Sphericity and Compound Symmetry

Why Compare These Covariance Assumptions?

In a repeated-measures study, the covariance among occasions affects the variability of within-subject contrasts. Sphericity places a specific restriction on that variability: every contrast of a given length must have the same variance. Compound symmetry is a more restrictive covariance pattern, requiring equal marginal variances and equal covariances between distinct occasions. Understanding the distinction matters because compound symmetry guarantees sphericity, but sphericity does not require compound symmetry.

Let $Y=(Y_1,\ldots,Y_p)'$ denote the vector of measurements on one subject across $p$ occasions, and let $\Sigma=\operatorname{Cov}(Y)$. A within-subject contrast has the form $c'Y$, where $c$ is a vector of coefficients satisfying $1'c=0$. The zero-sum condition removes a common level across occasions and leaves a comparison of occasions.

Definition: The covariance matrix $\Sigma$ is spherical on the within-subject contrast space if there is a constant $\lambda\geq 0$ such that $$ \operatorname{Var}(c'Y)=c'\Sigma c=\lambda c'c $$ for every $c\in\mathbb{R}^p$ with $1'c=0$. When $\lambda>0$, every nonzero within-subject contrast has positive variance. Equivalently, for any matrix $C$ whose columns form an orthonormal basis of the contrast space, $C'\Sigma C=\lambda I_{p-1}$.

The normalization matters: if a contrast is multiplied by a constant, its variance is multiplied by the square of that constant. Sphericity therefore says that contrast variance is proportional to squared coefficient length, not that all contrasts, regardless of scale, have identical variance.

Definition: A covariance matrix has compound symmetry if all occasions have the same variance and every pair of distinct occasions has the same covariance. It can be written $$ \Sigma=\sigma^2\big[(1-\rho)I_p+\rho\,11'\big], $$ where $\sigma^2>0$ is the common marginal variance and $\rho$ is the common correlation. For positive definiteness, $-1/(p-1)<\rho<1$. The endpoints give positive-semidefinite, possibly singular, matrices.

Why Compound Symmetry Implies Sphericity

Under compound symmetry, the variance of every occasion is $\sigma^2$, and the covariance of any two different occasions is $\sigma^2\rho$. For a contrast, the shared covariance component disappears because its coefficients sum to zero.

Theorem: If $\Sigma=\sigma^2[(1-\rho)I_p+\rho\,11']$, then $\Sigma$ is spherical on the contrast space, with contrast variance multiplier $\lambda=\sigma^2(1-\rho)$.

Proof. Take any contrast vector $c$ satisfying $1'c=0$. Substitution of the compound-symmetry form gives

\begin{aligned} c'\Sigma c &=\sigma^2\big[(1-\rho)c'c+\rho\,c'11'c\big]\\ &=\sigma^2\big[(1-\rho)c'c+\rho(1'c)^2\big]\\ &=\sigma^2(1-\rho)c'c. \end{aligned}

This is precisely the sphericity condition, with $\lambda=\sigma^2(1-\rho)$. For a positive-definite compound-symmetry matrix, $\rho<1$, so this multiplier is positive. The same calculation applies at the positive-semidefinite endpoints, where the contrast variance can be zero in a degenerate case. $\square$

Worked Example: Contrast Variance Under Compound Symmetry

Suppose there are four occasions with compound-symmetry covariance, common variance $\sigma^2=9$, and common correlation $\rho=0.4$. Find the variance of a unit-length contrast and the variance of the difference between occasions 1 and 2.

For every contrast $c$, the theorem gives $\operatorname{Var}(c'Y)=9(1-0.4)c'c=5.4c'c$. Thus any unit-length contrast has variance $5.4$.

The difference between occasions 1 and 2 uses $c=(1,-1,0,0)'$, which has $1'c=0$ and $c'c=2$. Its variance is $5.4(2)=10.8$. Directly, the two variances are 9 and the covariance is $9(0.4)=3.6$, so

\operatorname{Var}(Y_1-Y_2)=9+9-2(3.6)=10.8.

The direct calculation agrees with the contrast-space result. A unit-length contrast has coefficients scaled down by $1/\sqrt{2}$, so its variance is half the variance of this unscaled pairwise difference.

What Sphericity Requires—and What It Leaves Unrestricted

Sphericity constrains the covariance only after projecting into the contrast space. Define the projection matrix $P=I_p-J/p$, where $J=11'$. It is symmetric, satisfies $P1=0$, and projects vectors onto the space of zero-sum contrasts. The condition $C'\Sigma C=\lambda I_{p-1}$ says that covariance is isotropic within that space. It does not require equal variance in the direction of the overall mean, nor does it prohibit covariance between the mean direction and contrast directions.

Theorem: A symmetric matrix $\Sigma$ is spherical on the contrast space with multiplier $\lambda$ if and only if $$ P\Sigma P=\lambda P. $$ Equivalently, in an orthonormal coordinate system consisting of the mean direction $1/\sqrt p$ and any orthonormal contrast basis, its contrast-by-contrast block is $\lambda I_{p-1}$; the mean-direction variance and mean-to-contrast covariances are not fixed by sphericity.

Proof. Let $C$ have orthonormal columns spanning the contrast space. Then $P=CC'$. If $\Sigma$ is spherical, $C'\Sigma C=\lambda I_{p-1}$, and hence

P\Sigma P=CC'\Sigma CC'=C(C'\Sigma C)C' =C(\lambda I_{p-1})C'=\lambda P.

Conversely, suppose $P\Sigma P=\lambda P$. Since $C$ is a contrast basis, $PC=C$. Therefore

C'\Sigma C=C'P\Sigma PC=C'(\lambda P)C=\lambda C'C=\lambda I_{p-1}.

Thus the contrast covariance is spherical. To see why the mean direction can still be associated with contrasts, write $\Sigma$ in the orthonormal coordinates $[1/\sqrt p,\ C]$. The condition fixes its lower-right block at $\lambda I_{p-1}$, but places no such restriction on the off-diagonal block or the upper-left entry. Any such block matrix must, of course, still be positive semidefinite to qualify as a covariance matrix. $\square$

Worked Example: Spherical but Not Compound Symmetric

Consider the covariance matrix

\Sigma= \begin{pmatrix} 1.5&0.25&0.25\\ 0.25&1&0\\ 0.25&0&1 \end{pmatrix}.

First verify that it is a valid positive-definite covariance matrix. Its lower-right $2\times2$ block is the identity, and the Schur complement of that block is $1.5-0.25^2-0.25^2=1.375>0$. Thus $\Sigma$ is positive definite.

This matrix can be written $\Sigma=I_3+1a'+a1'$, where $a=(0.25,0,0)'$. For any contrast $c$, $1'c=0$, and therefore

\begin{aligned} c'\Sigma c &=c'c+(c'1)(a'c)+(c'a)(1'c)\\ &=c'c. \end{aligned}

It is spherical with $\lambda=1$. It is not compound symmetric: its diagonal entries are $1.5,1,1$, so the marginal variances are not equal; its off-diagonal entries are also not all equal. The unequal marginal variances do not contradict sphericity because the zero-sum contrast removes the added terms involving $1$.

More generally, the theorem’s projection condition allows spherical covariance matrices that are not compound symmetric. In mean-and-contrast coordinates, a spherical covariance has a contrast block $\lambda I_{p-1}$, while the mean-to-contrast covariances can be nonzero. Compound symmetry is a special case with no such cross-covariances and with additional structure in the mean direction. This explains the logical relationship: compound symmetry is sufficient for sphericity, but is not necessary.

Worked Example: Detecting a Sphericity Violation

Consider

\Sigma= \begin{pmatrix} 4&1&0\\ 1&3&1\\ 0&1&2 \end{pmatrix}.

This is positive definite: its leading principal minors are $4$, $4(3)-1=11$, and $\det(\Sigma)=18$, all positive. For three occasions, use the orthonormal contrast vectors $q_1=(1,-1,0)'/\sqrt{2}$ and $q_2=(1,1,-2)'/\sqrt{6}$. Their contrast variances are

q_1'\Sigma q_1=\frac{4+3-2}{2}=\frac52, \qquad q_2'\Sigma q_2=\frac{13}{6}.

These variances differ, since $5/2=15/6\ne13/6$. In addition,

q_1'\Sigma q_2=\frac{3}{\sqrt{12}}\ne0.

The contrast covariance matrix is therefore not a scalar multiple of the identity, so this covariance is not spherical. It is not compound symmetric either: its marginal variances and pairwise covariances are unequal. Checking positive definiteness alone does not establish either covariance structure.

Practical Interpretation and a Common Mistake

Sphericity is concerned with the spread of contrasts, not with every entry of the original covariance matrix. It allows unequal marginal variances and unequal pairwise covariances, provided the covariance restricted to the contrast space has the required form. Compound symmetry, by contrast, is a direct pattern imposed on the entries of $\Sigma$: all diagonal entries match, and all off-diagonal entries match. The distinction is especially useful when assessing whether an equal-correlation model is being assumed unnecessarily.

Common mistake: “Sphericity means all occasions have equal variance and all pairs have equal covariance, so it is the same as compound symmetry.” This reasoning confuses a sufficient covariance pattern with the condition on contrasts. Equal variances and equal covariances do imply sphericity, as the derivation shows, but sphericity only requires $C'\Sigma C$ to be a scalar identity. The worked spherical example has unequal variances and covariances while satisfying that contrast condition.

A second useful check is to respect contrast scaling. The pairwise difference $Y_i-Y_j$ has coefficient vector of squared length 2, so under sphericity its variance is $2\lambda$, not $\lambda$. It is the variance per unit squared contrast length that is common. Keeping this normalization explicit prevents an apparent disagreement between pairwise-difference calculations and orthonormal contrast coordinates.

Key takeaway: Compound symmetry imposes equal variances and equal covariances and therefore yields spherical contrast covariance. Sphericity is weaker: it fixes the covariance within the contrast space but leaves other directions of the full covariance matrix less constrained.

Check Your Understanding

Use the contrast-space definition to answer these questions.

Under compound symmetry with variance $\sigma^2$ and correlation $\rho$, what is the variance of a contrast $c'Y$ in terms of $c'c$?
Why does a common covariance component make no contribution to a zero-sum contrast?
Can marginal variances differ under sphericity? Explain what sphericity constrains instead.
For a contrast vector with squared length 2, what variance does sphericity imply when its multiplier is $\lambda$?
What feature of the contrast covariance matrix would show that a covariance matrix is not spherical?

Sphericity and Compound Symmetry

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