3
$\begingroup$

Consider gaussian variables belonging to a gaussian distribution with expected value $\mu$ and stadard deviation $\sigma$ (e.g. repeated measures). The mean values of different sets of these variables (e.g. different set of measures) belong to another gaussian distribution with same expected value but stadard deviation $\sigma / \sqrt{n}$ (n is the number of measures in each set).


Consider now the same case with poissonian distribution: for each set the distributrion is poissonian with expected value $\lambda$ and variance $\lambda$ . Are the means distributed still as poissonian with same expected value $\lambda$ and variance $\lambda/n$?

In other words is the error on mean value of poissonian distribution equal to $\sqrt{\lambda /n}$ and is this error the squared root of the variance of a poisson distribution the mean values belong to?

$\endgroup$
5
$\begingroup$

Nope! Poisson RVs only assume integral values, but the mean of two integers may be non-integral. The sum of Poisson RVs is Poisson. The sum is also the sufficient statistic for a Poisson distribution. So the mean is a scaled Poisson distribution, but that creates an unnecessary step: if you are trying to characterize the distribution of the sufficient statistic for the purpose of inference and estimation, focus on the sum. The variance of the scaled Poisson would be as you stated since $\text{var}(n^{-1}\sum X ) = n^{-2} \sum \text{var}(X) = n^{-1} \text{var}(X)$, which is actually a general result for IID random variables.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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