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 $X$?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.