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inequality becomes an equality i θ t x i θ x displaystyle i bigl theta t x bigr i theta x example edit 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 edit 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 f x θ h x g θ t x 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 6 and the theorem is sometimes referred to as the halmos savage factorization theorem 7 the proofs below handle special cases but an alternative general proof along the same lines can be given 8 in many simple cases the probability density function is fully specified by θ displaystyle theta and t x displaystyle t x and h x 1 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 edit 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 edit due to hogg and craig 9 let x 1 x 2 x n 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 y 1 u 1 x 1 x 2 x n be a statistic whose pdf is g 1 y 1 θ what we want to prove is that y 1 u 1 x 1 x 2 x n is a sufficient statistic for θ if and only if for some function h i 1 n f x i θ g 1 u 1 x 1 x 2 x n θ h x 1 x 2 x n 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 i 1 n f x i θ g 1 u 1 x 1 x 2 x n θ h x 1 x 2 x n 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 x 1 x 2 x n for i 1 n having inverse functions x i w i y 1 y 2 y n for i 1 n and jacobian j w i y j displaystyle j left w_ i y_ j right thus i 1 n f w i y 1 y 2 y n θ j g 1 y 1 θ h w 1 y 1 y 2 y n w n y 1 y 2 y n 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 y 1 y 2 y n θ of y 1 u 1 x 1 x n y n u n x 1 x n in the right hand member g 1 y 1 θ displaystyle g_ 1 y_ 1 theta is the pdf of y 1 displaystyle y_ 1 so that h w 1 w n j displaystyle h w_ 1 dots w_ n j is the quotient of g y 1 y n θ displaystyle g y_ 1 dots y_ n theta and g 1 y 1 θ displaystyle g_ 1 y_ 1 theta that is it is the conditional pdf h y 2 y n y 1 θ displaystyle h y_ 2 dots y_ n mid y_ 1 theta of y 2 y n displaystyle y_ 2 dots y_ n given y 1 y 1 displaystyle y_ 1 y_ 1 but h x 1 x 2 x n displaystyle h x_ 1 x_ 2 dots x_ n and thus h w 1 y 1 y n w n y 1 y n 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 θ displaystyle theta since θ displaystyle theta was not introduced in the transformation and accordingly not in the jacobian j displaystyle j it follows that h y 2 y n y 1 θ displaystyle h y_ 2 dots y_ n mid y_ 1 theta does not depend upon θ displaystyle theta and that y 1 displaystyle y_ 1 is a sufficient statistics for θ displaystyle theta the converse is proven by taking g y 1 y n θ g 1 y 1 θ h y 2 y n y 1 displaystyle g y_ 1 dots y_ n theta g_ 1 y_ 1 theta h y_ 2 dots y_ n mid y_ 1 where h y 2 y n y 1 displaystyle h y_ 2 dots y_ n mid y_ 1 does not depend upon θ displaystyle theta because y 2 y n displaystyle y_ 2 y_ n depend only upon x 1 x n displaystyle x_ 1 x_ n which are independent on θ displaystyle theta when conditioned by y 1 displaystyle y_ 1 a sufficient statistics by hypothesis now divide both members by the absolute value of the non vanishing jacobian j displaystyle j and replace y 1 y n displaystyle y_ 1 dots y_ n by the functions u 1 x 1 x n u n x 1 x n displaystyle u_ 1 x_ 1 dots x_ n dots u_ n x_ 1 dots x_ n in x 1 x n displaystyle x_ 1 dots x_ n this yields g u 1 x 1 x n u n x 1 x n θ j g 1 u 1 x 1 x n θ h u 2 u n u 1 j 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 j displaystyle j is the jacobian with y 1 y n displaystyle y_ 1 dots y_ n replaced by their value in terms x 1 x n displaystyle x_ 1 dots x_ n the left hand member is necessarily the joint pdf f x 1 θ f x n θ displaystyle f x_ 1 theta cdots f x_ n theta of x 1 x n displaystyle x_ 1 dots x_ n since h y 2 y n y 1 displaystyle h y_ 2 dots y_ n mid y_ 1 and thus h u 2 u n u 1 displaystyle h u_ 2 dots u_ n mid u_ 1 does not depend upon θ displaystyle theta then h x 1 x n h u 2 u n u 1 j displaystyle h x_ 1 dots x_ n frac h u_ 2 dots u_ n mid u_ 1 j is a function that does not depend upon θ displaystyle theta another proof edit 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 x t x displaystyle x t x by f θ x t displaystyle f_ theta x t since t displaystyle t is a deterministic function of x displaystyle x we have f θ x t f θ x displaystyle f_ theta x t f_ theta x as long as t t x displaystyle t t x and zero otherwise therefore f θ x f θ x t f θ x t f θ t f x t f θ t 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 f θ x a x b θ t displaystyle f_ theta x a x b_ theta t with a x f x t x displaystyle a x f_ x mid t x and b θ t f θ t displaystyle b_ theta t f_ theta t conversely if f θ x a x b θ t displaystyle f_ theta x a x b_ theta t we have f θ t x t x t f θ x t x t x t f θ x x t x t a x b θ t x t x t a x b θ t 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 t displaystyle t let f x t x displaystyle f_ x mid t x denote the conditional probability density of x displaystyle x given t x displaystyle t x then we can derive an explicit expression for this f x t x f θ x t f θ t f θ x f θ t a x b θ t x t x t a x b θ t a x x t x t a x 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 θ displaystyle theta and thus t displaystyle t is a sufficient statistic 10 minimal sufficiency edit 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 11 s x is sufficient and 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 f θ x f θ y displaystyle frac f_ theta x f_ theta y is independent of θ displaystyle longleftrightarrow 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 12 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 p θ 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 13 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 l x θ i l x θ 0 displaystyle left frac l x mid theta _ i l x mid theta _ 0 right for i 1 k displaystyle i 1 k is a minimal sufficient statistic if the parameter space is discrete θ 0 θ k displaystyle left theta _ 0 theta _ k right examples edit bernoulli distribution edit if x 1 x n are independent bernoulli distributed random variables with expected value p then the sum t x x 1 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 pr x x pr x 1 x 1 x 2 x 2 x n x n 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 p x 1 1 p 1 x 1 p x 2 1 p 1 x 2 p x n 1 p 1 x n 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 p x i 1 p n x i p t x 1 p n t x 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 edit see also german tank problem if x 1 x n are independent and uniformly distributed on the interval 0 θ then t x max x 1 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 x 1 x n because the observations are independent the pdf can be written as a product of individual densities f θ x 1 x n 1 θ 1 0 x 1 θ 1 θ 1 0 x n θ 1 θ n 1 0 min x i 1 max x i θ 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 x i 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 n 1 n t x 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 edit if x 1 x n displaystyle x_ 1 x_ n are independent and uniformly distributed on the interval α β displaystyle alpha beta where α displaystyle alpha and β displaystyle beta are unknown parameters then t x 1 n min 1 i n x i max 1 i n x i 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 α β displaystyle alpha beta to see this consider the joint probability density function of x 1 n x 1 x n 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 f x 1 n x 1 n i 1 n 1 β α 1 α x i β 1 β α n 1 α x i β i 1 n 1 β α n 1 α min 1 i n x i 1 max 1 i n x i β 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 h x 1 n 1 g α β x 1 n 1 β α n 1 α min 1 i n x i 1 max 1 i n x i β 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 h x 1 n displaystyle h x_ 1 n does not depend on the parameter α β displaystyle alpha beta and g α β x 1 n displaystyle g_ alpha beta x_ 1 n depends only on x 1 n displaystyle x_ 1 n through the function t x 1 n min 1 i n x i max 1 i n x i 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 t x 1 n min 1 i n x i max 1 i n x i 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 α β displaystyle alpha beta poisson distribution edit if x 1 x n are independent and have a poisson distribution with parameter λ then the sum t x x 1 x n is a sufficient statistic for λ to see this consider the joint probability distribution pr x x p x 1 x 1 x 2 x 2 x n x n 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 e λ λ x 1 x 1 e λ λ x 2 x 2 e λ λ x n x n 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 e n λ λ x 1 x 2 x n 1 x 1 x 2 x n 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 cr...