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Let $\mathbf{A}$ be an $m\times n$ random matrix with entries $A_{ij}$ being jointly Gaussian. Suppose all of these variables are independent of the random vector $\mathbf{X} = (X_1,\ldots,X_n)^\top$ which follows a multivariate Gaussian distribution.

Does the random vector $\mathbf{Y} = \mathbf{A}\mathbf{X}$ follow any particular distribution? Maybe some sort of modified $\chi^2$-distribution? I'm mostly interested in some expression for the covariance matrix of $\mathbf{Y}$ that can hopefully be expressed in terms of the the expected values and covariances associated with $\mathbf{A}$ and $\mathbf{X}$. For simplicity, it is okay to assume $\mathbf{X}$ to have mean zero.

Any guidance is appreciated.

EDIT 1: For instance, I realized that, for $m=n=1$, we have \begin{align*} \text{var}(AX) &= \mathbb{E}(A^2)\mathbb{E}(X^2) - \mathbb{E}^2(A)\mathbb{E}^2(X)\\ &= (\text{var}(A)+\mathbb{E}^2(A))(\text{var}(X)+\mathbb{E}^2(X)) - \mathbb{E}^2(A)\mathbb{E}^2(X). \end{align*}.

EDIT 2: While not exactly what I wanted, I found that denoting $\mathbf{A}_1,\ldots,\mathbf{A}_n$ the columns of $\mathbf{A}$, then: \begin{align*} \text{cov}(\mathbf{A}\mathbf{X}) &= \text{cov}\left(\sum_{i=1}^n X_i\mathbf{A}_i,\sum_{j=1}^n X_j\mathbf{A}_j\right)\\ &= \sum_{i=1}^n\sum_{j=1}^n \text{cov}(X_i\mathbf{A}_i,X_j\mathbf{A}_j)\\ &= \sum_{i=1}^n\sum_{j=1}^n \mathbb{E}\{(X_i\mathbf{A}_i)(X_j\mathbf{A}_j)^\top\} &\text{(*)}\\ &= \sum_{i=1}^n\sum_{j=1}^n \mathbb{E}\{X_iX_j\mathbf{A}_i\mathbf{A}_j^\top\}\\ &= \sum_{i=1}^n\sum_{j=1}^n \mathbb{E}(X_iX_j)\mathbb{E}(\mathbf{A}_i\mathbf{A}_j^\top) &\text{(**)}\\ &= \sum_{i=1}^n\sum_{j=1}^n \text{cov}(X_i,X_j)[\text{cov}(\mathbf{A}_i,\mathbf{A}_j) + \mathbb{E}(\mathbf{A}_i)\mathbb{E}(\mathbf{A}_j)^\top]. \\ \end{align*}

(*) $\mathbb{E}(X_i\mathbf{A}_i) = \mathbb{E}(X_i)\mathbb{E}(\mathbf{A}_i)$, from the independence assumption on $\mathbf{A}$ and $\mathbf{X}$, and hence $\mathbb{E}(X_i\mathbf{A}_i) = \mathbf{0}$ from the simplifying assumption that $\mathbb{E}(\mathbf{X}) = \mathbf{0}$ (since it makes no difference in my original problem).

(**) Again, independence of $\mathbf{A}$ and $\mathbf{X}$.

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    $\begingroup$ In some extremely general sense the marginals are linear combinations of interdependent $\chi^2(2)$ distributions, but there's no hope they even remotely resemble a $\chi^2$: they are just as likely to have negative values as positive ones. For a glimmer of what this might look like, and a reference, see stats.stackexchange.com/questions/48378. $\endgroup$
    – whuber ♦
    Commented Feb 3, 2018 at 19:46
  • $\begingroup$ Yes, that's a great point about it not being non-negative. But still, do you believe there is any hope to obtain some form of closed-form expression for the covariance matrix? (in terms of what we \emph{know}: the expectations and covariances between the different components of $\mathbf{A}$ and $\mathbf{X}$). Thanks! $\endgroup$
    – Orlando
    Commented Feb 3, 2018 at 19:48
  • $\begingroup$ Not closed form, no. Possibly under very special assumptions about the distributions of $A$ and $X$ that could be done, but not generally: the correlations make it extremely messy. $\endgroup$
    – whuber ♦
    Commented Feb 3, 2018 at 19:50
  • $\begingroup$ That's a shame, but thank you anyway! $\endgroup$
    – Orlando
    Commented Feb 3, 2018 at 19:52
  • $\begingroup$ Please take a look at my new edit, if you are interested. Hopefully I did not commit any mistake. $\endgroup$
    – Orlando
    Commented Feb 3, 2018 at 20:22

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