# Can a probability distribution have infinite standard deviation?

I believe $p[x]$ is a probability distribution, where

$$p[x] = \frac{1}{\pi (1+x^2)}$$

since it's positive everywhere and integrates to 1 on $-\infty, \infty$.

The mean is 0 by symmetry, even though integrating $xp[x]$ on $-\infty, \infty$ does not converge. This is "suspicious" since $p[x]$ is supposed to be a probability distribution, but reasonable because $xp[x]$ is $O(1/x)$ which is known to diverge.

The bigger problem is in computing the standard deviation. Since $x^2 p[x]$ also diverges, since $x^2 p[x]$ is $O(1)$.

If this isn't a probability distribution, why not? If it is, is its standard deviation infinite?

The cumulative distribution function is $\arctan[x]/\pi$ if that helps.

Someone mentioned this might be a gamma distribution, but that isn't clear to me.

-
@user1566: I formatted your equations using LaTex. Would you double check that I didn't introduce any errors? –  csgillespie Nov 9 '10 at 21:31
Thanks, the problem is solved, so no longer a biggie, but, yes, everything looks OK. –  barrycarter Nov 9 '10 at 22:11
The mean of a Cauchy is not zero. In fact, it doesn't exist. Thus, neither does any of its central moments. –  cardinal Apr 24 '11 at 18:55