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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.

flawr
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    You can make any distribution that satisfies your criteria symmetric by independently sampling from it or its negation with probability 1/2 each. – Mees de Vries May 02 '21 at 22:08
  • @MeesdeVries That is a nice construction, thanks! Is it clear though that these properties still hold in this construction (mean exists, variance does not)? – flawr May 03 '21 at 21:07

2 Answers2

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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.

Dave
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  • 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: $\begin{align*} \\ \mu &= \int_{-\infty}^\infty x f(x) dx \\ &= \int_{-\infty}^0 x f(x)dx + \int_0^\infty x f(x)dx \\&= -\int_{\infty}^0 (-x)f(-x)dx + \int_0^\infty x f(x)dx \\&= -\int_0^\infty x f(-x)dx + \int_0^\infty x f(x)dx \\&= -\int_0^\infty x f(x)dx + \int_0^\infty x f(x)dx = 0. \end{align*}$ – flawr May 01 '21 at 23:35
  • Yes - note that symmetry becomes irrelevant when you put an $x^2$ in the integrand, or consider $|f|$ instead of $f$. – rubikscube09 May 01 '21 at 23:49
  • 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. – Dave May 01 '21 at 23:56
  • @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. – flawr May 02 '21 at 08:43
  • I know it's a little bit nitpicky, but maybe *symmetry around the mean* is not the best way to describe it, if you e.g. consider a shifted Cauchy distribution:) – flawr May 02 '21 at 09:50
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    The statistical definition of symmetrical distribution is the same as the mathematical one: see https://stats.stackexchange.com/a/29010/919 and (for an even more general definition) https://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 $-\infty+\infty=0.$ – whuber May 02 '21 at 12:15
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The symmetric α-stable distribution with $\alpha \in (1,2)$.

user76284
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