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  "body": "\n## Definition\n\nLet $\\set{P_\\theta : \\theta \\in \\Theta}$ be a [[statistical model|JZ6R7J36K5]] on a measurable space $(S, \\mathcal{S})$, and let $X$ be the identity random element of $S$.\nLet $T = T(X)$ be a [[statistic|036756NFD6]] with values in a measurable space.\nThe statistic $T$ is *sufficient* for $\\theta$ if the following holds for every $A \\in \\mathcal{S}$.\nThere is a version of the conditional probability $P_\\theta(X \\in A \\mid T)$ that is the same for all $\\theta \\in \\Theta$.\nThat is, there is a function $k_A(T)$ with $k_A(T) = P_\\theta(X \\in A \\mid T)$ $P_\\theta$-almost surely, for every $\\theta$, as a [[conditional expectation|0YAMJRK04E]] of $\\mathbf{1}\\set{X \\in A}$ given $T$.\n\n## Remarks\n\nExample: let $X = (X_1, \\dots, X_n)$ be i.i.d. Bernoulli($p$), $p \\in (0, 1)$, and $T = X_1 + \\dots + X_n$.\nFor $x \\in \\set{0, 1}^n$ with $x_1 + \\dots + x_n = t$, one has $P_p(X = x \\mid T = t) = p^t(1 - p)^{n - t} / \\left(\\binom{n}{t} p^t (1 - p)^{n - t}\\right) = 1/\\binom{n}{t}$, which does not depend on $p$; so $T$ is sufficient.\nNon-example: for $n \\ge 2$, the statistic $X_1$ is not sufficient, since given $X_1$ the variable $X_2$ is still Bernoulli($p$), and its law depends on $p$.\nThe whole sample $X$ is always sufficient, and any one-to-one measurable function of a sufficient statistic with measurable inverse is sufficient.\n",
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  "name": "Sufficient statistic",
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