Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
@MeesdeVries That is a nice construction, thanks! Is it clear though that these properties still hold in this construction (mean exists, variance does not)?
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:)
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: $\begin{align*} \\ \mu &= \int_{-\infty}^\infty x f(x) dx \\ &= \int_{-\infty}^0 x f(x)dx + \int_0^\infty x f(x)dx \\&= -\int_{\infty}^0 (-x)f(-x)dx + \int_0^\infty x f(x)dx \\&= -\int_0^\infty x f(-x)dx + \int_0^\infty x f(x)dx \\&= -\int_0^\infty x f(x)dx + \int_0^\infty x f(x)dx = 0. \end{align*}$
What I forgot to add is that at the very start of the encoder I ususally used a convolution with a larger kernel (like 5x5 or 7x7) to increase the number of channels to e.g. something between 32 and 128, and again one at the end of the decoder to reduce the number of channels again. thanks a lot for taking the time to look in to this!
