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From Wikipedia:

If $X_{1},\dots, X_{n}$ are independent and have a Poisson distribution with parameter $\lambda$, then the sum $T(X) = X_{1} + ... + X_{n}$ is a sufficient statistic for $\lambda$.

Why the sufficient statistics for $\lambda$ is not $\frac{1}{n}\sum{X_{i}}$ instead?

I struggle to understand how the sum of the observations can say anything about the mean ($\lambda$) of the distribution that generated the observations. The vectors of observations 3 6 2 2 and 3 6 2 2 0 2 2 0 7 2 have different sums but they are generated from the same distribution with $\lambda = 3$.

I suspect my question is a duplicate of Sufficient statistic for Poisson in wiki? by I'm afraid I still don't get it.

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    $\begingroup$ One is a bijective transform of the other, hence they are equivalent. $\endgroup$
    – Xi'an
    Feb 25, 2021 at 16:49

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If $n$ is known, both are sufficient. More generally, any statistic which is in 1-1 correspondence with a sufficient statistic, is sufficient. What matters is the partition different values of the statistic induce in the sample space, not the particular values of the sufficient statistic.

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    $\begingroup$ What matters is the partition different values of the statistic induce in the sample space, not the particular values of the sufficient statistic. But then a sufficient statistics for the mean $\theta$ of a normal distribution could also be $\sum{X_{i}}$ instead of $\frac{1}{n}\sum{X_{i}}$ as reported on Wikipedia? I surmise the answer is no... but why? $\endgroup$
    – dariober
    Feb 25, 2021 at 10:04
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    $\begingroup$ Right. If $n$ is known, both are sufficient for the mean of a normal. (Both give the same information, you can always obtain one from the other.) $\endgroup$
    – F. Tusell
    Feb 25, 2021 at 11:54
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    $\begingroup$ (Thanks for your patience) You say If n is known. My understanding is/was that a sufficient statistic gives you everything you need to know to estimate the parameter of interest. For me, this means the $n$ cannot be assumed to be known. Or is $n$ "allowed" because it's a property of the sample rather than a statistic about the population? $\endgroup$
    – dariober
    Feb 25, 2021 at 12:18
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    $\begingroup$ If $n$ is not known, then neither $\bar{X}$ nor $\sum X_i$ are sufficient for the mean. If $n$ is known, then either are sufficient. It may be clearer if you take a specific situation. Suppose $n=100$. Someone tells you that $\bar{X}=3.18$. Then, you automatically know that $\sum X_i=318$. On the other hand, if they told you that $\sum X_i=318$, then you would know $\bar{X}$. They both have the same information from the sample because once you know one, you know the other. $\endgroup$
    – John L
    Feb 25, 2021 at 14:57
  • $\begingroup$ @JohnL - I feel dumb here - If you tell me that $\bar{X} = 3.18$ then I don't need to know what $n$ was. On the other hand, $\sum{X_{i}} = 318$ is meaningless without knowing $n$. If I'm allowed to communicate just one number for the estimate of $\mu$, then $\bar{X}$ is not only better but is also everything the receiver of my communication needs to know. The same cannot be said about the $\sum{X_{i}}$ (That's my informal understanding of sufficient statistic). $\endgroup$
    – dariober
    Feb 25, 2021 at 17:49

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