factorization criterion
Let 𝑿=(X1,…,Xn) be a random vector whose
coordinates are observations, and whose probability (density
)
function is, f(𝒙∣θ) where θ is an
unknown parameter. Then a statistic
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(𝒙∣θ)=g(T(𝑿),θ)h(𝒙). |
Applications.
-
1.
In view of the above statement, let’s show that the sample mean ˉX of n independent
observations from a normal distribution
N(μ,σ2) is a sufficient statistic for the unknown mean μ. Since the Xi’s are independent random variables
, then the probability density function
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) = n∏i=1f(x∣μ)=n∏i=11√2πσ2exp[-(xi-μ)22σ2] (1) = 1√(2π)nσ2nexp[n∑i=1-(xi-μ)22σ2] (2) = 1√(2π)nσ2nexp[-12σ2n∑i=1x2i]exp[μσ2n∑i=1xi-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)=ˉX is a sufficient statistic.
-
2.
Similarly, the above shows that the sample variance s2 is not a sufficient statistic for σ2 if μ is unknown.
-
3.
But, if μ is a known constant, then the statistic
T(X1,…,Xn)=1n-1n∑i=1(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 |