Product of two independent Student distributions What is the product of two independent student t distributions?
In which case does this product product result in another t distribution?
 A: When $X$ and $Y$ are independent random variables with densities $f_X$ and $f_Y,$ the density of their product can be found with a change of variables as
$$f_{XY}(z) = \int_{\mathbb R} f_X(x) f_Y(z/x)\,\frac{\mathrm{d}x}{|x|}.$$
Ignoring normalizing constants (we'll consider these at the end), for two Student t densities with $\nu$ and $\mu$ degrees of freedom this integrand is proportional to
$$h(x,z) = \left(1 + \frac{x^2}{\mu}\right)^{-(\mu+1)/2}\,  \left(1 + \frac{z^2}{x^2\nu}\right)^{-(\nu+1)/2}\,\frac{1}{|x|}.$$
Let's find a lower bound for $f(z)$ when $z$ is small.  To do so, we may restrict the region of integration and we may replace the integrand by anything that never exceeds it.
Let $z$ be positive but less than $1$ and consider the integration region where $x^2\nu$ ranges between $z^2$ and $1.$  To get an appreciation for what's going on, here (with $\mu=\nu=1$) are plots of $h(x,z)$ for $|z| = 1$ (blue), $1/2, 1/4,$ and $1/8$ (red).

You can see that as $|z|$ approaches $0,$ there's more and more area pushed into this region.  That's no surprise: we would expect the largest area (which corresponds to the highest density of the product) to be at the center of the product distribution, which (by symmetry) must be $0.$  But how large does it get?
Over the region $x^2/\nu \in[z^2, 1]$ the first factor of $h$ is smallest when $x$ is smallest and the second factor is smallest when $x$ is largest, whence throughout this region
$$\begin{aligned}
h(x,z) &\ge \left(1 + \frac{z^2}{\mu\nu}\right)^{-(\mu+1)/2}\,  \left(1 + \frac{z^2}{1}\right)^{-(\nu+1)/2}\,\frac{1}{|x|} \\
&\ge \left(1 + \frac{1}{\mu\nu}\right)^{-(\mu+1)/2}\,  \left(1 + 1\right)^{-(\nu+1)/2}\,\frac{1}{|x|}.
\end{aligned}$$
The second inequality is a consequence of $z^2 \le 1.$
The factors before $1/|x|$ are constant (but nonzero), depending only on $\mu$ and $\nu,$ so again let's consider them later and ignore them now. As $x$ varies over just the positive part of this region it runs from $z/\sqrt{\nu}$ to $1/\sqrt{\nu},$ giving a lower bound proportional to
$$\int_{z/\sqrt{\nu}}^{1/\sqrt{\nu}} \frac{\mathrm{d}x}{|x|} = -\log z.$$
As $z\to 0,$ this lower bound diverges.  Consequently, no matter what the constants of proportionality are that we ignored, $f_{XY}(z)$ diverges at $0.$
Here, to illustrate, is a histogram from a simulation of ten million products (with $\nu=\mu=1/2$).  Almost a million of those products are represented.  The red curve is the negative logarithm.  Clearly it approximates the density well near zero.

However, any Student t distribution with (say) $\kappa \gt 0$ degrees of freedom has a value proportional to $(1 + 0^2/\kappa)^{-(\kappa+1)/2} = 1$ at the origin, which is finite.  Consequently, the product of two independent Student t distributions is never (even remotely like) a Student t distribution.

The product density can be found analytically as a polynomial combination of Riemann hypergeometric functions.  Since this product is never a Student t distribution, though, I did not see any point into providing further details.
