factorization criterion

Let 𝑿=(X1,…,Xn) be a random vector whose coordinatesPlanetmathPlanetmath are observations, and whose probability (densityPlanetmathPlanetmath) function is, f(𝒙∣θ) 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=g⁢h where g is a function of T⁢(𝑿) and θ, and h is a function of 𝒙. In symbol, we have



  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(𝒙∣μ), being the joint probability density function of each of the Xi, is the product of the individual density functions f(x∣μ):

    f(𝒙∣μ) = ∏i=1nf(x∣μ)=∏i=1n12⁢π⁢σ2exp[-(xi-μ)22⁢σ2] (1)
    = 1(2⁢π)n⁢σ2⁢n⁢exp⁡[∑i=1n-(xi-μ)22⁢σ2] (2)
    = 1(2⁢π)n⁢σ2⁢n⁢exp⁡[-12⁢σ2⁢∑i=1nxi2]⁢exp⁡[μσ2⁢∑i=1nxi-n⁢μ22⁢σ2] (3)
    = h⁢(𝒙)⁢exp⁡[n⁢μσ2⁢T⁢(𝒙)-n⁢μ22⁢σ2] (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


    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