The Wishart Determinant Distribution

Why the Wishart Determinant Has a Product Law

A Wishart matrix summarizes multivariate scatter, and its determinant measures the volume scale of that scatter. Its entries are dependent, so finding a determinant distribution by treating those entries separately is not a useful route. Bartlett decomposition gives a direct one: after separating the scale matrix, the determinant becomes a product of independent chi-square random variables.

We work in the sampling setting: $W$ has a $p$-dimensional Wishart distribution with integer degrees of freedom $\nu\ge p$ and positive-definite scale matrix $\Sigma$. In this setting, $W$ is positive definite with probability one, so its determinant is positive. Write $|A|$ for the determinant of a square matrix $A$.

Definition: Let $W\sim W_p(\nu,\Sigma)$, where $\Sigma$ is positive definite and $\nu\ge p$ is an integer. The normalized Wishart determinant is $D=|W|/|\Sigma|$. It removes the determinant of the scale matrix, leaving a distribution that depends only on $p$ and $\nu$.

Distribution From Bartlett Decomposition

Let $\Sigma=LL'$ be a Cholesky factorization. Bartlett decomposition represents $W$ in distribution as $LAA'L'$, where $A$ is lower triangular. Its diagonal entries are independent, with $A_{ii}^2\sim\chi^2_{\nu-i+1}$ for $i=1,\ldots,p$; its entries below the diagonal are independent standard normal variables and are independent of those diagonal entries. Because the determinant of a triangular matrix is the product of its diagonal entries, only the diagonal factors enter $|W|$.

Theorem: If $W\sim W_p(\nu,\Sigma)$, with $\Sigma$ positive definite and integer $\nu\ge p$, then $$ |W|\ \overset{d}{=}\ |\Sigma|\prod_{i=1}^{p} Q_i, \qquad Q_i\ \text{independent},\qquad Q_i\sim\chi^2_{\nu-i+1}. $$ In particular, the normalized determinant $D$ has the distribution of the product $\prod_{i=1}^p Q_i$.

Here $\overset{d}{=}$ denotes equality in distribution. To verify the determinant identity, use $|L|^2=|\Sigma|$ and $|A|=\prod_i A_{ii}$:

|LAA'L'|=|L|\,|A|\,|A'|\,|L'| =|\Sigma|\,|A|^2 =|\Sigma|\prod_{i=1}^p A_{ii}^2.

The independence and chi-square distributions of the $A_{ii}^2$ then give the stated product law. This product representation specifies the distribution even when it has no convenient elementary closed-form density.

Worked Example: A One-Dimensional Wishart Determinant

Take $p=1$, $\nu=6$, and scale matrix $\Sigma=[3]$. Then $|\Sigma|=3$, and the determinant is the single entry of $W$. The product representation has just one factor:

$|W|\overset{d}{=}3Q$, where $Q\sim\chi^2_6$.

A $\chi^2_6$ variable has density $f_Q(q)=q^2e^{-q/2}/16$ for $q>0$. Applying the scaling transformation $x=3q$, so $q=x/3$ and $dq/dx=1/3$, gives

f_{|W|}(x)=\frac{1}{3}f_Q(x/3) =\frac{1}{3}\frac{(x/3)^2e^{-x/6}}{16} =\frac{x^2e^{-x/6}}{432},\qquad x>0.

The mean is $3E[Q]=3(6)=18$, and the variance is $3^2\operatorname{Var}(Q)=9(12)=108$. This example is the ordinary scaled chi-square law; in higher dimensions, multiple independent factors are multiplied.

Density of the Product

A product of independent continuous random variables can be described by a change of variables. Set $r_i=\nu-i+1$, so $Q_i$ has density $f_i(q)=q^{r_i/2-1}e^{-q/2}/(2^{r_i/2}\Gamma(r_i/2))$ for $q>0$. To obtain the density of $D=\prod_i Q_i$, retain $q_1,\ldots,q_{p-1}$ as coordinates and replace $q_p$ by $d=\prod_i q_i$. For fixed $q_1,\ldots,q_{p-1}$, the inverse transformation is $q_p=d/(q_1\cdots q_{p-1})$, with Jacobian factor $1/(q_1\cdots q_{p-1})$.

Density formula: For $d>0$, the normalized determinant has density $$ f_D(d)=\int_{(0,\infty)^{p-1}} \left\{\prod_{i=1}^{p-1}f_i(q_i)\right\} f_p\!\left(\frac{d}{\prod_{i=1}^{p-1}q_i}\right) \frac{1}{\prod_{i=1}^{p-1}q_i}\, dq_1\cdots dq_{p-1}. $$ For $p=1$, this reduces to $f_D(d)=f_1(d)$. For $|W|=|\Sigma|D$, the density is $f_{|W|}(x)=f_D(x/|\Sigma|)/|\Sigma|$ for $x>0$.

This formula follows from the ordinary change-of-variables formula for independent variables with densities. The integration ranges over positive $q_i$ because each chi-square factor is positive almost surely. In many applications, the product representation or its moments are more useful than evaluating this integral.

Real-Order Moments

The chi-square density is a gamma density, so its real-order moments can be obtained by integrating that density. For $Q\sim\chi^2_r$, and real $t$ with $t>-r/2$,

E[Q^t] =\frac{1}{2^{r/2}\Gamma(r/2)} \int_0^\infty q^{t+r/2-1}e^{-q/2}\,dq =2^t\frac{\Gamma(r/2+t)}{\Gamma(r/2)}.

For the integral to be finite near zero, its power must satisfy $t+r/2-1>-1$, which is exactly $t>-r/2$. It is finite at infinity because of the exponential term. Independence lets us multiply the moments of the Bartlett factors.

Theorem: For real $t$ satisfying $t>-(\nu-p+1)/2$, $$ E[|W|^t] =|\Sigma|^t\,2^{pt} \prod_{i=1}^{p} \frac{\Gamma\!\left((\nu-i+1)/2+t\right)} {\Gamma\!\left((\nu-i+1)/2\right)}. $$ If $t\le-(\nu-p+1)/2$, this moment is infinite.

To prove the condition, the degrees of freedom in the product are $\nu,\nu-1,\ldots,\nu-p+1$. Each factor has a finite $t$th moment precisely when $t>-(\nu-i+1)/2$. The most restrictive of these inequalities comes from the smallest degree of freedom, $\nu-p+1$. For such $t$, independence and the product representation give

E[|W|^t] =|\Sigma|^t E\left[\prod_{i=1}^p Q_i^t\right] =|\Sigma|^t\prod_{i=1}^p E[Q_i^t],

and substituting the chi-square moment formula proves the result. If the condition fails, the factor with $\nu-p+1$ degrees of freedom has an infinite $t$th moment. Since all other factors are positive and independent, their product cannot make that moment finite. $\square$

Mean and Variance of the Determinant

For $Q\sim\chi^2_r$, $E[Q]=r$ and $E[Q^2]=r(r+2)$. Applying these identities to the independent factors gives the determinant’s first two moments. The resulting formulas are useful both for calculation and for checking a simulated Wishart sample.

Corollary: If $W\sim W_p(\nu,\Sigma)$, with integer $\nu\ge p$ and positive-definite $\Sigma$, then $$ E[|W|]=|\Sigma|\prod_{i=1}^p(\nu-i+1) $$ and $$ \operatorname{Var}(|W|) =|\Sigma|^2\left\{ \prod_{i=1}^p(\nu-i+1)(\nu-i+3) -\left[\prod_{i=1}^p(\nu-i+1)\right]^2 \right\}. $$

Indeed, multiplying the individual means gives the first formula. For the second moment, the $i$th factor contributes $E[Q_i^2]=(\nu-i+1)(\nu-i+3)$. Thus $E[|W|^2]$ is $|\Sigma|^2$ times the first product inside braces. Subtracting $E[|W|]^2$ yields the variance formula. $\square$

Worked Example: Mean and Variance for a Two-Dimensional Wishart Matrix

Let $W\sim W_2(5,\Sigma)$, where $\Sigma=\begin{pmatrix}2&0.5\\0.5&1\end{pmatrix}$. Its scale determinant is $|\Sigma|=2(1)-(0.5)^2=1.75=7/4$. The determinant distribution is $|W|\overset{d}{=}(7/4)Q_1Q_2$, with independent $Q_1\sim\chi^2_5$ and $Q_2\sim\chi^2_4$.

The mean follows from the factor means:

E[|W|]=\frac{7}{4}(5)(4)=35.

For the second moment, $E[Q_1^2]=5(7)=35$ and $E[Q_2^2]=4(6)=24$. Therefore

E[|W|^2]=\left(\frac{7}{4}\right)^2(35)(24) =\frac{49}{16}(840)=\frac{5145}{2}.

Subtracting the squared mean gives $\operatorname{Var}(|W|)=5145/2-35^2=5145/2-2450/2=2695/2=1347.5$. The nonzero off-diagonal scale entries affect this calculation through $|\Sigma|$; no separate distribution for those entries is required.

Worked Example: A Three-Factor Determinant and Its Moments

Take $W\sim W_3(4,\Sigma)$, with $\Sigma=\operatorname{diag}(2,1,1/2)$. Here $|\Sigma|=2\cdot1\cdot(1/2)=1$. Bartlett decomposition gives $|W|\overset{d}{=}Q_1Q_2Q_3$, where the independent factors have degrees of freedom $4,3,2$, respectively.

The mean is $4\cdot3\cdot2=24$. For the second moment, the three factor contributions are $4(6)=24$, $3(5)=15$, and $2(4)=8$. Hence $E[|W|^2]=24\cdot15\cdot8=2880$, and $\operatorname{Var}(|W|)=2880-24^2=2304$.

This case also illustrates the boundary for negative moments. The smallest degrees of freedom are $2$, so $E[|W|^t]$ is finite only for $t>-1$. In particular, the mean reciprocal $E[1/|W|]$, which corresponds to $t=-1$, is infinite, despite $|W|$ being positive almost surely.

Common Pitfall: Multiplying Dependent Entries

A common mistake is to treat the entries of $W$ as independent when trying to find the determinant distribution. For a $2\times2$ matrix, $|W|=W_{11}W_{22}-W_{12}^2$. The Wishart entries are generally dependent, and this difference is not a product of independent chi-square variables. The correct reasoning uses Bartlett decomposition: it transforms the whole matrix into triangular factors, and the determinant depends only on their independent squared diagonal entries.

Another error is to use $\nu$ degrees of freedom for every factor. The degrees of freedom decrease across the diagonal: $\nu,\nu-1,\ldots,\nu-p+1$. That decrease matters especially for negative moments, whose existence is determined by the smallest factor. The determinant distribution is therefore not simply a single chi-square law, except when $p=1$.

Key takeaway: In the integer-degree-of-freedom sampling setting, $|W|/|\Sigma|$ is a product of independent chi-square variables with degrees of freedom $\nu,\nu-1,\ldots,\nu-p+1$. Its real-order moments follow by multiplying gamma-function moments, and negative moments exist only above the boundary set by the final factor.

Check Your Understanding

Use the Bartlett product representation and moment conditions to answer these questions.

For $W\sim W_4(7,\Sigma)$, list the degrees of freedom of the independent chi-square factors in $|W|/|\Sigma|$.
State the condition on real $t$ for $E[|W|^t]$ to be finite when $W\sim W_3(5,\Sigma)$.
For $W\sim W_2(6,\Sigma)$, express $E[|W|]$ in terms of $|\Sigma|$ and calculate its numerical multiplier.
Explain why the entries of a Wishart matrix cannot generally be treated as independent when finding the determinant law.
For $W\sim W_3(4,\Sigma)$, is $E[1/|W|]$ finite? Justify your answer using the smallest chi-square degree of freedom.