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Is there a symmetric continuous distribution that has a finite mean, but no variance?

What I've found so far: For instance the Pareto distribution satisfies everything but the symmetry, so I was wondering, can we also construct a symmetric distribution?

It is easy to see that if the mean exists for a symmetric distribution, it must be zero. So if the variance $\sigma^2$ did not exist, the integral $\int_{-\infty}^\infty (x-\mu)^2 f(x) dx= 2\int_0^\infty x^2 f(x)$ would have to diverge. But I did not manage to construct such an $f$ with a finite mean.

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    $\begingroup$ You can make any distribution that satisfies your criteria symmetric by independently sampling from it or its negation with probability 1/2 each. $\endgroup$
    – Mees de Vries
    Commented May 2, 2021 at 22:08
  • $\begingroup$ @MeesdeVries That is a nice construction, thanks! Is it clear though that these properties still hold in this construction (mean exists, variance does not)? $\endgroup$
    – flawr
    Commented May 3, 2021 at 21:07

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The $t$ distribution with two degrees of freedom satisfies your criteria. A $t$ distribution has a mean of zero when the degrees of freedom are greater than $1$ (otherwise undefined) but has no variance until the degrees of freedom exceeds $2$.

All $t$ distributions are continuous and symmetric.

I disagree that the mean of such a distribution has to be zero, however, to satisfy your criteria. A shifted $t$ distribution satisfies your criteria but can be centered anywhere.

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  • $\begingroup$ Thanks a lot! Regarding the mean having to be zero: If the PDF is symmetric in the sense that $f(x) = f(-x)$ for all $x$, then the mean should be zero: $\begin{align*} \\ \mu &= \int_{-\infty}^\infty x f(x) dx \\ &= \int_{-\infty}^0 x f(x)dx + \int_0^\infty x f(x)dx \\&= -\int_{\infty}^0 (-x)f(-x)dx + \int_0^\infty x f(x)dx \\&= -\int_0^\infty x f(-x)dx + \int_0^\infty x f(x)dx \\&= -\int_0^\infty x f(x)dx + \int_0^\infty x f(x)dx = 0. \end{align*}$ $\endgroup$
    – flawr
    Commented May 1, 2021 at 23:35
  • $\begingroup$ Yes - note that symmetry becomes irrelevant when you put an $x^2$ in the integrand, or consider $|f|$ instead of $f$. $\endgroup$
    – rubikscube09
    Commented May 1, 2021 at 23:49
  • $\begingroup$ Yes, it depends on your definition of symmetric. You mean that the PDF is an even function. My definition relaxes the assumptions and just means symmetry about the mean. I would expect most statisticians to use my definition (that’s why I use it), but if your field or your class uses your definition, go for it. $\endgroup$
    – Dave
    Commented May 1, 2021 at 23:56
  • $\begingroup$ @Dave Thanks for clarifying, I just used the function-notion of symmetry, I wasn't aware of the one used in stats, but I guess the idea is the same. $\endgroup$
    – flawr
    Commented May 2, 2021 at 8:43
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    $\begingroup$ The statistical definition of symmetrical distribution is the same as the mathematical one: see stats.stackexchange.com/a/29010/919 and (for an even more general definition) stats.stackexchange.com/a/185709/919. In particular, it does not require a mean to exist. The definition employed in the comments is overly limited in that it requires the distribution to have a density with finite absolute first moment. Without that latter assumption, the manipulations in the first comment are incorrect, because they assert that $-\infty+\infty=0.$ $\endgroup$
    – whuber ♦
    Commented May 2, 2021 at 12:15
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The symmetric α-stable distribution with $\alpha \in (1,2)$.

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    $\begingroup$ Thanks, I didn't know this class of distributions! $\endgroup$
    – flawr
    Commented May 1, 2021 at 23:30

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