Sufficient statistic

Mathematical definition

A statistic t = T(X) is sufficient for underlying parameter θ precisely if the conditional probability distribution of the data X, given the statistic t = T(X), does not depend on the parameter θ.

Alternatively, one can say the statistic T(X) is sufficient for θ if, for all prior distributions on θ, the mutual information between θ and T(X) equals the mutual information between θ and X. In other words, the data processing inequality becomes an equality:

${\displaystyle I(\theta ;T(X))=I(\theta ;X)}$

Example

As an example, the sample mean is sufficient for the (unknown) mean μ of a normal distribution with known variance. Once the sample mean is known, no further information about μ can be obtained from the sample itself. On the other hand, for an arbitrary distribution the median is not sufficient for the mean: even if the median of the sample is known, knowing the sample itself would provide further information about the population mean. For example, if the observations that are less than the median are only slightly less, but observations exceeding the median exceed it by a large amount, then this would have a bearing on one's inference about the population mean.

Fisher–Neyman factorization theorem

Fisher's factorization theorem or factorization criterion provides a convenient characterization of a sufficient statistic. If the probability density function is ƒ(x;θ), where θ is a parameter, then T is sufficient for θ if and only if nonnegative functions g and h can be found such that

${\displaystyle f(x;\theta )=h(x)g(\theta ,T(x)),}$

i.e., the density ƒ can be factored into a product such that one factor, h, does not depend on θ and the other factor, which does depend on θ, depends on x only through T(x). A general proof of this was given by Halmos and Savage and the theorem is sometimes referred to as the Halmos–Savage factorization theorem. The proofs below handle special cases, but an alternative general proof along the same lines can be given. In many simple cases the probability density function is fully specified by ${\displaystyle \theta }$ and ${\displaystyle T(x)}$, and ${\displaystyle h(x)=1}$ (see Examples).

It is easy to see that if F(t) is a one-to-one function and T is a sufficient statistic, then F(T) is a sufficient statistic. In particular we can multiply a sufficient statistic by a nonzero constant and get another sufficient statistic.

Likelihood principle interpretation

An implication of the theorem is that when using likelihood-based inference, two sets of data yielding the same value for the sufficient statistic T(X) will always yield the same inferences about θ. By the factorization criterion, the likelihood's dependence on θ is only in conjunction with T(X). As this is the same in both cases, the dependence on θ will be the same as well, leading to identical inferences.

Proof

Due to Hogg and Craig. Let ${\displaystyle X_1,X_2,\dots ,X_n}$, denote a random sample from a distribution having the pdf f(x, θ) for ι < θ < δ. Let Y₁ = u₁(X₁, X₂, ..., X_n) be a statistic whose pdf is g₁(y₁; θ). What we want to prove is that Y₁ = u₁(X₁, X₂, ..., X_n) is a sufficient statistic for θ if and only if, for some function H,

${\displaystyle \prod _{i=1}^{n}f(x_i;\theta )=g_1\left[u_1(x_1,x_2,\dots ,x_n);\theta \right]H(x_1,x_2,\dots ,x_n).}$

First, suppose that

${\displaystyle \prod _{i=1}^{n}f(x_i;\theta )=g_1\left[u_1(x_1,x_2,\dots ,x_n);\theta \right]H(x_1,x_2,\dots ,x_n).}$

We shall make the transformation y_i = u_i(x₁, x₂, ..., x_n), for i = 1, ..., n, having inverse functions x_i = w_i(y₁, y₂, ..., y_n), for i = 1, ..., n, and Jacobian ${\displaystyle J=\left[w_i/y_j\right]}$. Thus,

${\displaystyle \prod _{i=1}^{n}f\left[w_i(y_1,y_2,\dots ,y_n);\theta \right]=|J|g_1(y_1;\theta )H\left[w_1(y_1,y_2,\dots ,y_n),\dots ,w_n(y_1,y_2,\dots ,y_n)\right].}$

The left-hand member is the joint pdf g(y₁, y₂, ..., y_n; θ) of Y₁ = u₁(X₁, ..., X_n), ..., Y_n = u_n(X₁, ..., X_n). In the right-hand member, ${\displaystyle g_1(y_1;\theta )}$ is the pdf of ${\displaystyle Y_1}$, so that ${\displaystyle H[w_1,\dots ,w_n]|J|}$ is the quotient of ${\displaystyle g(y_1,\dots ,y_n;\theta )}$ and ${\displaystyle g_1(y_1;\theta )}$; that is, it is the conditional pdf ${\displaystyle h(y_2,\dots ,y_n\mid y_1;\theta )}$ of ${\displaystyle Y_2,\dots ,Y_n}$ given ${\displaystyle Y_1=y_1}$.

But ${\displaystyle H(x_1,x_2,\dots ,x_n)}$, and thus ${\displaystyle H\left[w_1(y_1,\dots ,y_n),\dots ,w_n(y_1,\dots ,y_n))\right]}$, was given not to depend upon ${\displaystyle \theta }$. Since ${\displaystyle \theta }$ was not introduced in the transformation and accordingly not in the Jacobian ${\displaystyle J}$, it follows that ${\displaystyle h(y_2,\dots ,y_n\mid y_1;\theta )}$ does not depend upon ${\displaystyle \theta }$ and that ${\displaystyle Y_1}$ is a sufficient statistics for ${\displaystyle \theta }$.

The converse is proven by taking:

${\displaystyle g(y_1,\dots ,y_n;\theta )=g_1(y_1;\theta )h(y_2,\dots ,y_n\mid y_1),}$

where ${\displaystyle h(y_2,\dots ,y_n\mid y_1)}$ does not depend upon ${\displaystyle \theta }$ because ${\displaystyle Y_2...Y_n}$ depend only upon ${\displaystyle X_1...X_n}$, which are independent on ${\displaystyle \mathrm{\Theta }}$ when conditioned by ${\displaystyle Y_1}$, a sufficient statistics by hypothesis. Now divide both members by the absolute value of the non-vanishing Jacobian ${\displaystyle J}$, and replace ${\displaystyle y_1,\dots ,y_n}$ by the functions ${\displaystyle u_1(x_1,\dots ,x_n),\dots ,u_n(x_1,\dots ,x_n)}$ in ${\displaystyle x_1,\dots ,x_n}$. This yields

${\displaystyle \frac{g\left[u_1(x_1,\dots ,x_n),\dots ,u_n(x_1,\dots ,x_n);\theta \right]}{|J^{\ast }|}=g_1\left[u_1(x_1,\dots ,x_n);\theta \right]\frac{h(u_2,\dots ,u_n\mid u_1)}{|J^{\ast }|}}$

where ${\displaystyle J^{\ast }}$ is the Jacobian with ${\displaystyle y_1,\dots ,y_n}$ replaced by their value in terms ${\displaystyle x_1,\dots ,x_n}$. The left-hand member is necessarily the joint pdf ${\displaystyle f(x_1;\theta )\cdots f(x_n;\theta )}$ of ${\displaystyle X_1,\dots ,X_n}$. Since ${\displaystyle h(y_2,\dots ,y_n\mid y_1)}$, and thus ${\displaystyle h(u_2,\dots ,u_n\mid u_1)}$, does not depend upon ${\displaystyle \theta }$, then

${\displaystyle H(x_1,\dots ,x_n)=\frac{h(u_2,\dots ,u_n\mid u_1)}{|J^{\ast }|}}$

is a function that does not depend upon ${\displaystyle \theta }$.

Another proof

A simpler more illustrative proof is as follows, although it applies only in the discrete case.

We use the shorthand notation to denote the joint probability density of ${\displaystyle (X,T(X))}$ by ${\displaystyle f_{\theta }(x,t)}$. Since ${\displaystyle T}$ is a deterministic function of ${\displaystyle X}$, we have ${\displaystyle f_{\theta }(x,t)=f_{\theta }(x)}$, as long as ${\displaystyle t=T(x)}$ and zero otherwise. Therefore:

${\displaystyle \begin{array}{cc}{f}_{\theta }(x)& =f_{\theta }(x,t)\\ & =f_{\theta }(x\mid t)f_{\theta }(t)\\ & =f(x\mid t)f_{\theta }(t)\end{array}}$

with the last equality being true by the definition of sufficient statistics. Thus ${\displaystyle f_{\theta }(x)=a(x)b_{\theta }(t)}$ with ${\displaystyle a(x)=f_{X\mid t}(x)}$ and ${\displaystyle b_{\theta }(t)=f_{\theta }(t)}$.

Conversely, if ${\displaystyle f_{\theta }(x)=a(x)b_{\theta }(t)}$, we have

${\displaystyle \begin{array}{cc}{f}_{\theta }(t)& =\sum _{x:T(x)=t}{f}_{\theta }(x,t)\\ & =\sum _{x:T(x)=t}{f}_{\theta }(x)\\ & =\sum _{x:T(x)=t}a(x)b_{\theta }(t)\\ & =\left(\sum _{x:T(x)=t}a(x)\right)b_{\theta }(t).\end{array}}$

With the first equality by the definition of pdf for multiple variables, the second by the remark above, the third by hypothesis, and the fourth because the summation is not over ${\displaystyle t}$.

Let ${\displaystyle f_{X\mid t}(x)}$ denote the conditional probability density of ${\displaystyle X}$ given ${\displaystyle T(X)}$. Then we can derive an explicit expression for this:

${\displaystyle \begin{array}{cc}{f}_{X\mid t}(x)& =\frac{f_{\theta }(x,t)}{f_{\theta }(t)}\\ & =\frac{f_{\theta }(x)}{f_{\theta }(t)}\\ & =\frac{a(x)b_{\theta }(t)}{\left(\sum _{x:T(x)=t}a(x)\right)b_{\theta }(t)}\\ & =\frac{a(x)}{\sum _{x:T(x)=t}a(x)}.\end{array}}$

With the first equality by definition of conditional probability density, the second by the remark above, the third by the equality proven above, and the fourth by simplification. This expression does not depend on ${\displaystyle \theta }$ and thus ${\displaystyle T}$ is a sufficient statistic.

Minimal sufficiency

A sufficient statistic is minimal sufficient if it can be represented as a function of any other sufficient statistic. In other words, S(X) is minimal sufficient if and only if

1. S(X) is sufficient, and
2. if T(X) is sufficient, then there exists a function f such that S(X) = f(T(X)).

Intuitively, a minimal sufficient statistic most efficiently captures all possible information about the parameter θ.

A useful characterization of minimal sufficiency is that when the density f_θ exists, S(X) is minimal sufficient if

${\displaystyle \frac{f_{\theta }(x)}{f_{\theta }(y)}}$ is independent of θ :${\displaystyle ⟺}$ S(x) = S(y)

This follows as a consequence from Fisher's factorization theorem stated above.

A case in which there is no minimal sufficient statistic was shown by Bahadur, 1954. However, under mild conditions, a minimal sufficient statistic does always exist. In particular, in Euclidean space, these conditions always hold if the random variables (associated with ${\displaystyle P_{\theta }}$ ) are all discrete or are all continuous.

If there exists a minimal sufficient statistic, and this is usually the case, then every complete sufficient statistic is necessarily minimal sufficient (note that this statement does not exclude a pathological case in which a complete sufficient exists while there is no minimal sufficient statistic). While it is hard to find cases in which a minimal sufficient statistic does not exist, it is not so hard to find cases in which there is no complete sufficient statistic.

The collection of likelihood ratios ${\displaystyle \left\{\frac{L(X\mid {\theta }_i)}{L(X\mid {\theta }_0)}\right\}}$ for ${\displaystyle i=1,...,k}$, is a minimal sufficient statistic if the parameter space is discrete ${\displaystyle \left\{{\theta }_0,...,{\theta }_k\right\}}$.

Examples

Bernoulli distribution

If X₁, ...., X_n are independent Bernoulli-distributed random variables with expected value p, then the sum T(X) = X₁ + ... + X_n is a sufficient statistic for p (here 'success' corresponds to X_i = 1 and 'failure' to X_i = 0; so T is the total number of successes)

This is seen by considering the joint probability distribution:

${\displaystyle Pr\{X=x\}=Pr\{X_1=x_1,X_2=x_2,\dots ,X_n=x_n\}.}$

Because the observations are independent, this can be written as

${\displaystyle p^{x_1}(1-p{)}^{1-x_1}{p}^{x_2}(1-p{)}^{1-x_2}\cdots p^{x_n}(1-p{)}^{1-x_n}}$

and, collecting powers of p and 1 − p, gives

${\displaystyle p^{\sum x_i}(1-p{)}^{n-\sum x_i}=p^{T(x)}(1-p{)}^{n-T(x)}}$

which satisfies the factorization criterion, with h(x) = 1 being just a constant.

Note the crucial feature: the unknown parameter p interacts with the data x only via the statistic T(x) = Σ x_i.

As a concrete application, this gives a procedure for distinguishing a fair coin from a biased coin.

Uniform distribution

If X₁, ...., X_n are independent and uniformly distributed on the interval [0,θ], then T(X) = max(X₁, ..., X_n) is sufficient for θ — the sample maximum is a sufficient statistic for the population maximum.

To see this, consider the joint probability density function of X (X₁,...,X_n). Because the observations are independent, the pdf can be written as a product of individual densities

${\displaystyle \begin{array}{cc}{f}_{\theta }(x_1,\dots ,x_n)& =\frac{1}{\theta }{\mathbf{1}}_{\{0\le x_1\le \theta \}}\cdots \frac{1}{\theta }{\mathbf{1}}_{\{0\le x_n\le \theta \}}\\ & =\frac{1}{{\theta }^n}{\mathbf{1}}_{\{0\le min\{x_i\}\}}{\mathbf{1}}_{\{max\{x_i\}\le \theta \}}\end{array}}$

where 1_{...} is the indicator function. Thus the density takes form required by the Fisher–Neyman factorization theorem, where h(x) = 1_{{min{xi}≥0}}, and the rest of the expression is a function of only θ and T(x) = max{x_i}.

In fact, the minimum-variance unbiased estimator (MVUE) for θ is

${\displaystyle \frac{n+1}{n}T(X).}$

This is the sample maximum, scaled to correct for the bias, and is MVUE by the Lehmann–Scheffé theorem. Unscaled sample maximum T(X) is the maximum likelihood estimator for θ.

Uniform distribution (with two parameters)

If ${\displaystyle X_1,...,X_n}$ are independent and uniformly distributed on the interval ${\displaystyle [\alpha ,\beta ]}$ (where ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are unknown parameters), then ${\displaystyle T(X_1^n)=\left(\underset{1\le i\le n}{min}{X}_i,\underset{1\le i\le n}{max}{X}_i\right)}$ is a two-dimensional sufficient statistic for ${\displaystyle (\alpha ,\beta )}$.

To see this, consider the joint probability density function of ${\displaystyle X_1^n=(X_1,\dots ,X_n)}$. Because the observations are independent, the pdf can be written as a product of individual densities, i.e.

${\displaystyle \begin{array}{cc}{f}_{X_1^n}(x_1^n)& =\prod _{i=1}^n\left(\frac{1}{\beta -\alpha }\right){\mathbf{1}}_{\{\alpha \le x_i\le \beta \}}={\left(\frac{1}{\beta -\alpha }\right)}^n{\mathbf{1}}_{\{\alpha \le x_i\le \beta ,\mathrm{\forall }i=1,\dots ,n\}}\\ & ={\left(\frac{1}{\beta -\alpha }\right)}^n{\mathbf{1}}_{\{\alpha \le \underset{1\le i\le n}{min}{X}_i\}}{\mathbf{1}}_{\{\underset{1\le i\le n}{max}{X}_i\le \beta \}}.\end{array}}$

The joint density of the sample takes the form required by the Fisher–Neyman factorization theorem, by letting

${\displaystyle \begin{array}{c}h(x_1^n)=1,g_{(\alpha ,\beta )}(x_1^n)={\left(\frac{1}{\beta -\alpha }\right)}^n{\mathbf{1}}_{\{\alpha \le \underset{1\le i\le n}{min}{X}_i\}}{\mathbf{1}}_{\{\underset{1\le i\le n}{max}{X}_i\le \beta \}}.\end{array}}$

Since ${\displaystyle h(x_1^n)}$ does not depend on the parameter ${\displaystyle (\alpha ,\beta )}$ and ${\displaystyle g_{(\alpha ,\beta )}(x_1^n)}$ depends only on ${\displaystyle x_1^n}$ through the function ${\displaystyle T(X_1^n)=\left(\underset{1\le i\le n}{min}{X}_i,\underset{1\le i\le n}{max}{X}_i\right),}$

the Fisher–Neyman factorization theorem implies ${\displaystyle T(X_1^n)=\left(\underset{1\le i\le n}{min}{X}_i,\underset{1\le i\le n}{max}{X}_i\right)}$ is a sufficient statistic for ${\displaystyle (\alpha ,\beta )}$.

Poisson distribution

If X₁, ...., X_n are independent and have a Poisson distribution with parameter λ, then the sum T(X) = X₁ + ... + X_n is a sufficient statistic for λ.

To see this, consider the joint probability distribution:

${\displaystyle Pr(X=x)=P(X_1=x_1,X_2=x_2,\dots ,X_n=x_n).}$

Because the observations are independent, this can be written as

${\displaystyle \frac{e^{-\lambda }{\lambda }^{x_1}}{x_1!}\cdot \frac{e^{-\lambda }{\lambda }^{x_2}}{x_2!}\cdots \frac{e^{-\lambda }{\lambda }^{x_n}}{x_n!}}$

which may be written as

${\displaystyle e^{-n\lambda }{\lambda }^{(x_1+x_2+\cdots +x_n)}\cdot \frac{1}{x_1!x_2!\cdots x_n!}}$

which shows that the factorization criterion is satisfied, where h(x) is the reciprocal of the product of the factorials. Note the parameter λ interacts with the data only through its sum T(X).

Normal distribution

If ${\displaystyle X_1,\dots ,X_n}$ are independent and normally distributed with expected value ${\displaystyle \theta }$ (a parameter) and known finite variance ${\displaystyle {\sigma }^2,}$ then

${\displaystyle T(X_1^n)=\overline{x}=\frac{1}{n}\sum _{i=1}^n{X}_i}$

is a sufficient statistic for ${\displaystyle \theta .}$

To see this, consider the joint probability density function of ${\displaystyle X_1^n=(X_1,\dots ,X_n)}$. Because the observations are independent, the pdf can be written as a product of individual densities, i.e.

${\displaystyle \begin{array}{cc}{f}_{X_1^n}(x_1^n)& =\prod _{i=1}^n\frac{1}{\sqrt{2\pi {\sigma }^2}}\exp \left(-\frac{(x_i-\theta {)}^2}{2{\sigma }^2}\right)\\ & =(2\pi {\sigma }^2{)}^{-\frac{n}{2}}\exp \left(-\sum _{i=1}^n\frac{(x_i-\theta {)}^2}{2{\sigma }^2}\right)\\ & =(2\pi {\sigma }^2{)}^{-\frac{n}{2}}\exp \left(-\sum _{i=1}^n\frac{{\left(\left(x_i-\overline{x}\right)-\left(\theta -\overline{x}\right)\right)}^2}{2{\sigma }^2}\right)\\ & =(2\pi {\sigma }^2{)}^{-\frac{n}{2}}\exp \left(-\frac{1}{2{\sigma }^2}\left(\sum _{i=1}^n(x_i-\overline{x}{)}^2+\sum _{i=1}^n(\theta -\overline{x}{)}^2-2\sum _{i=1}^n(x_i-\overline{x})(\theta -\overline{x})\right)\right)\\ & =(2\pi {\sigma }^2{)}^{-\frac{n}{2}}\exp \left(-\frac{1}{2{\sigma }^2}\left(\sum _{i=1}^n(x_i-\overline{x}{)}^2+n(\theta -\overline{x}{)}^2\right)\right)& & \sum _{i=1}^n(x_i-\overline{x})(\theta -\overline{x})=0\\ & =(2\pi {\sigma }^2{)}^{-\frac{n}{2}}\exp \left(-\frac{1}{2{\sigma }^2}\sum _{i=1}^n(x_i-\overline{x}{)}^2\right)\exp \left(-\frac{n}{2{\sigma }^2}(\theta -\overline{x}{)}^2\right)\end{array}}$

The joint density of the sample takes the form required by the Fisher–Neyman factorization theorem, by letting

${\displaystyle \begin{array}{cc}h(x_1^n)& =(2\pi {\sigma }^2{)}^{-\frac{n}{2}}\exp \left(-\frac{1}{2{\sigma }^2}\sum _{i=1}^n(x_i-\overline{x}{)}^2\right)\\ g_{\theta }(x_1^n)& =\exp \left(-\frac{n}{2{\sigma }^2}(\theta -\overline{x}{)}^2\right)\end{array}}$

Since ${\displaystyle h(x_1^n)}$ does not depend on the parameter ${\displaystyle \theta }$ and ${\displaystyle g_{\theta }(x_1^n)}$ depends only on ${\displaystyle x_1^n}$ through the function

${\displaystyle T(X_1^n)=\overline{x}=\frac{1}{n}\sum _{i=1}^n{X}_i,}$

the Fisher–Neyman factorization theorem implies ${\displaystyle T(X_1^n)}$ is a sufficient statistic for ${\displaystyle \theta }$.

If ${\displaystyle {\sigma }^2}$ is unknown and since ${\displaystyle s^2=\frac{1}{n-1}\sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}$, the above likelihood can be rewritten as

${\displaystyle \begin{array}{c}{f}_{X_1^n}(x_1^n)=(2\pi {\sigma }^2{)}^{-n/2}\exp \left(-\frac{n-1}{2{\sigma }^2}{s}^2\right)\exp \left(-\frac{n}{2{\sigma }^2}(\theta -\overline{x}{)}^2\right).\end{array}}$

The Fisher–Neyman factorization theorem still holds and implies that ${\displaystyle (\overline{x},s^2)}$ is a joint sufficient statistic for ${\displaystyle (\theta ,{\sigma }^2)}$.

Exponential distribution

If ${\displaystyle X_1,\dots ,X_n}$ are independent and exponentially distributed with expected value θ (an unknown real-valued positive parameter), then ${\displaystyle T(X_1^n)=\sum _{i=1}^n{X}_i}$ is a sufficient statistic for θ.

To see this, consider the joint probability density function of ${\displaystyle X_1^n=(X_1,\dots ,X_n)}$. Because the observations are independent, the pdf can be written as a product of individual densities, i.e.

${\displaystyle \begin{array}{cc}{f}_{X_1^n}(x_1^n)& =\prod _{i=1}^n\frac{1}{\theta }{e}^{\frac{-1}{\theta }{x}_i}=\frac{1}{{\theta }^n}{e}^{\frac{-1}{\theta }\sum _{i=1}^n{x}_i}.\end{array}}$

The joint density of the sample takes the form required by the Fisher–Neyman factorization theorem, by letting

${\displaystyle \begin{array}{c}h(x_1^n)=1,g_{\theta }(x_1^n)=\frac{1}{{\theta }^n}{e}^{\frac{-1}{\theta }\sum _{i=1}^n{x}_i}.\end{array}}$

Since ${\displaystyle h(x_1^n)}$ does not depend on the parameter ${\displaystyle \theta }$ and ${\displaystyle g_{\theta }(x_1^n)}$ depends only on ${\displaystyle x_1^n}$ through the function ${\displaystyle T(X_1^n)=\sum _{i=1}^n{X}_i}$

the Fisher–Neyman factorization theorem implies ${\displaystyle T(X_1^n)=\sum _{i=1}^n{X}_i}$ is a sufficient statistic for ${\displaystyle \theta }$.

Gamma distribution

If ${\displaystyle X_1,\dots ,X_n}$ are independent and distributed as a ${\displaystyle \mathrm{\Gamma }(\alpha ,\beta )}$, where ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are unknown parameters of a Gamma distribution, then ${\displaystyle T(X_1^n)=\left(\prod _{i=1}^n{X}_i,\sum _{i=1}^n{X}_i\right)}$ is a two-dimensional sufficient statistic for ${\displaystyle (\alpha ,\beta )}$.

To see this, consider the joint probability density function of ${\displaystyle X_1^n=(X_1,\dots ,X_n)}$. Because the observations are independent, the pdf can be written as a product of individual densities, i.e.

${\displaystyle \begin{array}{cc}{f}_{X_1^n}(x_1^n)& =\prod _{i=1}^n\left(\frac{1}{\mathrm{\Gamma }(\alpha ){\beta }^{\alpha }}\right)x_i^{\alpha -1}{e}^{(-1/\beta )x_i}\\ & ={\left(\frac{1}{\mathrm{\Gamma }(\alpha ){\beta }^{\alpha }}\right)}^n{\left(\prod _{i=1}^n{x}_i\right)}^{\alpha -1}{e}^{\frac{-1}{\beta }\sum _{i=1}^n{x}_i}.\end{array}}$

The joint density of the sample takes the form required by the Fisher–Neyman factorization theorem, by letting

${\displaystyle \begin{array}{c}h(x_1^n)=1,g_{(\alpha ,\beta )}(x_1^n)={\left(\frac{1}{\mathrm{\Gamma }(\alpha ){\beta }^{\alpha }}\right)}^n{\left(\prod _{i=1}^n{x}_i\right)}^{\alpha -1}{e}^{\frac{-1}{\beta }\sum _{i=1}^n{x}_i}.\end{array}}$

Since ${\displaystyle h(x_1^n)}$ does not depend on the parameter ${\displaystyle (\alpha ,\beta )}$ and ${\displaystyle g_{(\alpha ,\beta )}(x_1^n)}$ depends only on ${\displaystyle x_1^n}$ through the function ${\displaystyle T(x_1^n)=\left(\prod _{i=1}^n{x}_i,\sum _{i=1}^n{x}_i\right),}$

the Fisher–Neyman factorization theorem implies ${\displaystyle T(X_1^n)=\left(\prod _{i=1}^n{X}_i,\sum _{i=1}^n{X}_i\right)}$ is a sufficient statistic for ${\displaystyle (\alpha ,\beta ).}$

Rao–Blackwell theorem

Sufficiency finds a useful application in the Rao–Blackwell theorem, which states that if g(X) is any kind of estimator of θ, then typically the conditional expectation of g(X) given sufficient statistic T(X) is a better (in the sense of having lower variance) estimator of θ, and is never worse. Sometimes one can very easily construct a very crude estimator g(X), and then evaluate that conditional expected value to get an estimator that is in various senses optimal.

Exponential family

According to the Pitman–Koopman–Darmois theorem, among families of probability distributions whose domain does not vary with the parameter being estimated, only in exponential families is there a sufficient statistic whose dimension remains bounded as sample size increases. Intuitively, this states that nonexponential families of distributions on the real line require nonparametric statistics to fully capture the information in the data.

Less tersely, suppose ${\displaystyle X_n,n=1,2,3,\dots }$ are independent identically distributed real random variables whose distribution is known to be in some family of probability distributions, parametrized by ${\displaystyle \theta }$, satisfying certain technical regularity conditions, then that family is an exponential family if and only if there is a ${\displaystyle {\mathbb{R}}^m}$-valued sufficient statistic ${\displaystyle T(X_1,\dots ,X_n)}$ whose number of scalar components ${\displaystyle m}$ does not increase as the sample size n increases.

This theorem shows that the existence of a finite-dimensional, real-vector-valued sufficient statistics sharply restricts the possible forms of a family of distributions on the real line.

When the parameters or the random variables are no longer real-valued, the situation is more complex.

Other types of sufficiency

Bayesian sufficiency

An alternative formulation of the condition that a statistic be sufficient, set in a Bayesian context, involves the posterior distributions obtained by using the full data-set and by using only a statistic. Thus the requirement is that, for almost every x,

${\displaystyle Pr(\theta \mid X=x)=Pr(\theta \mid T(X)=t(x)).}$

More generally, without assuming a parametric model, we can say that the statistics T is predictive sufficient if

${\displaystyle Pr(X^{\prime }=x^{\prime }\mid X=x)=Pr(X^{\prime }=x^{\prime }\mid T(X)=t(x)).}$

It turns out that this "Bayesian sufficiency" is a consequence of the formulation above, however they are not directly equivalent in the infinite-dimensional case. A range of theoretical results for sufficiency in a Bayesian context is available.

Linear sufficiency

A concept called "linear sufficiency" can be formulated in a Bayesian context, and more generally. First define the best linear predictor of a vector Y based on X as ${\displaystyle \hat{E}[Y\mid X]}$. Then a linear statistic T(x) is linear sufficient if

${\displaystyle \hat{E}[\theta \mid X]=\hat{E}[\theta \mid T(X)].}$