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Lin described two equivalent characterizations of the multivariate t-distribution, viz.

  1. As a normal vector divided by an independent chi. That is, $t = Z / \sqrt{\chi^2/v}$, where $Z$ is multivariate normal, zero mean, with covariance $I$, independent of the chi-square in the denominator, which has $v$ degrees of freedom.
  2. As the symmetric square root of a Wishart times a normal (independent of each other). That is $t = \left(V^{1/2}\right)^{-1} Z$, where $V^{1/2}$ is a symmetric square root of $V$, which is Wishart with matrix $I$ and degrees of freedom $v+p-1$, independent of $Z$, which is multivariate normal, zero mean with covariance $I$.

These two characterizations are equivalent in that they have the same density. (Lin's description is more general, allowing for a mean vector and non-identity covariances; I opted for simplicity.)

My question is whether there is a similar equivalence for the case of Wishart to an arbitrary power. That is, let $V$ be Wishart with matrix $I$ and $v+p-1$ degrees of freedom, and let $\gamma$ be some fixed real. Let $V^\gamma$ be the symmetric matrix defined in the 'usual way' (matrix power: take eigenvalues to $\gamma$). Consider $y = \left(V^{\gamma}\right)^{-1} Z$, where $Z$ is multivariate normal, zero mean, covariance $I$, independent of $V$. Is there an alternative characterization for random variables like $y$ that is like the first (say, a normal divided by some weird random variable?).

I seem to recall that a Wishart could be decomposed as something like $Q\Lambda Q'$, where $Q$ is drawn uniformly from the orthogonal group, independent of the diagonal matrix $\Lambda$, but I cannot find the reference now. Couldn't something like this yield an alternative characterization? (Lin does not prove his equivalence, by the way, so I am not sure how to approach the problem.)

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