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factorization criterion
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(Theorem)
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Let
 be a random vector whose coordinates are observations, and whose probability (density) function is,
 where  is an unknown parameter. Then a statistic
 for  is a sufficient statistic iff  can be expressed as a product of (or factored into) two functions  ,  where  is a function of
 and  , and  is a function of
 . In symbol, we have
Applications.
- In view of the above statement, let's show that the sample mean
 of  independent observations from a normal distribution
 is a sufficient statistic for the unknown mean  . Since the  's are independent random variables, then the probability density function
 , being the joint probability density function of each of the  , is the product of the individual density functions
 :
where  is the last exponential expression and  is the rest of the expression in  . By the factorization criterion,
 is a sufficient statistic.
- Similarly, the above shows that the sample variance
 is not a sufficient statistic for  if  is unknown.
- But, if
 is a known constant, then the statistic
is sufficient for  by observing in  above, and letting
 and
 be all of expression  .
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"factorization criterion" is owned by CWoo.
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(view preamble)
| Other names: |
factorization theorem, Fisher-Neyman factorization theorem |
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Cross-references: sufficient, sample variance, expression, exponential, density functions, probability density function, random variables, mean, normal distribution, independent, sample mean, product, iff, sufficient statistic, statistic, parameter, function, density, observations, coordinates, random vector
There is 1 reference to this entry.
This is version 1 of factorization criterion, born on 2005-02-16.
Object id is 6761, canonical name is FactorizationCriterion.
Accessed 4047 times total.
Classification:
| AMS MSC: | 62B05 (Statistics :: Sufficiency and information :: Sufficient statistics and fields) |
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Pending Errata and Addenda
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![$\displaystyle \prod_{i=1}^n f(x\mid\mu)= \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}}\exp\Big[-\frac{(x_i-\mu)^2}{2\sigma^2}\Big]$](http://images.planetmath.org:8080/cache/objects/6761/l2h/img25.png "$\displaystyle \prod_{i=1}^n f(x\mid\mu)= \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}}\exp\Big[-\frac{(x_i-\mu)^2}{2\sigma^2}\Big]$")
![$\displaystyle \frac{1}{\sqrt{(2\pi)^n\sigma^{2n}}}\exp\Big [\sum_{i=1}^{n}-\frac{(x_i-\mu)^2}{2\sigma^2}\Big]$](http://images.planetmath.org:8080/cache/objects/6761/l2h/img27.png "$\displaystyle \frac{1}{\sqrt{(2\pi)^n\sigma^{2n}}}\exp\Big [\sum_{i=1}^{n}-\frac{(x_i-\mu)^2}{2\sigma^2}\Big]$")
![$\displaystyle \frac{1}{\sqrt{(2\pi)^n\sigma^{2n}}}\exp\Big [\frac{-1}{2\sigma^2... ...ig] \exp\Big[\frac{\mu}{\sigma^2}\sum_{i=1}^n x_i-\frac{n\mu^2}{2\sigma^2}\Big]$](http://images.planetmath.org:8080/cache/objects/6761/l2h/img29.png "$\displaystyle \frac{1}{\sqrt{(2\pi)^n\sigma^{2n}}}\exp\Big [\frac{-1}{2\sigma^2... ...ig] \exp\Big[\frac{\mu}{\sigma^2}\sum_{i=1}^n x_i-\frac{n\mu^2}{2\sigma^2}\Big]$")
![$\displaystyle h(\boldsymbol{x}) \exp\Big[\frac{n\mu}{\sigma^2}T(\boldsymbol{x})-\frac{n\mu^2}{2\sigma^2}\Big]$](http://images.planetmath.org:8080/cache/objects/6761/l2h/img31.png "$\displaystyle h(\boldsymbol{x}) \exp\Big[\frac{n\mu}{\sigma^2}T(\boldsymbol{x})-\frac{n\mu^2}{2\sigma^2}\Big]$")
