# When does a distribution not have a mean or a variance?

I believe I read today a phrase which went something like this:

If a distribution has a mean and a variance ...

So I guess that means some distributions do not have means or variances?

I fiend that a bit difficult to understand - for example:

I could get data which is said to follow such a mean-less and variance-less distribution and as an estimation compute the mean and variance of such a data set.

could someone help shed some light on what I am missing?

• The Cauchy distribution is a famous distribution that has neither a mean nor a variance (no finite moments in general). This post explains very nicely why not. You can, of course compute the sample mean and variance. Sep 15 '13 at 21:37
• I just read the post - fascinating, fascinating Sep 15 '13 at 21:40
• The mean and/or variance don't exist when the limits implied by the improper integrals don't exist; e.g. for the mean, $\lim_{a,b\rightarrow\infty}\int_{-a}^b x\, dF$ has to exist for the mean to exist (for a continuous density $dF = f(x)dx$). For a non-Cauchy example, consider the discrete distribution $P(X=x)= \frac{6}{\pi^2x^2};\,x=1,2,3,...$, a particular case of the zeta distribution; in this case neither variance nor mean exist (it's the case that if the mean doesn't exist, the variance won't either, but the mean can exist when the variance doesn't). Sep 16 '13 at 0:53
• The wiki article en.wikipedia.org/wiki/… explains well what are you estimating by calculating the sample mean, and how "good" is this estimation. Sep 16 '13 at 23:55
• See this related question. The sample mean and variance will always exist (the sample consists of a finite collection of numbers, for which the sample quantities will be well-defined), it's the things that they estimate that are not finite (either infinite or undefined), and as such the sample estimates can't really convey useful information about the population. Feb 16 '17 at 22:00