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Let $A\in\mathbb{R}^{k\times d}$ be matrix with i.i.d. $\mathcal{N}(0,1/k)$ entries with $k<d$, and let $B=A^{\top}A$. I would like to compute the distribution of $Bx$ where $x\in\mathbb{R}^{d}$ is a fixed vector.

The matrix $B$ essentially projects onto a $k$-dimensional subspace of $\mathbb{R}^{d}$ (it has rank $k$ almost surely). However, computing the distribution explicitly using the expression for entries of $Bx$ is tricky since the entries are dependent. Any hints on how to proceed?

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  • $\begingroup$ Seems like it would depend a lot on what $x$ exactly is. Consider the case that $k=1$. Then if $x = e_1$ then $Bx$ will depend a lot on just one random gaussian versus the case where $x$ is the all ones vector. $\endgroup$ – Sandeep Silwal Jul 16 at 14:02
  • $\begingroup$ @SandeepSilwal Yes, the distribution of $Bx$ will be parameterized by $x$. The mean $\mathbb{E}[Bx]$ is clearly $x$, but it is not clear what the distribution is. $\endgroup$ – nemo Jul 17 at 6:54
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    $\begingroup$ Just for fun, I made a few simulations for different x's. The first indications are that the solution seems far from easy to obtain. Some marginals of the vector obtained look Gaussian, some other absolutely don't and I did not even start digging into dependence patterns. Is there any setting in which you naturally encounter this question ? $\endgroup$ – Gilles Mordant Jul 18 at 10:42
  • $\begingroup$ @GillesMordant Thanks for taking the effort to perform a simulation! I was studying random projections and this came to mind. Usually, one would map $x\in\mathbb{R}^d$ to a smaller vector $Ax\in\mathbb{R}^{k}$. But $A^{\top}Ax$ also ``effectively'' does the same thing: its range is a $k$-dimensional subspace of $\mathbb{R}^{d}$. I was curious to see what the points in the range would be distributed like. $\endgroup$ – nemo Jul 18 at 11:06
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    $\begingroup$ It could also be possible that your question can be rewritten as a function of an order two non-decoupled Gaussian chaos, which has been studied in the literature. I however never came across distributional results. I hope this is not a misleading track... $\endgroup$ – Gilles Mordant Jul 18 at 11:35

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