$x$ is a $n\times 1$ random vector,$A$ is a $m\times n$ matrix. Given that $x$'s linear transform $z = Ax$ follows a multivariate normal distribution: $$ z = Ax \sim N(\mu_z,\Sigma_z) $$

The question is, what is the distribution of $x$? Is it also multivariate normal? If yes, what are the corresponding mean and covariance matrix, in terms of $A$, $\mu_z$ and $\Sigma_z$?


Thanks for @whuber 's comment. It seems that I have overcomplicated the problem. The problem appear as some kind of statistical topic but is actually a matter of whether an arbitary linear transform has a unique inverse, and of course only when the linear transform matrix $A$ is invertible. If $A$ is not invertible, then we can't find a uniquely defined distribution for $x$.

A simple example, if: $$ x=\begin{pmatrix} x_1\\ x_2 \end{pmatrix}, A=\begin{pmatrix} 1 & 1 \end{pmatrix} $$ then $$ Ax = x_1+x_2 \sim N(\mu_z,\Sigma_z) $$ Even if $x_1$ and $x_2$ each follows an independent Gaussian with mean $\mu_{x_1},\Sigma_{x_1}$ and $\mu_{x_2},\Sigma_{x_2}$, there are still infinite possible instances of $\mu_{x_1},\Sigma_{x_1},\mu_{x_2},\Sigma_{x_2}$ that meet the condition $$ \mu_{x_1} + \mu_{x_2} = \mu_z \\ \Sigma_{x_1}+\Sigma_{x_2} = \Sigma_z $$

  • $\begingroup$ If $A$ is invertible, this question comes down to computing the mean and variance of a linear transformation of a multivariate Normal variable (which is answered in many threads). If $A$ is not invertible, you cannot find the distribution of $x$ from the information given. Please explain, then, what you assume about $A$ and $x.$ $\endgroup$
    – whuber
    Commented Jul 20, 2022 at 15:16
  • $\begingroup$ Hi @whuber, thanks for the comment. There is no assumption made on $A$ and $x$. Your are right that it's quite clear when $A$ is invertible. I was wondering if $x$ can have some sort of closed form distribution when $A$ meet certain criterias (other than invertible). $\endgroup$ Commented Jul 20, 2022 at 19:54
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    $\begingroup$ Not at all. To see why not, consider the very simplest case where $A$ is the zero matrix: multiplying by $A$ destroys all information about $x.$. In the general case, $A$ decomposes as a direct sum of an invertible matrix and a zero matrix, so the same insight applies. $\endgroup$
    – whuber
    Commented Jul 20, 2022 at 22:32
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    $\begingroup$ That says there's always information loss about $x$ when $A$ is not invertible, which makes it impossible to uniquely maps back to $x$'s original distribution. Thanks for the answer @whuber ! $\endgroup$ Commented Jul 21, 2022 at 6:52


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