The fisher-neyman factorization theorem (ticket #4340)

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Determining a sufficient statistic directly from the definition can be a tedious process. The following result can simplify this process byallowing one to spot a sufficient statistic directly from the functional form of the density or mass function.

Fisher-neyman factorization theorem

Let $f x$ be the density or mass function for the random vector $x$ , parametrized by the vector . The statistic $t T x$ is sufficient for if and only if there exist functions $a x$ (not depending on ) and $b t$ such that $f x a x b t$ for all possible values of $x$ .

In an earlier example we computed a sufficient statistic for a binary communication source (independent Bernoulli trials) from thedefinition. Using the above result, this task becomes substantially easier.

Bernoulli trials revisited

Suppose $x_{n} Bernoulli$ are IID, $n n 1,, N$ . Denote $x x_{1} x_{n}$ . Then

f x n 1 N x_{n} 1 1 x_{n} k 1 N k a x b k

where

k n 1 N x_{n}

a x 1

, and

b k k 1 N k

. By the Fisher-Neyman factorization theorem,

k

is sufficient for

The next example illustrates the appliction of the theorem to a continuous random variable.

Normal data with unknown mean

Consider a normally distributed random sample $x_{1},, x_{N} 1$ , IID, where is unknown. The joint pdf of $x x_{1} x_{n}$ is $f x n 1 N f x_{n} 12 N 2 -1 2 n 1 N x_{n} 2$ We would like to rewrite $f x$ is the form of $a x b t$ , where $dim t N$ . At this point we require a trick-one that is commonly used when manipulating normal densities, and worthremembering. Define $x 1 N n 1 N x_{n}$ , the sample mean. Then

f x 12 N 2 -1 2 n 1 N x_{n} x x 2 12 N 2 -1 2 n 1 N x_{n} x 2 2 x_{n} x x x 2

Now observe

n 1 N x_{n} x x x n 1 N x_{n} x x x x 0

so the middle term vanishes. We are left with

f x 12 N 2 -1 2 n 1 N x_{n} x 2 -1 2 n 1 N x 2

where

a x 12 N 2 -1 2 n 1 N x_{n} x 2

b t -1 2 n 1 N x 2

, and

t x

. Thus, the sample mean is a one-dimensional sufficient statistic for the mean.

Proof of theorem

First, suppose $t T x$ is sufficient for . By definition, $f T x t x$ is independent of . Let $f x t$ denote the joint density or mass function for $(X, T (X))$ . Observe $f x f x t$ . Then

f x f x t f t x f t a x b t

where

a x f t x

and

b t f t

. We prove the reverse implication for the discrete case only. The continuous case follows a similar argument, butrequires a bit more technical work ( Scharf, pp.82 ; Kay, pp.127 ).

Suppose the probability mass function for $x$ can be written $f x a x b x$ where $t T x$ . The probability mass function for $t$ is obtained by summing $f x t$ over all $x$ such that $T x t$ :

f t x T x t f x t x T x t f x x T x t a x b t

Therefore, the conditional mass function of

x

, given

t

, is

f t x f x t f t f x f t a x x T x t a x

This last expression does not depend on

, so

t

is a sufficient statistic for

. This completes the proof.

From the proof, the Fisher-Neyman factorization gives us a formula for the conditionalprobability of

x

given

t

. In the discrete case we have

f t x a x x T x t a x

An analogous formula holds for continuous random variables ( Scharf, pp.82 ).

Further examples

The following exercises provide additional examples where the Fisher-Neyman factorization may be used toidentify sufficient statistics.

Uniform measurements

Suppose $x_{1},, x_{N}$ are independent and uniformly distributed on the interval $_{1}_{2}$ . Find a sufficient statistic for $_{1}_{2}$ .

Express the likelihood

f x

in terms of indicator functions.

Poisson

Suppose $x_{1},, x_{N}$ are independent measurements of a Poisson random variable with intensity parameter : $x x 0, 1, 2, f x x x$

Find a sufficient statistic $t$ for .

What is the conditional probability mass function of $x$ , given $t$ , where $x x_{1} x_{N}$ ?

Normal with unknown mean and variance

Consider $x_{1},, x_{N} 2$ , IID, where $_{1}$ and $_{2} 2$ are both unknown. Find a sufficient statistic for $_{1}_{2}$ .

Use the same trick as in .

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Source: OpenStax, Pdf generation problem modules. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10514/1.4

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