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Let $W = X^TX$ denote a standard Wishart matrix, i.e., where $X$ is a Gaussian random matrix with iid standard Normal entries.

In this case we can write $W = U D U^T$, where $U$ is orthogonal and $D$ is the diagonal matrix of nonnegative eigenvalues.

My question is: are $(U, D)$ independent? If so, would there be a place that would contain proofs of such facts?

Of course, this is obvious in the case that $X$ is a single Gaussian vector, as then $D$ corresponds to the squared norm and $U$ corresponds to the direction, which are well known to be independent.

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A proof is on page 80-81 and 90 of Forrester, the probability distribution function of $W=X^\top X$ is $\propto e^{-\tfrac{1}{2}\operatorname{tr}W}(\operatorname{det}W)^{(n-m-1)/2}$, for an $n\times m$ matrix $X$, so only dependent on the eigenvalues of $W$. The $m\times m$ matrix of eigenvectors of $W$ is thus uniformly distributed in the orthogonal group, independently of the eigenvalues.

You can also see this directly from the fact that $P(X)\propto e^{-\tfrac{1}{2}\operatorname{tr}X^\top X}$, so invariant under $X\mapsto XO$ with orthogonal $O$.

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    $\begingroup$ Ah of course. Obviously should have noticed the density does NOT depend on the orthogonal matrix...thanks very much. $\endgroup$
    – Drew Brady
    Commented Jun 24 at 21:42
  • $\begingroup$ By "the probability distribution function" do you mean the probability density function? $\endgroup$
    – Michael Hardy
    Commented Jun 25 at 0:38
  • $\begingroup$ yes, I think it is an established terminology: en.wikipedia.org/wiki/Probability_distribution $\endgroup$
    – Carlo Beenakker
    Commented Jun 25 at 5:15

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