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It is well known that affine transformations of Gaussian vectors are also Gaussian.

Let $Y \sim \mathcal{N}( \mathbf{0}, I)$ be an $n$ dimensional iid Gaussian random vector. Let $A$ be an $n \times m$ matrix that does not have a left inverse. Let $A X \stackrel{d}{=} Y$. What can we say about the distribution of $X$?

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We need $m>n$ and $A$ full rank. Then, one option is $X=A^T (AA^T)^{-1}W$, where $W\sim \mathcal N (0, I_n)$. In particular, this means that $X$ has a degenerate Gaussian distribution as it's covariance matrix $A^T (AA^T)^{-2}A^T$ is singular.

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  • $\begingroup$ While I think this answer has some value (+1), I think there are two problems (unless I am mistaken, please let me know): (1) The question specifies that $A$ does not have a left-inverse, and you have used the Moore-Penrose pseudo-inverse. (2) I think you have $m > n$ where you should have $ n < m$. That is, the existence of a left-inverse requires more rows than columns for linear independence. EDIT: Actually it seems that you have used a right-inverse, so I don't understand the answer at all? $\endgroup$ – Student May 31 at 5:31
  • $\begingroup$ Yes that is a right inverse. The condition $m>n$ and full rank is needed because we cannot "create randomness" just by linear manipulations: if $X$ has less entries than $Y$, $Y$ cannot be standard normal (I think). The idea is then to put a standard normal inside of $X$ such that a later multiplication by $A$ reveals the standard normal. $\endgroup$ – Riccardo Sven Risuleo May 31 at 7:12

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