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 $\mathbb{A}$ ?


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented 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. $\endgroup$ Commented Jul 19, 2021 at 13:40
  • 1
    $\begingroup$ this is so completely different from the posted question, I would just start a fresh new question. $\endgroup$ Commented Jul 19, 2021 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.