Good day, esteemed students of mathematics!
I have been trying to prove that the convolution of $2q$ Gaussian probability distributions is another $q$ Gaussian probability distribution with the same value of $q$ (which I'm not certain is true). After taking the inverse Fourier transform of the product of the $F$ transforms of $2q$ Gaussians I obtained this
$$f(\omega) = \int_{-\infty}^{\infty}\big[se^{k}+\sigma e^{-k}\big]^{2\nu} \bigg[1+\frac {\omega^2}{(se^k+\sigma e^{-k})(se^{-k}+\sigma e^k)}\bigg]^{2\nu+\frac{1}{2}}dk$$
with $\nu>0$, $s>0$ and $\sigma>0$. $\sigma$ and $s$ are the width parameters of the $2q$ Gaussians I am convolving and $\nu$ is related to the parameter $q$ by
$$\frac{1}{q-1} = \nu + \frac{1}{2}$$
Can somebody prove that $f(\omega)$ can be expressed as a $q$ Gaussian? Or prove that it cannot.
The definition of a $q$ Gaussian can be found here.
Cheers, Allen.