Sufficient statistic

As an example, the sample mean is sufficient for the 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.

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.

Due to Hogg and Craig.[5] Let ${\displaystyle X_{1},X_{2},\ldots ,X_{n}}$, 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 ${\displaystyle \Theta }$ 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

${\displaystyle {\frac {g\left[u_{1}(x_{1},\dots ,x_{n}),\dots ,u_{n}(x_{1},\dots ,x_{n});\theta \right]}{|J^{*}|}}=g_{1}\left[u_{1}(x_{1},\dots ,x_{n});\theta \right]{\frac {h(u_{2},\dots ,u_{n}\mid u_{1})}{|J^{*}|}}}$

where ${\displaystyle J^{*}}$ 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

${\displaystyle H(x_{1},\dots ,x_{2})={\frac {h(u_{2},\dots ,u_{n}\mid u_{1})}{|J^{*}|}}}$

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

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 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:

${\displaystyle {\begin{aligned}f_{\theta }(x)&=f_{\theta }(x,t)\\[5pt]&=f_{\theta }(x\mid t)f_{\theta }(t)\\[5pt]&=f(x\mid t)f_{\theta }(t)\end{aligned}}}$

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

${\displaystyle {\begin{aligned}f_{\theta }(t)&=\sum _{x:T(x)=t}f_{\theta }(x,t)\\[5pt]&=\sum _{x:T(x)=t}f_{\theta }(x)\\[5pt]&=\sum _{x:T(x)=t}a(x)b_{\theta }(t)\\[5pt]&=\left(\sum _{x:T(x)=t}a(x)\right)b_{\theta }(t).\end{aligned}}}$

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:

${\displaystyle {\begin{aligned}f_{X\mid t}(x)&={\frac {f_{\theta }(x,t)}{f_{\theta }(t)}}\\[5pt]&={\frac {f_{\theta }(x)}{f_{\theta }(t)}}\\[5pt]&={\frac {a(x)b_{\theta }(t)}{\left(\sum _{x:T(x)=t}a(x)\right)b_{\theta }(t)}}\\[5pt]&={\frac {a(x)}{\sum _{x:T(x)=t}a(x)}}.\end{aligned}}}$

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.[6]

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:

${\displaystyle \Pr\{X=x\}=\Pr\{X_{1}=x_{1},X_{2}=x_{2},\ldots ,X_{n}=x_{n}\}.}$

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.

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

${\displaystyle {\begin{aligned}f_{\theta }(x_{1},\ldots ,x_{n})&={\frac {1}{\theta }}\mathbf {1} _{\{0\leq x_{1}\leq \theta \}}\cdots {\frac {1}{\theta }}\mathbf {1} _{\{0\leq x_{n}\leq \theta \}}\\[5pt]&={\frac {1}{\theta ^{n}}}\mathbf {1} _{\{0\leq \min\{x_{i}\}\}}\mathbf {1} _{\{\max\{x_{i}\}\leq \theta \}}\end{aligned}}}$

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.

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 ${\displaystyle T(X_{1}^{n})=\left(\min _{1\leq i\leq n}X_{i},\max _{1\leq i\leq n}X_{i}\right)}$ is a two-dimensional sufficient statistic for ${\displaystyle (\alpha \,,\,\beta )}$.

To see this, consider the joint probability density function of ${\displaystyle X_{1}^{n}=(X_{1},\ldots ,X_{n})}$. Because the observations are independent, the pdf can be written as a product of individual densities, i.e.

${\displaystyle {\begin{aligned}f_{X_{1}^{n}}(x_{1}^{n})&=\prod _{i=1}^{n}\left({1 \over \beta -\alpha }\right)\mathbf {1} _{\{\alpha \leq x_{i}\leq \beta \}}=\left({1 \over \beta -\alpha }\right)^{n}\mathbf {1} _{\{\alpha \leq x_{i}\leq \beta ,\,\forall \,i=1,\ldots ,n\}}\\&=\left({1 \over \beta -\alpha }\right)^{n}\mathbf {1} _{\{\alpha \,\leq \,\min _{1\leq i\leq n}X_{i}\}}\mathbf {1} _{\{\max _{1\leq i\leq n}X_{i}\,\leq \,\beta \}}.\end{aligned}}}$

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

${\displaystyle {\begin{aligned}h(x_{1}^{n})=1,\quad g_{(\alpha ,\beta )}(x_{1}^{n})=\left({1 \over \beta -\alpha }\right)^{n}\mathbf {1} _{\{\alpha \,\leq \,\min _{1\leq i\leq n}X_{i}\}}\mathbf {1} _{\{\max _{1\leq i\leq n}X_{i}\,\leq \,\beta \}}.\end{aligned}}}$

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 ${\displaystyle T(X_{1}^{n})=\left(\min _{1\leq i\leq n}X_{i},\max _{1\leq i\leq n}X_{i}\right),}$

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

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:

${\displaystyle \Pr(X=x)=P(X_{1}=x_{1},X_{2}=x_{2},\ldots ,X_{n}=x_{n}).}$

Because the observations are independent, this can be written as

${\displaystyle {e^{-\lambda }\lambda ^{x_{1}} \over x_{1}!}\cdot {e^{-\lambda }\lambda ^{x_{2}} \over x_{2}!}\cdots {e^{-\lambda }\lambda ^{x_{n}} \over x_{n}!}}$

which may be written as

${\displaystyle e^{-n\lambda }\lambda ^{(x_{1}+x_{2}+\cdots +x_{n})}\cdot {1 \over x_{1}!x_{2}!\cdots x_{n}!}}$

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).

If ${\displaystyle X_{1},\ldots ,X_{n}}$ 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.

${\displaystyle {\begin{aligned}f_{X_{1}^{n}}(x_{1}^{n})&=\prod _{i=1}^{n}{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\exp \left(-{\frac {(x_{i}-\theta )^{2}}{2\sigma ^{2}}}\right)\\[6pt]&=(2\pi \sigma ^{2})^{-{\frac {n}{2}}}\exp \left(-\sum _{i=1}^{n}{\frac {(x_{i}-\theta )^{2}}{2\sigma ^{2}}}\right)\\[6pt]&=(2\pi \sigma ^{2})^{-{\frac {n}{2}}}\exp \left(-\sum _{i=1}^{n}{\frac {\left(\left(x_{i}-{\overline {x}}\right)-\left(\theta -{\overline {x}}\right)\right)^{2}}{2\sigma ^{2}}}\right)\\[6pt]&=(2\pi \sigma ^{2})^{-{\frac {n}{2}}}\exp \left(-{1 \over 2\sigma ^{2}}\left(\sum _{i=1}^{n}(x_{i}-{\overline {x}})^{2}+\sum _{i=1}^{n}(\theta -{\overline {x}})^{2}-2\sum _{i=1}^{n}(x_{i}-{\overline {x}})(\theta -{\overline {x}})\right)\right)\\[6pt]&=(2\pi \sigma ^{2})^{-{\frac {n}{2}}}\exp \left(-{1 \over 2\sigma ^{2}}\left(\sum _{i=1}^{n}(x_{i}-{\overline {x}})^{2}+n(\theta -{\overline {x}})^{2}\right)\right)&&\sum _{i=1}^{n}(x_{i}-{\overline {x}})(\theta -{\overline {x}})=0\\[6pt]&=(2\pi \sigma ^{2})^{-{\frac {n}{2}}}\exp \left(-{1 \over 2\sigma ^{2}}\sum _{i=1}^{n}(x_{i}-{\overline {x}})^{2}\right)\exp \left(-{\frac {n}{2\sigma ^{2}}}(\theta -{\overline {x}})^{2}\right)\end{aligned}}}$

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

${\displaystyle {\begin{aligned}h(x_{1}^{n})&=(2\pi \sigma ^{2})^{-{\frac {n}{2}}}\exp \left(-{1 \over 2\sigma ^{2}}\sum _{i=1}^{n}(x_{i}-{\overline {x}})^{2}\right)\\[6pt]g_{\theta }(x_{1}^{n})&=\exp \left(-{\frac {n}{2\sigma ^{2}}}(\theta -{\overline {x}})^{2}\right)\end{aligned}}}$

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

${\displaystyle {\begin{aligned}f_{X_{1}^{n}}(x_{1}^{n})=(2\pi \sigma ^{2})^{-n/2}\exp \left(-{\frac {n-1}{2\sigma ^{2}}}s^{2}\right)\exp \left(-{\frac {n}{2\sigma ^{2}}}(\theta -{\overline {x}})^{2}\right).\end{aligned}}}$

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

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.

${\displaystyle {\begin{aligned}f_{X_{1}^{n}}(x_{1}^{n})&=\prod _{i=1}^{n}{1 \over \theta }\,e^{{-1 \over \theta }x_{i}}={1 \over \theta ^{n}}\,e^{{-1 \over \theta }\sum _{i=1}^{n}x_{i}}.\end{aligned}}}$

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

${\displaystyle {\begin{aligned}h(x_{1}^{n})=1,\,\,\,g_{\theta }(x_{1}^{n})={1 \over \theta ^{n}}\,e^{{-1 \over \theta }\sum _{i=1}^{n}x_{i}}.\end{aligned}}}$

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 ${\displaystyle \theta }$.

If ${\displaystyle X_{1},\dots ,X_{n}}$ are independent and distributed as a ${\displaystyle \Gamma (\alpha \,,\,\beta )}$, 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.

${\displaystyle {\begin{aligned}f_{X_{1}^{n}}(x_{1}^{n})&=\prod _{i=1}^{n}\left({1 \over \Gamma (\alpha )\beta ^{\alpha }}\right)x_{i}^{\alpha -1}e^{(-1/\beta )x_{i}}\\[5pt]&=\left({1 \over \Gamma (\alpha )\beta ^{\alpha }}\right)^{n}\left(\prod _{i=1}^{n}x_{i}\right)^{\alpha -1}e^{{-1 \over \beta }\sum _{i=1}^{n}x_{i}}.\end{aligned}}}$

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

${\displaystyle {\begin{aligned}h(x_{1}^{n})=1,\,\,\,g_{(\alpha \,,\,\beta )}(x_{1}^{n})=\left({1 \over \Gamma (\alpha )\beta ^{\alpha }}\right)^{n}\left(\prod _{i=1}^{n}x_{i}\right)^{\alpha -1}e^{{-1 \over \beta }\sum _{i=1}^{n}x_{i}}.\end{aligned}}}$

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 ${\displaystyle (\alpha \,,\,\beta ).}$

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

${\displaystyle \Pr(X'=x'\mid X=x)=\Pr(X'=x'\mid T(X)=t(x)).}$

It turns out that this "Bayesian sufficiency" is a consequence of the formulation above,[10] however they are not directly equivalent in the infinite-dimensional case.[11] A range of theoretical results for sufficiency in a Bayesian context is available.[12]

A concept called "linear sufficiency" can be formulated in a Bayesian context,[13] and more generally.[14] 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[15] if

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