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Suppose

\begin{equation} A_1 \sim \mathcal{W}(A_1; \Psi_1, v_1)\\ A_2 \sim \mathcal{W}(A_2; \Psi_2, v_2), \end{equation}

where $\mathcal{W}$ is the Wishart distribution and $A_i$, and $\Psi_i$ are PSD matrices. Does

\begin{equation} A_1 + A_2 \end{equation} have a closed-form distribution?

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    $\begingroup$ It does have a closed form (Wishart) only if the scale matrices coincide. $\endgroup$
    – utobi
    Jan 9 at 5:31
  • $\begingroup$ Hey do you mean the matrices commute? ie $\Psi_1\Psi_2 = \Psi_2\Psi_1$ ? I cant find a definition of coinciding matrices. tx $\endgroup$ Jan 9 at 6:01
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    $\begingroup$ I mean that they should be equal. $\endgroup$
    – utobi
    Jan 9 at 6:03

1 Answer 1

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Depending on what you mean by closed-form, the answer is to some extent, yes.

As per my comment, if $\Psi_1=\Psi_2$, then it can be shown that $A = A_1+A_2$ follows a Wishart distribution with scale matrix $\Psi_1$ and degrees of freedom $\nu_1+\nu_1$.

On the other, if $\Psi_1\neq \Psi_2$ (but they are both square and with $p$ rows), the distribution is not Whishart anymore. Indeed, in this case, the pdf of $A$ is

$ \left\{2^{\frac{1}{2}(\nu_1+\nu_2)p}\Gamma_p\left(\frac{1}{2}(\nu_1+\nu_2)\right)\det(\Psi_1)^{\frac{1}{2}\nu_1}\det(\Psi_2)^{\frac{1}{2}\nu_2}\right\}^{-1}\text{etr}\left(-\frac{1}{2}\Psi_1^{-1}A\right)\det(A)^{\frac{1}{2}(\nu_1+\nu_2-p-1)} {_1}F_1\left(\frac{1}{2}\nu_2;\frac{1}{2}(\nu_1+\nu_2);\frac{1}{2}(\Psi_1^{-1}-\Psi_2^{-1})A\right), A>0, $

where $\Gamma_p(a)= \int_{A>0}\text{etr}(-A)\det(A)^{a-\frac{1}{3}(p+1)}dA$, where $\text{Re}(a)>\frac{1}{2}(p-1)$ and the integral is over the space of $p\times p$ symmetric positive definite matrices, ${_1}F_1$ is the confluent hypergeometric function (of the first kind) and $\text{etr}(A) = \exp(\text{tr}A)$.

For details and proofs check Chapter 3 of Gupta and Nagar (1999) Matrix Variate Distributions, Monographs and Surveys in Pure and Applied Mathematics 104, Chapman & Hall/CRC.

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