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Heads up: my statistics knowledge is very limited, I am not a mathematician.

I've recently learned that given two random variables $x$ and $y$ sampled from the same normal distribution their product $xy$ is not normal but it instead belongs to a modified bessel function distribution of the second kind (I believe this is only for $\mu=0$ and $\sigma = 1$).

What are some well known probability distributions such that $xy$ belongs to the same distribution? I am specially interested in symmetric distributions around the mean.

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Suppose that $X,Y$ are random variables and consider $Z = XY$ \begin{align} F_Z(z) & = P(Z \le z) \\ & = P(XY \le z) \\ & = P(X \le z/Y | Y > 0)P(Y>0) + P(X \ge z/Y | Y < 0) P(Y<0) \\ & = \int_0^\infty \int_{-\infty}^{z/y} f_{X,Y}(x,y) \ dx dy + \int_{-\infty}^0 \int_{z/y}^\infty f_{X,Y}(x,y) \ dx dy \\ f_Z(z) & = \int_0^\infty f_{X,Y}(z/y,y) \frac{1}{y} \ dy - \int_{-\infty}^0 f_{X,Y}(z/y,y) \frac{1}{y} \ dy \\ f_Z(z) & = \int_{-\infty}^\infty f_{X,Y}(z/y,y)\frac{1}{|y|} \ dy. \end{align} Now suppose that $X,Y$ are i.i.d. with common density function $f$ such that $f(x) = f(-x)$, then the above reduces to $$ f_Z(z) = \int_{-\infty}^\infty \frac{f(z/y) f(y)}{|y|} \ dy $$ What you require involves solving the integral equation $$ f(z) = \int_{-\infty}^\infty \frac{f(z/y) f(y)}{|y|} \ dy = 2\int_0^\infty \frac{f(z/y)f(y)}{y} \ dy, $$ which to me is not obvious how to do.

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If $X$ and $Y$ are allowed to be discrete random variables rather than only continuous random variables such as normal random variables, then when $X$ and $Y$ are independent Bernoulli random variables with parameters $p$ and $q$ respectively, $XY$ is also a Bernoulli random variable with albeit with a different parameter $pq$.

If you want symmetric distributions, take $X$ and $Y$ to be independent Bernoulli random variables with parameter $\frac 12$. Then, $W = (-1)^X$ and $Z = (-1)^Y$ are independent random variables taking on values $\pm 1$ with equal probability $\frac 12$ (i.e. have symmetric distributions). Then, $WZ$ also takes on values $\pm 1$ with equal probability $\frac 12$.

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