factorization criterion


Let 𝑿=(X1,,Xn) be a random vector whose coordinatesPlanetmathPlanetmath are observations, and whose probability (densityPlanetmathPlanetmath) function is, f(𝒙θ) where θ is an unknown parameter. Then a statisticMathworldMathworldPlanetmath T(𝑿) for θ is a sufficient statistic iff f can be expressed as a product of (or factored into) two functions g,h, f=gh where g is a function of T(𝑿) and θ, and h is a function of 𝒙. In symbol, we have

f(𝒙θ)=g(T(𝑿),θ)h(𝒙).

Applications.

  1. 1.

    In view of the above statement, let’s show that the sample mean X¯ of n independentPlanetmathPlanetmath observations from a normal distributionMathworldPlanetmath N(μ,σ2) is a sufficient statistic for the unknown mean μ. Since the Xi’s are independent random variablesMathworldPlanetmath, then the probability density functionMathworldPlanetmath f(𝒙μ), being the joint probability density function of each of the Xi, is the product of the individual density functions f(xμ):

    f(𝒙μ) = i=1nf(xμ)=i=1n12πσ2exp[-(xi-μ)22σ2] (1)
    = 1(2π)nσ2nexp[i=1n-(xi-μ)22σ2] (2)
    = 1(2π)nσ2nexp[-12σ2i=1nxi2]exp[μσ2i=1nxi-nμ22σ2] (3)
    = h(𝒙)exp[nμσ2T(𝒙)-nμ22σ2] (4)
    = h(𝒙)g(T(𝒙),μ) (5)

    where g is the last exponential expression and h is the rest of the expression in (3). By the factorization criterion, T(𝑿)=X¯ is a sufficient statistic.

  2. 2.

    Similarly, the above shows that the sample variance s2 is not a sufficient statistic for σ2 if μ is unknown.

  3. 3.

    But, if μ is a known constant, then the statistic

    T(X1,,Xn)=1n-1i=1n(Xi-μ)2

    is sufficient for σ2 by observing in (2) above, and letting h(𝒙)=1 and g(T,σ2) be all of expression (2).

Title factorization criterion
Canonical name FactorizationCriterion
Date of creation 2013-03-22 15:02:48
Last modified on 2013-03-22 15:02:48
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 4
Author CWoo (3771)
Entry type Theorem
Classification msc 62B05
Synonym factorization theorem
Synonym Fisher-Neyman factorization theorem