# linear transform of a random variable follows multivariate normal, what is the distribution before the transform?

$$x$$ 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 $$\Sigma_z$$?

## Update:

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 $$\mu_{x_1} + \mu_{x_2} = \mu_z \\ \Sigma_{x_1}+\Sigma_{x_2} = \Sigma_z$$

• 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 $x.$
– whuber
Commented Jul 20, 2022 at 15:16
• 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). Commented Jul 20, 2022 at 19:54
• 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.
– whuber
Commented Jul 20, 2022 at 22:32
• 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 ! Commented Jul 21, 2022 at 6:52