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I've been working on some problems, the question asked me if the mean of a sample is a sufficient statistic for poisson distribution. I've already proved that $T(X)=\Sigma_{i=1}^n X_i$ is a sufficient statistic. Can I just conclude that the mean is also a sufficient statistic since we are just dividing by $n$?

And can we say the same about completeness? If $T(X)=\Sigma_{i=1}^n X_i$ is complete then the mean is also complete?

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  • $\begingroup$ Yes by factorization theorem and definition of complete statistic $\endgroup$
    – bdeonovic
    Apr 1 '21 at 21:04
  • $\begingroup$ Two statistics $S$ and $T$ are equivalent if there exists a one-to-one function $f$ such that $S = f(T)$. If two statistics are equivalent and one is a sufficient statistic, then so is the other. $\endgroup$ Apr 1 '21 at 21:13
  • $\begingroup$ @bdeonovic thank you for answering, can you check my other question stats.stackexchange.com/questions/517792/…? $\endgroup$
    – Hijaw
    Apr 1 '21 at 21:14
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Using wikipedia's intuitive definition:

Roughly, given a set $\mathbf {X}$ of independent identically distributed data conditioned on an unknown parameter $\theta$ , a sufficient statistic is a function $T(\mathbf {X} )$ whose value contains all the information needed to compute any estimate of the parameter.

If $T(X)$ contains all the information, $T(X)/n$ contains all the information as well.

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$ f(X|mean=y,\theta)=f(X|\sum X_i =ny, \theta)=f(X| \sum X_i=ny) =f(X| mean =y) $

So by definiton, yes

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