24

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?

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
shabbychef
  • 10,388
  • 7
  • 50
  • 93
  • Is Bartlett decomposition of any help? – StasK May 02 '12 at 21:51
  • 2
    @StasK is there a Bartlett-type decomposition for the inverse-Wishart? I only know the one for the Wishart. – shabbychef May 02 '12 at 22:11
  • 1
    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). – cardinal May 03 '12 at 00:41
  • @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. – shabbychef May 03 '12 at 04:51
  • Maybe there is an approach via the eigenvalues and eigenvectors of $A$, perhaps using the Marchenko Pastur distribution. – shabbychef May 03 '12 at 17:20

0 Answers0