Does the product of two t-distribution formulas with same degrees of freedom simplify?

$T_v(x; \mu_1, \Sigma_1)T_v(x; \mu_2, \Sigma_2) =\ ?...$

In the normal case it simplifies to:

$\mathcal{N}(x; \mu_1, \Sigma_2)\mathcal{N}(x; \mu_1, \Sigma_2) \propto \mathcal{N}(x; \mu_3, \Sigma_3)$
$\Sigma_3 = (\Sigma_1^{-1} + \Sigma_2^{-1})^{-1} \ \ \ $ and $\ \ \ \mu_3 = \Sigma_3\Sigma_1^{-1}\mu_1 + \Sigma_3\Sigma_2^{-1}\mu_2$


  • $\begingroup$ Please explain what you mean by a distribution "formula:" would that refer to a CDF, a pdf, a characteristic function, a random variable, or possibly something else? $\endgroup$
    – whuber
    Commented Feb 3 at 14:32
  • $\begingroup$ Just multiplying the PDFs by each-other, I acknowledge this does not give a valid PDF. But It will be proportional to one. I am not interested in PDF of: $x_1 \times x_2$ st $x_1 \sim t$ and $x_2 \sim t$. $\endgroup$ Commented Feb 4 at 5:04
  • $\begingroup$ What have you done to check? Have you, for instance, performed the calculation for univariate distributions where the algebra is the simplest? If you don't want to do any algebra, graph a few examples. E.g., take $\mu_1=0,$ $\mu_3=10.$ If significant simplification is possible, the graph will be unimodal. $\endgroup$
    – whuber
    Commented Feb 4 at 14:54
  • $\begingroup$ For my application I could simplify it to where $\mu_1 = \mu_2$. I have been approximating it by sampling and approximating by a mixture of model of 2 $t$-distributions. Seems to work fine enough, though an analytic solution would be nice. $\endgroup$ Commented Feb 5 at 3:39
  • 1
    $\begingroup$ By using partial fractions I believe you can indeed express that case as a mixture of t distributions provided $\nu$ is a positive odd integer. That suggests a mixture representation would likely be a good approximation for other values of $\nu,$ too. $\endgroup$
    – whuber
    Commented Feb 5 at 14:35


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