3
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ One is a bijective transform of the other, hence they are equivalent. $\endgroup$
    – Xi'an
    Commented Feb 25, 2021 at 16:49

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ 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
    Commented Feb 25, 2021 at 10:04
  • 1
    $\begingroup$ 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
    Commented Feb 25, 2021 at 11:54
  • 1
    $\begingroup$ (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
    Commented Feb 25, 2021 at 12:18
  • 1
    $\begingroup$ 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
    Commented 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
    Commented Feb 25, 2021 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.