The Sample Correlation From A Bivariate Wishart Matrix

From Wishart Scatter to Sample Correlation

A sample correlation is a ratio of three entries of a scatter matrix. Even when two population variables are independent, that ratio is random: its numerator fluctuates around zero, while its denominator is itself estimated from the data. A two-dimensional Wishart model makes it possible to derive the full sampling distribution of this ratio, including its behavior when the sample size is very small.

Let $W$ be a $2\times2$ Wishart matrix with $\nu$ degrees of freedom and diagonal scale matrix $\operatorname{diag}(\sigma_1^2,\sigma_2^2)$, where $\sigma_1,\sigma_2>0$. Under zero population correlation, its off-diagonal scale entry is zero. The sample correlation formed from the scatter entries is

R=\frac{W_{12}}{\sqrt{W_{11}W_{22}}}.

For $n$ independent bivariate normal observations with unknown mean, the centered scatter matrix has $\nu=n-1$ degrees of freedom. If the mean is known and the scatter is formed about that known mean, then $\nu=n$. The distribution below is stated in terms of $\nu$, so the appropriate sample-size convention must be chosen before applying it.

Definition: For $\nu>1$, let $W\sim W_2(\nu,\Sigma)$ with $\Sigma=\operatorname{diag}(\sigma_1^2,\sigma_2^2)$, where both diagonal entries are positive. The Wishart sample correlation is $R=W_{12}/\sqrt{W_{11}W_{22}}$. Since multiplying either variable by a positive constant does not change its correlation, its distribution is the same as when $\Sigma=I_2$.

The condition $\nu>1$ is the nonsingularity condition for a two-dimensional Wishart density. The case $\nu=1$ must be handled separately: the Wishart matrix then has rank one, and the ratio is supported only at $-1$ and $1$.

Deriving the Density

It suffices to take $\Sigma=I_2$. Write $W_{11}=x$, $W_{22}=z$, and $W_{12}=y$. The Wishart density, with respect to $dx\,dy\,dz$ on positive-definite matrices, is

f_{x,y,z}(x,y,z) = \frac{(xz-y^2)^{(\nu-3)/2}e^{-(x+z)/2}} {2^\nu\Gamma_2(\nu/2)}, \qquad x>0,\quad z>0,\quad y^2<xz,

where $\Gamma_2(a)=\pi^{1/2}\Gamma(a)\Gamma(a-1/2)$ is the multivariate gamma function in dimension two. Transform to $r=y/\sqrt{xz}$, so $y=r\sqrt{xz}$ and $-1<r<1$. The absolute Jacobian of $(x,z,r)\mapsto(x,z,y)$ is $\sqrt{xz}$. Consequently, the transformed density is

\frac{(xz)^{(\nu-2)/2}(1-r^2)^{(\nu-3)/2}e^{-(x+z)/2}} {2^\nu\Gamma_2(\nu/2)}.

The powers of $x$ and $z$ each have exponent $\nu/2-1$. Integrating each over the positive half-line gives $2^{\nu/2}\Gamma(\nu/2)$. Thus the joint density of $x,z,r$, after integrating out $x$ and $z$, is

f_R(r) = \frac{\Gamma(\nu/2)} {\sqrt{\pi}\,\Gamma((\nu-1)/2)} (1-r^2)^{(\nu-3)/2}, \qquad -1<r<1.

It is zero outside this interval. The constant follows because $\Gamma_2(\nu/2)=\sqrt{\pi}\Gamma(\nu/2)\Gamma((\nu-1)/2)$. The density is integrable precisely when $(\nu-3)/2>-1$, or $\nu>1$, agreeing with the nonsingularity condition.

Theorem: If $W\sim W_2(\nu,\operatorname{diag}(\sigma_1^2,\sigma_2^2))$ with $\nu>1$ and $\sigma_1,\sigma_2>0$, then $R=W_{12}/\sqrt{W_{11}W_{22}}$ has density $$ f_R(r)= \frac{\Gamma(\nu/2)} {\sqrt{\pi}\,\Gamma((\nu-1)/2)} (1-r^2)^{(\nu-3)/2},\qquad -1<r<1. $$

Small Degrees of Freedom: Three Worked Cases

The exponent $(\nu-3)/2$ determines whether the density is large or small near the endpoints. The smallest nonsingular degrees of freedom illustrate three different shapes.

Worked Example: Three Degrees of Freedom Give a Uniform Correlation

Set $\nu=3$. The normalizing constant is \[ \frac{\Gamma(3/2)}{\sqrt{\pi}\Gamma(1)} = \frac{\sqrt{\pi}/2}{\sqrt{\pi}} = \frac12, \] and the power of $1-r^2$ is zero. Therefore $R$ is uniform on $(-1,1)$, with density $f_R(r)=1/2$.

As a check, the density integrates to $(1/2)\times2=1$. Its mean is zero by symmetry. Its second moment is \[ E[R^2]=\int_{-1}^{1}r^2\frac12\,dr =\frac12\left[\frac{r^3}{3}\right]_{-1}^{1} =\frac12\left(\frac13-\left(-\frac13\right)\right) =\frac13. \] The integral without the density factor is $\int_{-1}^{1}r^2\,dr=2/3$, or approximately $0.6667$; multiplying by the uniform density $1/2$ gives the expectation $1/3$, not $2/3$.

Worked Example: Two Degrees of Freedom and Endpoint Concentration

For $\nu=2$, the density becomes \[ f_R(r)=\frac{1}{\pi\sqrt{1-r^2}},\qquad -1<r<1. \] It diverges as $r$ approaches either endpoint, but the divergence is integrable. To verify normalization, set $r=\sin t$; then $dr=\cos t\,dt$ and $\sqrt{1-r^2}=\cos t$ for $-\pi/2<t<\pi/2$. Hence the integral of the density is \[ \frac1\pi\int_{-\pi/2}^{\pi/2}dt=1. \]

For example, the probability that $|R|>0.8$ is \[ P(|R|>0.8) =\frac{2}{\pi}\int_{0.8}^{1}\frac{dr}{\sqrt{1-r^2}} =\frac{2}{\pi}\left(\frac{\pi}{2}-\arcsin(0.8)\right) =\frac{2}{\pi}\arccos(0.8) \approx 0.410. \] Thus a large absolute sample correlation is not especially surprising with only two Wishart degrees of freedom, even when the population correlation is zero.

Worked Example: One Degree of Freedom Is Singular

Now let $\nu=1$. The density formula cannot be used: its exponent would be $-1$, and the Wishart matrix is singular rather than having a density on the positive-definite matrices. Write $W=uu^\mathsf{T}$, where $u=(U,V)^\mathsf{T}$ and $U,V$ are independent standard normal variables under the zero-correlation model. Then $W_{11}=U^2$, $W_{22}=V^2$, and $W_{12}=UV$. With probability one, neither $U$ nor $V$ is zero, so \[ R=\frac{UV}{|U||V|}=\operatorname{sign}(UV). \]

The signs of independent centered normal variables are independent, with probability $1/2$ for each sign. Therefore $P(R=1)=P(R=-1)=1/2$. This is a two-point distribution, not a continuous density. For centered observations, $\nu=1$ corresponds to $n=2$; each centered variable has only one direction of variation, so their sample correlation must be exactly $1$ or $-1$.

Moment and Shape Checks

The density is symmetric about zero, so $E[R]=0$. Its second moment can be derived for every $\nu>1$, giving a useful check on calculations and simulations.

Result: For $\nu>1$, $R^2$ has a beta distribution with parameters $1/2$ and $(\nu-1)/2$. In particular, $E[R^2]=1/\nu$, and since $E[R]=0$, $\operatorname{Var}(R)=1/\nu$.

To prove the result, let $T=R^2$. For $0<t<1$, the two inverse values are $r=\sqrt{t}$ and $r=-\sqrt{t}$; each has absolute derivative $1/(2\sqrt{t})$. By symmetry, their contributions to the density of $T$ are equal. Thus

f_T(t)=\frac{f_R(\sqrt{t})}{\sqrt{t}} = \frac{\Gamma(\nu/2)} {\sqrt{\pi}\Gamma((\nu-1)/2)} t^{-1/2}(1-t)^{(\nu-3)/2}, \qquad 0<t<1.

A beta density with parameters $a,b>0$ is proportional to $t^{a-1}(1-t)^{b-1}$ on $(0,1)$. Here $a=1/2$ and $b=(\nu-1)/2$, both positive. The constant above is $1/B(a,b)$, so $T$ has the stated beta distribution. Using $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$,

E[T]=\frac{B(a+1,b)}{B(a,b)} =\frac{\Gamma(a+1)\Gamma(a+b)} {\Gamma(a)\Gamma(a+b+1)} =\frac{a}{a+b} =\frac{1}{\nu}.

The last equality uses $\Gamma(a+1)=a\Gamma(a)$ and $a+b=\nu/2$. Symmetry gives $E[R]=0$, and therefore $\operatorname{Var}(R)=E[R^2]-(E[R])^2=1/\nu$. This also confirms the uniform-case calculation: at $\nu=3$, the second moment is $1/3$. $\square$

For $1<\nu<3$, the exponent is negative and the density increases without bound near $r=\pm1$, while remaining integrable. At $\nu=3$, it is constant. For $\nu>3$, the density tends to zero at both endpoints. As $\nu$ grows, the variance $1/\nu$ decreases, so the distribution places progressively more mass near zero. These statements concern the density’s shape and variance; none says the correlation can exceed its natural bounds $[-1,1]$.

Common Pitfall: Confusing a Singular Law with a Density

A common mistake is to substitute $\nu=1$ into the continuous density formula, obtain a nonintegrable expression, and try to normalize it. The failure is not a defect in the calculation: the formula assumes a nonsingular two-dimensional Wishart matrix, which requires $\nu>1$. At $\nu=1$, the matrix has rank one and the correlation has point masses at the endpoints. At $\nu=0$, as can occur for a centered sample of size one, both centered sums of squares are zero and the sample correlation is undefined.

Another easy error is to confuse the unweighted integral $\int_{-1}^{1}r^2\,dr=2/3$ with the expectation under the uniform density. An expectation includes the density: for $R$ uniform on $(-1,1)$, $E[R^2]=(1/2)(2/3)=1/3$. Keeping the density factor visible is a simple way to catch the mistake.

Key takeaway: Under zero population correlation, the sample correlation from a two-dimensional Wishart matrix has density proportional to $(1-r^2)^{(\nu-3)/2}$ when $\nu>1$. The low-degree cases matter: $\nu=2$ gives an endpoint-singular density, $\nu=3$ gives a uniform law, and $\nu=1$ gives a two-point distribution.

Check Your Understanding

Use the degrees of freedom and the support of the correlation carefully.

For a centered bivariate normal sample of size $n=8$, what Wishart degrees of freedom apply, and what is the variance of the sample correlation under zero population correlation?
Why can the density for $\nu=2$ diverge at $r=\pm1$ and still integrate to one?
For $\nu=3$, calculate $P(|R|<1/2)$ from the uniform density.
Describe the distribution of $R$ when $\nu=1$, and explain why a continuous density formula does not apply.
For a uniform random variable on $(-1,1)$, distinguish $\int_{-1}^{1}r^2\,dr$ from $E[R^2]$.