Timeline for Symmetric distribution with finite Mean but no Variance
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 2, 2021 at 12:15 | comment | added | whuber♦ | 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 $-\infty+\infty=0.$ | |
| May 2, 2021 at 9:50 | comment | added | flawr | 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:) | |
| May 2, 2021 at 8:43 | comment | added | flawr | @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. | |
| May 1, 2021 at 23:56 | comment | added | Dave | 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. | |
| May 1, 2021 at 23:49 | comment | added | rubikscube09 | Yes - note that symmetry becomes irrelevant when you put an $x^2$ in the integrand, or consider $|f|$ instead of $f$. | |
| May 1, 2021 at 23:38 | vote | accept | flawr | ||
| May 1, 2021 at 23:35 | comment | added | flawr | 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: $\begin{align*} \\ \mu &= \int_{-\infty}^\infty x f(x) dx \\ &= \int_{-\infty}^0 x f(x)dx + \int_0^\infty x f(x)dx \\&= -\int_{\infty}^0 (-x)f(-x)dx + \int_0^\infty x f(x)dx \\&= -\int_0^\infty x f(-x)dx + \int_0^\infty x f(x)dx \\&= -\int_0^\infty x f(x)dx + \int_0^\infty x f(x)dx = 0. \end{align*}$ | |
| May 1, 2021 at 23:05 | history | answered | Dave | CC BY-SA 4.0 |