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Background:

Let $d\in \mathbb{N}$. Define the space of (real symmetric) positive definite matrices of size $d\times d$ as follows: \begin{align} \mathcal{S}_{++}^d := \big\{\mathbb{M}\in \mathbb{R}^{d\times d} : \text{$\mathbb{M}$ is symmetric and positive definite}\big\}. \label{eq:def:positive.definite.matrices} \end{align}

According to Wikipedia, given $\nu > d + 1$ and $\mathbb{V}\in \mathcal{S}_{++}^d$, the density function of the $\mathrm{Inv\hspace{0.2mm}Wishart}_d(\nu,\mathbb{V})$ distribution is defined by \begin{equation}\label{eq:Wishart.density} \begin{aligned} f_{\nu,\mathbb{V}}(\mathbb{X}) := \frac{|\frac{1}{2} \mathbb{V} \, \mathbb{X}^{-1}|^{\nu/2} \exp\big(-\frac{1}{2}\mathrm{tr}(\mathbb{V} \, \mathbb{X}^{-1})\big)}{|\mathbb{X}|^{(d + 1)/2} \pi^{d(d-1)/4} \prod_{i=1}^d \Gamma(\frac{1}{2} (\nu - (d + i)))}, \quad \mathbb{X}\in \mathcal{S}_{++}^d, \end{aligned} \end{equation} where $\nu$ is the number of degrees of freedom, $\mathbb{V}$ is the scale matrix, and \begin{equation} \Gamma(a) := \int_0^{\infty} t^{a - 1} e^{-t} d t, \quad a > 0, \end{equation} denotes the Euler gamma function.


Question :

If $\mathbb{X} \sim \mathrm{Inv\hspace{0.2mm}Wishart}_d(\nu,\mathbb{V})$, how can I normalize $\mathbb{X}$ to get a $\mathrm{Inv\hspace{0.2mm}Wishart}_d(\nu,\mathrm{I}_d)$ random matrix ? Namely, I think the question reduces to finding $\mathbb{S}\in \mathcal{S}_{++}^d$ such that $\mathbb{Y} = \mathbb{S}^{-1/2} \mathbb{X} \mathbb{S}^{-1/2} \sim \mathrm{Inv\hspace{0.2mm}Wishart}_d(\nu,\mathrm{I}_d)$.


What I know:

The mean and covariance matrix for the vectorization of $\mathbb{H}\sim \mathrm{Inv\hspace{0.2mm}Wishart}_d(\nu,\mathbb{V})$, namely \begin{equation}\label{eq:vectorization} \mathrm{vecp}(\mathbb{H}) := (\mathbb{H}_{11}, \mathbb{H}_{12}, \mathbb{H}_{22}, \dots, \mathbb{H}_{1d}, \mathbb{H}_{2d}, \dots, \mathbb{H}_{dd})^{\top}, \end{equation} ($\mathrm{vecp}$ is the operator that stacks the columns of the upper triangular portion of a symmetric matrix on top of each other) are well known to be: \begin{equation} \mathbb{E}[\mathrm{vecp}(\mathbb{H})] = \frac{\mathrm{vecp}(\mathbb{V})}{\nu - d - 1} \quad \text{(alternatively, $\mathbb{E}[\mathbb{H}] = \tfrac{1}{\nu - d - 1} \mathbb{V}$)} \end{equation} and (see Theorem~3.3.16~(ii) of the book Gupta & Nagar (1999) - Matrix Variate Distributions) \begin{equation}\label{eq:covariance.explicit.estimate} \mathbb{V}\mathrm{ar}(\mathrm{vecp}(\mathbb{H})) = \frac{2 \, B_d^{\top} \big[\mathrm{vec}(\mathbb{V}) \mathrm{vec}(\mathbb{V})^{\top} + (\nu - d - 1) (\mathbb{V} \otimes \mathbb{V})\big] B_d}{(\nu - d) (\nu - d - 1)^2 (\nu - d - 3)}, \quad (\text{not 100% sure this is correctly written}) \end{equation} where $\mathrm{I}_d$ is the identity matrix of order $d$, $B_d$ is a $d^{\hspace{0.2mm}2} \times \frac{1}{2} d(d + 1)$ transition matrix (see p.11 of the book Gupta & Nagar (1999) - Matrix Variate Distributions - for the precise definition), and $\otimes$ denotes the Kronecker product.


Side question :

Can we rewrite $\mathrm{vec}(\mathbb{V}) \mathrm{vec}(\mathbb{V})^{\top}$ or $\mathrm{vec}(\mathbb{V}) \mathrm{vec}(\mathbb{V})^{\top} + (\nu - d - 1) (\mathbb{V} \otimes \mathbb{V})$ as $\mathbb{A} \otimes \mathbb{A}$ for some positive definite matrix $\mathbb{A}$ ?

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Given that $$ f_{\nu,V}(X) = C_{\nu,V}\times|X|^{-(\nu+d+1)/2} \exp\left(-\tfrac{1}{2}\,\text{tr}\,(VX^{-1})\right), $$ with $V$ positive definite having square root $V^{1/2}$, it follows that the distribution of $Y=V^{-1/2}XV^{-1/2}$ has the desired form, $$ f_{\nu,I}(Y) = C_{\nu,I}\times|Y|^{-(\nu+d+1)/2} \exp\left(-\tfrac{1}{2}\,\text{tr}\,(Y^{-1})\right),$$ since $\text{tr}\, (VX^{-1})=\text{tr}\, Y^{-1}$ and $|X|=\text{constant}\times |Y|$. (The constants $C_{\nu,V}$ and $C_{\nu,I}$ being fixed by the normalization of the distribution. Also note that the measure $dY=|V|^{-1}dX$ differs from $dX$ by a constant, so all these constants can be absorbed into $C$.)

The transformation of the first moment can be checked easily, since $E(X)=V$ we have $(\nu-p-1)E(Y)=(\nu-p-1)V^{-1/2}E(X)V^{-1/2}=I$, as it should be.

The transformation of the variance is more complicated, but it cannot fail.

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  • $\begingroup$ This is good, however I just realised I didn't exactly ask the right question. How would you normalize inside Y to get Var(vecp(Y)) = Identity matrix. I think this question is harder. $\endgroup$ Jul 19, 2021 at 13:35
  • $\begingroup$ It should be possible to find a matrix $M = M(\nu)$ such that $M^{-1/2} (X - \frac{V}{\nu - d - 1}) M^{-1/2}$ tends in distribution (as $\nu\to \infty$) to a standard normal random matrix. $\endgroup$ Jul 19, 2021 at 13:40
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    $\begingroup$ this is so completely different from the posted question, I would just start a fresh new question. $\endgroup$ Jul 19, 2021 at 14:01

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