The Rho Parameter in SKAT-O

Why a Correlation Parameter Connects Burden and SKAT

Rare-variant sets can contain effects that point in different directions, or effects that mostly point in the same direction. A burden test gains efficiency when effects align, because it aggregates variant contributions; SKAT is more accommodating when effects vary in direction, because it aggregates squared contributions. SKAT-O connects these two forms of evidence through a parameter, $\rho$, that specifies a correlation pattern among variant effects.

The important interpretation is that $\rho$ describes a working model for the effects across variants. It does not change the genotypes, the observed linkage disequilibrium, or the null covariance of the score vector. Those features still affect the test’s null distribution, but they are distinct from the effect-correlation pattern indexed by $\rho$.

Definition: For a set of $m$ variants, let $\mathbf{1}$ be the $m$-vector of ones and $I_m$ the $m$-by-$m$ identity matrix. The SKAT-O equicorrelation matrix is $$ R_\rho=(1-\rho)I_m+\rho\mathbf{1}\mathbf{1}^{\mathsf T},\qquad 0\leq\rho\leq1. $$ Its diagonal entries are 1, and every off-diagonal entry is $\rho$. It represents equal marginal variances and a common pairwise correlation for the standardized variant effects.

One way to express the working effect model is $\boldsymbol{\beta}\sim N(\mathbf{0},\tau^2 W R_\rho W)$, where $\boldsymbol{\beta}$ is the vector of variant effects, $W$ is the diagonal matrix of fixed, nonnegative variant-weight factors, and $\tau^2$ is a variance scale. Equivalently, the standardized effects before applying the weights have covariance proportional to $R_\rho$. This is a model for alternatives, not a claim that the effects in a particular data set are known to have that distribution.

How Rho Changes the Effect Covariance

At $\rho=0$, $R_\rho=I_m$: the standardized effects are uncorrelated under the working model. At $\rho=1$, $R_\rho=\mathbf{1}\mathbf{1}^{\mathsf T}$: all standardized effects are perfectly positively correlated. Intermediate values specify partial positive correlation. For example, when $m=2$, the variance of the difference between the two standardized effects, each with variance $\tau^2$, is $2\tau^2(1-\rho)$. Increasing $\rho$ therefore concentrates variation in the shared, same-direction component and reduces variation in contrasts between variants.

The eigenstructure makes this change precise. The vector $\mathbf{1}$ is an eigenvector of $R_\rho$ with eigenvalue $1+(m-1)\rho$. Every vector orthogonal to $\mathbf{1}$ has eigenvalue $1-\rho$. Thus the common-direction component gains relative variance as $\rho$ increases, while the $m-1$ contrast directions lose variance. At $\rho=1$, the contrast eigenvalues are zero, so the matrix is singular: the model allows variation only along the common direction.

Worked Example: Reading the Correlation Pattern

Suppose a hypothetical three-variant set uses $\rho=0.4$. The correlation matrix for the standardized effects is

R_{0.4}= \begin{pmatrix} 1&0.4&0.4\\ 0.4&1&0.4\\ 0.4&0.4&1 \end{pmatrix}.

Each diagonal entry is 1, and each off-diagonal entry is 0.4, so each pair of standardized effects has correlation 0.4. The common-direction eigenvalue is $1+2(0.4)=1.8$; the two contrast eigenvalues are $1-0.4=0.6$. Their sum is $1.8+0.6+0.6=3$, which equals the trace of the matrix, the sum of its diagonal entries. Compared with $\rho=0$, more model variation is assigned to the common direction and less to contrasts. It is still a partial-correlation model, not a requirement that all three realized effects have the same sign.

For $\rho$ between 0 and 1, the eigenvalues just given are nonnegative, so $R_\rho$ is positive semidefinite and is a valid covariance structure. SKAT-O uses this nonnegative range to interpolate between independent standardized effects and perfect positive dependence. A broader mathematical equicorrelation family can include negative correlations under additional bounds, but those are not part of this SKAT-O parameterization.

From Effect Correlation to the SKAT-O Statistic

Let $U$ be the vector of variant score statistics from a covariate-adjusted null model, let $V$ be its covariance matrix under the null, and let $W$ contain the fixed variant-weight factors. For a fixed $\rho$, the SKAT-O quadratic statistic is

Q_\rho=U^{\mathsf T}W R_\rho W U.

Expanding $R_\rho$ gives the central interpolation identity:

Q_\rho=(1-\rho)U^{\mathsf T}W^2U+ \rho\,U^{\mathsf T}W\mathbf{1}\mathbf{1}^{\mathsf T}WU =(1-\rho)Q_{\mathrm{SKAT}}+\rho Q_{\mathrm{burden}},

where $Q_{\mathrm{SKAT}}=U^{\mathsf T}W^2U$, and $Q_{\mathrm{burden}}=(\mathbf{1}^{\mathsf T}WU)^2$. Thus $\rho=0$ gives the SKAT quadratic statistic and $\rho=1$ gives the squared weighted burden score. For an intermediate value, the statistic is a weighted average of these endpoint statistics. It remains nonnegative because it is a quadratic form with a positive-semidefinite matrix.

Worked Example: Calculating an Intermediate Statistic

Consider hypothetical score values $U=(2,-1,3)^{\mathsf T}$, weight factors $W=\operatorname{diag}(1,0.5,1.5)$, and $\rho=0.4$. The weighted score contributions are $WU=(2,-0.5,4.5)^{\mathsf T}$.

The SKAT endpoint is the sum of their squares:

Q_{\mathrm{SKAT}}=2^2+(-0.5)^2+4.5^2=4+0.25+20.25=24.5.

The weighted burden score is their sum, $2-0.5+4.5=6$, so its squared statistic is $Q_{\mathrm{burden}}=6^2=36$. Therefore,

Q_{0.4}=0.6(24.5)+0.4(36)=14.7+14.4=29.1.

The value 29.1 is between the two endpoint statistics, as the convex-combination formula requires. This calculation gives the statistic, not a p-value: a p-value also requires the null distribution, which depends on $V$, the weights, and the chosen $\rho$.

Null Distribution and a Short Derivation

For a fixed $\rho$, suppose the null model is correctly specified and the score vector is approximately multivariate normal with mean zero and covariance $V$: $U\sim N(\mathbf{0},V)$. Assume the weights and $\rho$ are fixed for this calculation. If $V$ is positive semidefinite, write $U=V^{1/2}Z$, where $Z\sim N(\mathbf{0},I)$. Define the symmetric positive-semidefinite matrix $A_\rho=V^{1/2}W R_\rho W V^{1/2}$. Then

Q_\rho=Z^{\mathsf T}A_\rho Z.

Diagonalize $A_\rho=P\Lambda P^{\mathsf T}$, where $P$ is orthogonal and $\Lambda$ has nonnegative eigenvalues $\lambda_1,\ldots,\lambda_m$. The rotated vector $P^{\mathsf T}Z$ still has independent standard normal components, because an orthogonal rotation preserves the standard multivariate normal distribution. Consequently,

Q_\rho=\sum_{j=1}^{m}\lambda_j(P^{\mathsf T}Z)_j^2 \ \sim\ \sum_{j=1}^{m}\lambda_j\chi^2_{1,j},

where the $\chi^2_{1,j}$ variables are independent chi-square random variables with one degree of freedom. This is a weighted mixture of chi-square variables, with weights determined by the eigenvalues of $A_\rho$. The mixture changes with $\rho$; the same observed quadratic statistic cannot be assigned a valid p-value by using an endpoint distribution if a different $\rho$ was used to compute it.

Worked Example: Effect Covariance at Two Variants

Suppose a hypothetical two-variant model has equal marginal effect variance $\tau^2=0.04$, equal weight factors of 1, and $\rho=0.75$. The effect covariance matrix is

\operatorname{Cov}(\boldsymbol{\beta}) =0.04 \begin{pmatrix} 1&0.75\\ 0.75&1 \end{pmatrix} = \begin{pmatrix} 0.04&0.03\\ 0.03&0.04 \end{pmatrix}.

The marginal standard deviation of either effect is $\sqrt{0.04}=0.2$, and their correlation is $0.03/\sqrt{0.04(0.04)}=0.75$. The variance of their difference is

\operatorname{Var}(\beta_1-\beta_2)=0.04+0.04-2(0.03)=0.02.

If $\rho=0$ with the same marginal variances, that contrast variance would be $0.04+0.04=0.08$. The higher $\rho$ reduces variation in differences and favors a shared-direction pattern in the alternative model. It does not prove that the variants have equal observed effects or that the variants are in linkage disequilibrium.

Assumptions, Diagnostics, and Interpretation

The fixed-$\rho$ test relies on an appropriate null score model and its covariance $V$. Covariate adjustment, phenotype type, and sample structure must be handled in constructing the scores and their covariance; a correlation parameter cannot repair a misspecified null model. The usual mixture-chi-square calculation also relies on a sufficiently accurate normal approximation for the score vector. With very sparse binary data, that approximation may require special calibration.

Weights should be specified independently of the observed association scores for the stated null calculation to apply directly. Variant inclusion, allele coding, and weight definitions should be checked: a coding or weight error can change the burden direction and the quadratic statistic. The eigenvalues used for the null distribution should be nonnegative up to numerical tolerance. A substantial negative eigenvalue can indicate a numerical or covariance-construction problem, not a meaningful negative mixture weight.

A useful conceptual diagnostic is to compare the weighted burden and SKAT endpoint statistics. A large burden score reflects coherent signed contributions after weighting; a large SKAT score can arise from large contributions that cancel in the burden sum. But neither endpoint comparison nor a selected intermediate $\rho$ identifies the true effect architecture. The test assesses association under a specified score construction; it does not estimate $\rho$ as a biological parameter.

How to Use the Endpoints and What They Do Not Say

Parameter value	Effect pattern represented	Statistic	Main implication
$\rho=0$	Uncorrelated standardized effects	$Q_{\mathrm{SKAT}}$	Variant contributions are squared, so opposite signs do not cancel.
$0<\rho<1$	Partially positively correlated standardized effects	$(1-\rho)Q_{\mathrm{SKAT}}+\rho Q_{\mathrm{burden}}$	Evidence from the common direction and variant-specific deviations is combined.
$\rho=1$	Perfectly positively correlated standardized effects	$Q_{\mathrm{burden}}$	Weighted contributions are aggregated before squaring, so opposing signs can cancel.

This comparison explains the advantage and the limitation of the interpolation. A burden endpoint can be powerful when effects align but can lose signal when positive and negative contributions cancel. The SKAT endpoint avoids that cancellation but may be less efficient when a common-direction pattern is strong. An intermediate $\rho$ changes the balance; it does not eliminate the need for an appropriate null distribution. Selecting among multiple $\rho$ values based on their observed p-values also requires accounting for that selection rather than interpreting the smallest fixed-$\rho$ p-value as if only one test had been run.

Common Mistake

Incorrect reasoning: “A high $\rho$ means the variants are in strong linkage disequilibrium, so the score covariance changes accordingly.” This confuses two different covariance structures. Linkage disequilibrium concerns dependence among genotypes and contributes to the score covariance $V$. The SKAT-O $\rho$ parameter specifies dependence among effects in the working alternative and changes the kernel $W R_\rho W$. Both can influence the null distribution, but $\rho$ is not an LD estimate.

A second mistake is to say that $\rho=1$ forces every realized estimated effect to be identical. It is the endpoint of the specified covariance model: the standardized random effects have no variation in contrast directions under that model. The observed estimates still contain sampling variation, and weights can make the corresponding unstandardized effects differ in scale.

Key takeaway: $R_\rho$ allocates effect variation between a common direction and variant-specific contrasts. Through $Q_\rho=(1-\rho)Q_{\mathrm{SKAT}}+\rho Q_{\mathrm{burden}}$, $\rho$ connects the variance-component and burden endpoints, while the score covariance $V$ separately determines how the statistic is calibrated under the null.

Check Your Understanding

Answer each question, then compare with the solutions below.

For $m=4$ and $\rho=0.25$, what are the common-direction eigenvalue and the three contrast eigenvalues of $R_\rho$?
If $Q_{\mathrm{SKAT}}=12$, $Q_{\mathrm{burden}}=20$, and $\rho=0.25$, calculate $Q_\rho$.
Explain why $\rho$ and linkage disequilibrium are not interchangeable.
What assumptions permit the fixed-$\rho$ null statistic to be represented as a weighted sum of independent $\chi^2_1$ variables?

Solutions. (1) The common-direction eigenvalue is $1+(4-1)(0.25)=1.75$. Each of the three contrast eigenvalues is $1-0.25=0.75$. (2) $Q_{0.25}=0.75(12)+0.25(20)=9+5=14$. (3) $\rho$ specifies correlation among standardized effects in the alternative model, whereas linkage disequilibrium is dependence among genotypes and affects $V$, the null score covariance. (4) The null model must yield an approximately mean-zero multivariate normal score vector with correctly specified positive-semidefinite covariance $V$; the weights and $\rho$ are fixed for the calculation. Writing $U=V^{1/2}Z$ and diagonalizing $V^{1/2}W R_\rho W V^{1/2}$ then gives the stated weighted chi-square representation.

Parameter value	Effect pattern represented	Statistic	Main implication
\(\rho=0\)	Uncorrelated standardized effects	\(Q_{\mathrm{SKAT}}\)	Variant contributions are squared, so opposite signs do not cancel.
\(0<\rho<1\)	Partially positively correlated standardized effects	\((1-\rho)Q_{\mathrm{SKAT}}+\rho Q_{\mathrm{burden}}\)	Evidence from the common direction and variant-specific deviations is combined.
\(\rho=1\)	Perfectly positively correlated standardized effects	\(Q_{\mathrm{burden}}\)	Weighted contributions are aggregated before squaring, so opposing signs can cancel.

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