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I am trying to find the normalizing constant for a probability distribution and ran into a difficult integral. When $X$ is an $p \times k$ matrix, $a>0,$ and $g>0,$ how can I compute

$$\int \left|I_k + \frac{g}{a}X^TX\right|^{-d/2} \text{exp}\left\{-\frac{1}{2}\text{tr}[aX^TX]\right\}dX$$

where the bars indicate the determinant?

I've wanted to change variables. For example, to send $X \to \left|I_k + \frac{g}{a}X^TX\right|$ but this isn't injective. Similarly, I've considered other transformations, e.g. using a singular value decomposition, but haven't had much success.

If $k=1$ then I can use a result like this one: Integration of radial functions? but I don't know of any similar formula that helps when $k>1.$

Edit: I claim the integrand is the kernel of a spherical matrix variate density in $X.$

From the Matrix Variate Distributions book referenced in my comment, a random matrix $X$ is said to have a spherical distribution if $X = \Gamma X \Lambda$ where the equality is in distribution and $\Gamma$ and $\Lambda$ are orthogonal matrices of appropriate dimension.

The claim follows from two observations:

  1. That $\text{tr}[a(\Gamma X \Lambda)^T\Gamma X \Lambda] = \text{tr}[aX^TX],$ which follows easily from the othogonality of $\Gamma$ and $\Lambda$ and the cyclic property of the trace
  2. That $\left|I_k + \frac{g}{a}(\Gamma X \Lambda)^T\Gamma X \Lambda\right| = \left|I_k + \frac{g}{a} \Lambda^T X^T X \Lambda \right| = \left|I_k + \frac{g}{a}X^TX\right|,$ which we get using orthogonality and the matrix determinant lemma.

The Gupta and Nagar book discusses densities, including normalizing constants, of spherical matrix variate distributions. So it is likely possible to calculate the integral by recognizing this kernel.

Alternatively, we might do some sort of matrix integration by parts in which we differentiate the determinant term with respect to $X$.

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    $\begingroup$ Welcome to MSE. :) As your user name suggests, it is tough! :D $\endgroup$ – H. R. Dec 23 '15 at 20:29
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    $\begingroup$ There is a section in the book Matrix Variate Distributions by A.K. Gupta and D. K. Nagar that appears to be relevant. I think this is a matrix variate spherical distribution, discussed in section 9.3 of that book. $\endgroup$ – tough_integral Dec 23 '15 at 23:35
  • $\begingroup$ Would it be correct to assume that $X$ is a real matrix of full rank? $\endgroup$ – Jonas Dahlbæk Jun 14 '16 at 17:40
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As you pointed out, the integrand is invariant under the action of the orthogonal group. Thus it suffices to integrate over diagonal matrices, and multiply the result by the volume of the orthogonal group.

Suppose $X=diag(x_1,\ldots,x_k)$. Then the integral becomes $$ \int_{\mathbb R} \prod_{i=1}^k\left(1+\frac{g}{a}x_i^2\right)^{-d/2}e^{-ax_i^2/2}\ dx_1\cdots dx_k=\left[\int_{\mathbb R}\left(1+\frac{g}{a}x^2\right)^{-d/2}e^{-ax^2/2}\ dx\right]^k. $$ The innermost integral may be expressed in terms of moments of the Gaussian distribution.

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