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Jun 17, 2023 at 21:01 answer added japalmer timeline score: 1
Sep 24, 2022 at 21:44 comment added John Jiang Any motivation for this question?
Feb 3, 2021 at 16:17 history edited kjetil b halvorsen
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Apr 13, 2017 at 12:44 history edited CommunityBot
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May 9, 2012 at 23:16 history notice removed CommunityBot
May 9, 2012 at 23:16 history bounty ended CommunityBot
May 3, 2012 at 17:20 comment added shabbychef Maybe there is an approach via the eigenvalues and eigenvectors of $A$, perhaps using the Marchenko Pastur distribution.
May 3, 2012 at 4:51 comment added shabbychef @cardinal I suspect you are right, and had tried the same trick, but instead assuming wlog that $\ell$ was the vector of all ones, hoping it would lead to the trace of $A^{-m}$. It did not, though.
May 3, 2012 at 0:41 comment added cardinal I think we can reduce the problem considerably (please check, though). Note that the reciprocal of your quantity of interest is completely invariant to orthogonal transformations of $\ell$, so without loss of generality we can take $\ell = e_1$. Thus, your question amounts to finding the marginal distribution of the reciprocal of (any) of the diagonal entries of $A^{-m}$. The standard proof in the $m = 1$ case effectively uses this fact and I think a connection between invariance under group actions of $O(n,\mathbb R)$ and independence can be drawn, but I haven't worked out the details (yet).
May 2, 2012 at 22:11 comment added shabbychef @StasK is there a Bartlett-type decomposition for the inverse-Wishart? I only know the one for the Wishart.
May 2, 2012 at 21:51 comment added StasK Is Bartlett decomposition of any help?
May 1, 2012 at 23:13 history tweeted twitter.com/#!/StackStats/status/197463837149102081
May 1, 2012 at 21:21 history notice added shabbychef Authoritative reference needed
May 1, 2012 at 21:21 history bounty started shabbychef
Apr 27, 2012 at 22:58 history asked shabbychef CC BY-SA 3.0