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In a related question, I had asked about the norm induced by an inverse Wishart matrix. I am interested in generalizing that result somewhat. Let $A\sim\mathcal{W}_p\left(I,n\right)$, a Wishart matrix with scale matrix $I$, the identity, and $n$ degrees of freedom. Let $l$ be some fixed $p$-dimensional vector. Consider $$h = \frac{l^{\top}l}{l^{\top} \left(A^{-1}\right)^{m} l},$$ where $m$ is some integer. When $m=1$, $h$ is evidently distributed as a Chi-square random variable (with $n-p+1$ degrees of freedom, I believe).

Are there known results on the distribution of $h$ when $m=2$? Other values?

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  • $\begingroup$ Is Bartlett decomposition of any help? $\endgroup$
    – StasK
    Commented May 2, 2012 at 21:51
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    $\begingroup$ @StasK is there a Bartlett-type decomposition for the inverse-Wishart? I only know the one for the Wishart. $\endgroup$
    – shabbychef
    Commented May 2, 2012 at 22:11
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    $\begingroup$ 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). $\endgroup$
    – cardinal
    Commented May 3, 2012 at 0:41
  • $\begingroup$ @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. $\endgroup$
    – shabbychef
    Commented May 3, 2012 at 4:51
  • $\begingroup$ Maybe there is an approach via the eigenvalues and eigenvectors of $A$, perhaps using the Marchenko Pastur distribution. $\endgroup$
    – shabbychef
    Commented May 3, 2012 at 17:20

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The distribution of the eigenvalues is known, https://dornsife.usc.edu/assets/sites/406/docs/505b/505project_Narae_Lee.pdf, which you can transform to the distribution of the inverse power of the eigenvalues.

The random vector with components $|l^T \mathbf{v}_i|/\|l\|$, where $\mathbf{v}_i$, $i=1,\ldots,p$ are the eigenvectors of $\mathbf{A}$, is uniformly distributed on the surface of the $p$ dimensional hypersphere in the positive orthant.

You should be able to calculate the distribution from by combining the two results.

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