Can a probability distribution have infinite standard deviation?

Question

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

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

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.

Answer 1

To answer your question title: Yes, a probability distribution can have infinite standard deviation (see below).

Your example is a special case of the Cauchy distribution whose mean or variance does not exist. Set the location parameter to 0 and the scale to 1 for the Cauchy to get to your pdf.