = \sum_{i=1}^n Y_i$ is sufficient for $p$. With $f(t) = t/n$, we can show that $\bar{Y}_n$ is also sufficient for $p$.

Factorization theorem

The statistic $T = f(Y_1, \cdots, Y_n)$ is sufficient for $\theta$ if and only if

$$ L(\theta; Y_1, \cdots, Y_n) = g(T, \theta) \times h(Y_1, \cdots, Y_n) $$

where $g(T, \theta)$ involves both $T$ and $\theta$, whereas $h(\cdot)$ has no $\theta$ involved.

Remark: this factorization is not unique! Often $h$ can be chosen as a constant $1$.

Take the $Y_i \overset{i.i.d.}{\sim} Bern(p)$ example again. We have $T = \sum_{i=1}^n Y_i$.

$$ \begin{aligned} L(p; Y_1, \cdots, Y_n) &= p^{\sum_{i=1}^n Y_i}(1-p)^{n-\sum_{i=1}^n Y_i} \\
&= \underbrace{p^T(1-p)^{n-T}}_{g} \\
&= g \times 1 \end{aligned} $$

So by the factorization theorem, $T$ is sufficient for $p$.

Normal distribution example

$Y \sim N(0, \sigma^2)$ and our sample size $n = 1$. We want to find if $|Y|$ is sufficient.

$$ \begin{aligned} L(\sigma^2; Y) &= \frac{1}{\sigma\sqrt{2\pi}}\exp\left( -\frac{Y^2}{2\sigma^2} \right) \\
&= \frac{1}{\sigma\sqrt{2\pi}}\exp\left( -\frac{|Y|^2}{2\sigma^2} \right) \\
&= g(|Y|, \sigma^2) \times 1 \end{aligned} $$

We can also check the sufficiency by directly using the definition. To find the distribution of $Y$ conditioning on $|Y|$, see that $(Y \mid |Y|)$ can only take two values.

$$ P(Y = y \mid |Y| = t) = \begin{cases} \frac{1}{2}, & y = t \\
\frac{1}{2}, & y = -t \end{cases} $$

There’s no $\sigma^2$ in this distribution.

Going back to the factorization theorem, when $Y_i$ is discrete, let $p(t)$ denote the marginal PMF of $T$. Assuming the factorization holds,

$$ \begin{aligned} p(y_1, \cdots, y_n \mid t) &= P(Y_1 = y_1, \cdots, Y_n = y_n \mid T = t) \\
&= \frac{P(Y_1 = y_1, \cdots, Y_n = y_n, T = t)}{p(t)} \\
&= \frac{g(t, \theta) h(y_1, \cdots, y_n)I{ f(y_1, \cdots, y_n) = t }}{p(t)} \\
&= g(t, \theta) \cdot h^*(y_1, \cdots, y_n; t) \end{aligned} $$

$$ \begin{gather*} \sum_{y_1, \cdots, y_n} p(y_1, \cdots, y_n \mid t) = 1 \Leftrightarrow \sum_{y_1, \cdots, y_n} g(t, \theta) h^*(y_1, \cdots, y_n; t) = 1 \\
g(t, \theta)\sum_{y_1, \cdots, y_n} h^*(y_1, \cdots, y_n; t) = 1 \\
g(t, \theta) = \frac{1}{\sum_{y_1, \cdots, y_n} h^*(y_1, \cdots, y_n; t)} \end{gather*} $$

The $g(t, \theta)$ function does not depend on $\theta$, therefore $P(Y_1 = y_1, \cdots, Y_n = y_n \mid T = t)$ does not depend on $\theta$.

Chi square distribution example

Suppose $Y_i$ are i.i.d. samples from PDF

$$ f(y; \nu) = \frac{1}{2^{\frac{\nu}{2}}\Gamma(\frac{\nu}{2})}y^{\frac{\nu}{2} - 1}e^{-\frac{y}{2}} I\{y > 0\} $$

which is the $\chi^2$ distribution with $\nu$ degrees of freedom, and $\nu = 1, 2, 3, \cdots$ is unknown. The likelihood function is

$$ L(\nu; Y_1, \cdots, Y_n) = \underbrace{\frac{1}{2^{\frac{n\nu}{2}}\Gamma(\frac{\nu}{2})^n}(Y_1\cdots Y_n)^{\frac{\nu}{2} - 1}}_{g} \underbrace{e^{-\frac{Y_1 + \cdots + Y_n}{2}}}_{h} $$

Thus $T = Y_1 Y_2 \cdots Y_n$ is the sufficient statistic for $\nu$.

Example with two parameters

In the factorization, both $T$ and $\theta$ can be vectors, which means the theorem can be applied when there are multiple parameters.

$Y_1, \cdots, Y_n \overset{i.i.d.}{\sim} Unif(a, b)$ where $a < b$. In this case $\vec\theta = (a, b)$. Find a sufficient statistic for $\theta$.

The likelihood function is given by

$$ \begin{aligned} L(a, b; Y_1, \cdots, Y_n) &= \left(\frac{1}{b-a}\right)^n I\left\{ a \leq \min(Y_1, \cdots, Y_n), b \geq \max(Y_1, \cdots, Y_n) \right\} \\
&= \left(\frac{1}{b-a}\right)^n I\left\{ a \leq Y_{(1)}, b \geq Y_{(n)} \right\} \end{aligned} $$

See example in previous page for details. The entire RHS would be the $g(\cdot)$ part as they all contain $a$ and $b$. So $T = (Y_{(1)}, Y_{(n)})$ is sufficient for $(a, b)$.

Proposition

The MLE is always a function of (or based on) a sufficient statistic, because the MLE automatically explores full information in the sample.

By the factorization theorem, if $T$ is sufficient, the likelihood function can be written as

$$ L(\theta) = g(T, \theta)h(Y_1, \cdots, Y_n) $$

In the MLE, maximizing $L(\theta)$ is the same as

$$ \arg\max_{\theta \in \Theta} g(T, \theta) $$

which depends on $T$ only.

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