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Sufficient statistic
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===Poisson distribution=== If ''X''<sub>1</sub>, ...., ''X''<sub>''n''</sub> are independent and have a [[Poisson distribution]] with parameter ''Ξ»'', then the sum ''T''(''X'') = ''X''<sub>1</sub> + ... + ''X''<sub>''n''</sub> is a sufficient statistic for ''Ξ»''. To see this, consider the joint probability distribution: :<math> \Pr(X=x)=P(X_1=x_1,X_2=x_2,\ldots,X_n=x_n). </math> Because the observations are independent, this can be written as :<math> {e^{-\lambda} \lambda^{x_1} \over x_1 !} \cdot {e^{-\lambda} \lambda^{x_2} \over x_2 !} \cdots {e^{-\lambda} \lambda^{x_n} \over x_n !} </math> which may be written as :<math> e^{-n\lambda} \lambda^{(x_1+x_2+\cdots+x_n)} \cdot {1 \over x_1 ! x_2 !\cdots x_n ! } </math> which shows that the factorization criterion is satisfied, where ''h''(''x'') is the reciprocal of the product of the factorials. Note the parameter Ξ» interacts with the data only through its sum ''T''(''X'').
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