Assume $M$ is an $N \times k$ Gaussian matrix, i.e., its entries are i.i.d. standard normal random variables, with $N>>k$. Take $D=\text{diag}(\lambda_1, \dotsc ,\lambda_N)$ for some fixed real scalars. I am interested in finding the p.d.f. of the $N \times k$ "unitary" matrix $Q$ from the QR decomposition of $DM$ (and possibly $D^2M$, etc.).

It is known that if $k=N$ and $D=I_N$, the identity matrix, then $Q$ is distributed with respect to the Haar measure on the Lie group of orthonormal matrices of order $N$. Can you provide any insight on the general case for $k<N$ and/or general $D$?

I also tried to look for the simplest case, i.e., $k=1$. Then the QR decomposition coincides with a simple normalization. I have found this result for common variance, i.e., the case $\lambda_1=\dotsc =\lambda_N$. Can this be easily generalized for the general case with different $\lambda_i$?

I attempted in the simplest case to scale the matrix $M$ (which is for $k=1$ just an $N$ dimensional random vector). Indeed, then the above-mentioned result is applicable and one gets $$DM=DUR, $$ where $UR$ is the QR decomposition of $M$ and the p.d.f. of entries of $U$ is known from the above. Nonetheless, I haven't found any easy way to connect the p.d.f. of $DU$ with the one of $Q$. Thanks in advance.

  • $\begingroup$ @whumber is the one to give credit for the special case. He/she also mentioned the p.d.f. in the comment for $N=2, k=1$, however I do not see the idea of how was this obtained and hence cannot try to generalize it for higher dimensions $N$. $\endgroup$ – michalOut Aug 4 '16 at 11:51
  • $\begingroup$ What are the $\lambda_i$? Are they arbitrary values or have they somehow been derived from $M$ itself (such as eigenvalues of $MM^\prime$)? $\endgroup$ – whuber Aug 4 '16 at 14:23
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    $\begingroup$ In general they are arbitrary. At the moment I am focusing on the case where $\lambda_1=1$ and $\lambda_i=\lambda^{i-1}$ for $i=2,3,\dotsc ,N$, i.e., they are geometrically decaying with a fixed parameter $\lambda \in (0,1)$. $\endgroup$ – michalOut Aug 4 '16 at 14:31
  • $\begingroup$ @whumber , could you please clarify the simplest case with $N=2$ and $k=1$? $\endgroup$ – michalOut Aug 7 '16 at 9:39

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