Steiger Directionality Tests

Why Compare Variance Explained?

Mendelian randomization (MR) uses genetic variants as instruments to investigate whether an exposure has a causal effect on an outcome. A directionality question is whether the genetic instruments account for more variation in the proposed exposure or in the outcome. If the genetic association is stronger with the exposure, that pattern is consistent with the proposed direction; if it is stronger with the outcome, it raises concern that the assumed direction may be reversed.

A Steiger directionality test formalizes this comparison using the amount of phenotype variance explained by a genetic variant, often expressed as a squared genotype–phenotype correlation. The comparison is about correlation magnitudes, not signed correlations. A negative genetic association can explain just as much variance as a positive association of the same magnitude.

Definition: For a genetic variant \(G\) and phenotype \(X\), let \(\rho_X=\operatorname{Corr}(G,X)\) in the population. The variance explained in \(X\) by a single variant in a simple linear association is \(R_X^2=\rho_X^2\). A Steiger comparison asks whether \(R_X^2\) exceeds \(R_Y^2\), equivalently whether \(|\rho_X|>|\rho_Y|\), for two phenotypes \(X\) and \(Y\).

The test is a directionality diagnostic, not a standalone proof of causation. Interpreting its result as evidence about causal order relies on the genetic associations being meaningfully comparable and on assumptions about measurement, sampling, and the instruments. In particular, a variant that affects the outcome through pathways other than the exposure can produce a variance comparison that does not represent the proposed causal chain.

Hypotheses and the Fisher-Transformation Test

Suppose \(r_X\) and \(r_Y\) are sample correlations between the same variant and phenotypes \(X\) and \(Y\), estimated in independent samples of sizes \(n_X\) and \(n_Y\). The hypotheses for a one-sided comparison favoring \(X\) are

H_0: |\rho_X|\leq|\rho_Y| \qquad\text{versus}\qquad H_a: |\rho_X|>|\rho_Y|.

Because \(R^2=\rho^2\), these hypotheses also compare the population variance explained. The Fisher transformation is \(z(r)=\operatorname{atanh}(r)=\frac{1}{2}\log\left(\frac{1+r}{1-r}\right)\). For nonzero correlations, the approximate sampling variance of \(z(r)\) is \(1/(n-3)\). Since the comparison is about magnitudes, use \(z(|r_X|)\) and \(z(|r_Y|)\), not \(z(r_X)\) and \(z(r_Y)\).

Test statistic: When the two phenotype samples are independent and the correlations are not near zero, an approximate one-sided statistic is

Z_{\text{Steiger}}= \frac{\operatorname{atanh}(|r_X|)-\operatorname{atanh}(|r_Y|)} {\sqrt{\frac{1}{n_X-3}+\frac{1}{n_Y-3}}}.

Under the boundary null \(|\rho_X|=|\rho_Y|\), this statistic is approximately standard normal. Large positive values favor \(X\); the one-sided p-value is \(P(Z\geq Z_{\text{Steiger}})\) for \(Z\sim N(0,1)\).

This approximation treats the two estimated correlations as independent. If the phenotype samples overlap, or if both correlations are estimated in the same individuals, their estimation errors are generally dependent; the denominator above is then not justified without accounting for that dependence. Near a population correlation of zero, taking an absolute value also creates a non-smooth boundary, so the usual Fisher approximation can be unreliable. Report the comparison and its uncertainty with these qualifications rather than treating every calculated p-value as definitive.

Why the Test Uses Correlation Magnitudes

The directionality target is \(R^2\), and the square removes the sign. Fisher transformation is increasing for positive inputs, so comparing \(\operatorname{atanh}(|r_X|)\) and \(\operatorname{atanh}(|r_Y|)\) preserves the ordering of the observed squared correlations. A signed comparison would answer a different question.

Worked Example: A Negative Association Can Explain More Variance

Suppose hypothetical data give \(r_X=-0.40\) and \(r_Y=0.20\). The variant's estimated variance explained is

r_X^2=(-0.40)^2=0.16,\qquad r_Y^2=(0.20)^2=0.04.

Thus the variant explains more sample variance in \(X\), even though its signed correlation with \(X\) is smaller than its correlation with \(Y\). The magnitude comparison is \(|r_X|=0.40>|r_Y|=0.20\). A test based on \(r_X-r_Y\) would compare signed correlations and could point in a different direction; it would not be a general test of which phenotype has more variance explained.

A short derivation connects regression output to a correlation estimate. In ordinary least squares with one predictor and an intercept, the regression test statistic satisfies \(t^2=F\), where \(F\) is the one-predictor regression statistic. The regression identity \(F=R^2/(1-R^2)\times(n-2)\), together with \(R^2=r^2\), gives

t^2=\frac{r^2}{1-r^2}(n-2) \quad\Longrightarrow\quad r^2=\frac{t^2}{t^2+n-2}.

The sign of \(r\) is the sign of the fitted slope. If the reported slope is \(\hat\beta\) and its standard error is \(SE\), then \(t=\hat\beta/SE\), so the corresponding squared-correlation estimate is

\widehat{R^2}= \frac{(\hat\beta/SE)^2}{(\hat\beta/SE)^2+n-2}.

This identity is exact for the usual single-predictor linear regression with an intercept and its ordinary least-squares standard error. With approximate sample degrees of freedom, it is common to use \(n\) in place of \(n-2\). Robust standard errors, logistic regression coefficients, and other association models do not in general obey this identity. For a set of variants, use a joint variance-explained estimate when appropriate; simply adding individual \(R^2\) values is only a reasonable approximation under conditions such as negligible linkage disequilibrium.

Worked Comparisons

Worked Example: Testing the Difference in Explained Variance

In a hypothetical analysis, a variant has \(r_X=-0.30\) in an exposure sample of \(n_X=500\), and \(r_Y=0.20\) in an independent outcome sample of \(n_Y=500\). The observed explained variances are \(0.30^2=0.0900\) and \(0.20^2=0.0400\), so the point estimates favor \(X\). To test the magnitude comparison, calculate

Z_{\text{Steiger}}= \frac{\operatorname{atanh}(0.30)-\operatorname{atanh}(0.20)} {\sqrt{1/497+1/497}} = \frac{0.30952-0.20273}{\sqrt{0.00402414}} = \frac{0.10679}{0.063436} = 1.683.

The one-sided standard-normal p-value is approximately \(0.046\). At the \(0.05\) level, this is nominal evidence that the magnitude of the genetic correlation, and hence variance explained, is greater for \(X\). It does not establish that \(X\) causes \(Y\), nor does it address whether the variant has other effects. The calculation assumes the two samples are independent and the Fisher approximation is adequate.

Worked Example: Estimating Explained Variance From Regression Output

Consider hypothetical single-variant linear regressions with an intercept. For the exposure, suppose \(n_X=10{,}000\), \(\hat\beta_X=0.08\) exposure standard deviations per allele, and \(SE_X=0.01\) in the same units. Then \(t_X=0.08/0.01=8\), and

\widehat{R_X^2}=\frac{8^2}{8^2+10{,}000-2} =\frac{64}{10{,}062} =0.00636.

For the outcome, suppose \(n_Y=12{,}000\), \(\hat\beta_Y=0.03\) outcome standard deviations per allele, and \(SE_Y=0.01\). Then \(t_Y=3\), giving

\widehat{R_Y^2}=\frac{3^2}{3^2+12{,}000-2} =\frac{9}{12{,}007} =0.000750.

The estimated variance explained is larger for the exposure: about \(0.636\%\), compared with \(0.0750\%\) for the outcome. This is a point-estimate comparison, not by itself the Fisher test above. The samples have different sizes, and a formal test requires appropriate uncertainty for the two correlation estimates. The calculation also assumes ordinary least-squares regression as specified; a beta and standard error from a different model cannot automatically be substituted.

Measurement Error, Scale, and Diagnostics

Worked Example: Unequal Reliability Can Reverse the Comparison

Suppose, hypothetically, that the true correlations of a variant with two underlying traits are \(0.30\) for \(X\) and \(0.25\) for \(Y\). The corresponding true explained variances are \(0.30^2=0.0900\) and \(0.25^2=0.0625\), favoring \(X\). Under a classical measurement-error model with error independent of the variant and trait, the observed correlation is attenuated by the square root of the measurement reliability. Suppose reliability is \(0.50\) for \(X\) and \(0.90\) for \(Y\). Then

r_{X,\text{obs}}=0.30\sqrt{0.50}=0.2121,\qquad r_{Y,\text{obs}}=0.25\sqrt{0.90}=0.2372.

The observed explained variances are \(0.2121^2=0.0450\) and \(0.2372^2=0.05625\). The observed comparison now favors \(Y\), despite the underlying comparison favoring \(X\). This illustrates why unequal measurement error can distort directionality. The calculation assumes classical, independent measurement error; more complicated error mechanisms need not follow this attenuation formula.

Phenotype definitions and scales matter as well. Changing a continuous phenotype by a linear rescaling with a nonzero slope leaves its correlation with a variant unchanged, including its magnitude. But a nonlinear transformation, categorization, or change from a continuous to a binary measurement can change the correlation and the variance explained. For a binary phenotype, the observed-scale correlation depends on prevalence and coding; it is not automatically the variance explained on an unobserved liability scale. A logistic-regression coefficient and standard error should not be inserted into the linear-regression identity without a justified conversion.

Before interpreting a result, check the data and analysis that produced both correlations. Confirm that the association estimates refer to the same variant and compatible allele coding, that each phenotype is measured on a clearly described scale, and that the sample sizes and uncertainty estimates correspond to the reported correlations. Also inspect whether one result is driven by very imprecise estimates, an unusual phenotype definition, sample selection, or incompatible association models. A directionality result can be sensitive to these features even when the arithmetic is correct.

Interpretation and Limitations

A result favoring \(X\) means that the estimated genetic association is stronger in magnitude with \(X\) on the scales and in the samples analyzed. Under the assumptions that make variance explained informative about direction, this supports \(X\) as the upstream phenotype. It does not show that \(X\) is causal, quantify a causal effect, or verify the instrumental-variable assumptions. In particular, horizontal pleiotropy, reverse causation, selection, unequal measurement error, or differences in phenotype scale may all complicate the interpretation.

A non-significant comparison is not evidence that the two directions are equally plausible. It may reflect weak associations, small samples, or imprecise variance-explained estimates. Likewise, a statistically significant comparison is not a guarantee that the preferred direction is correct. Treat the test as one diagnostic among the evidence relevant to an MR analysis, and report the estimated \(R^2\) values, the test assumptions, and the scale and sample for each phenotype.

Wrong reasoning: “The signed correlation with \(X\) is lower, so the variant explains less variance in \(X\).”
Correction: Compare squared correlations, or equivalently correlation magnitudes. The sign indicates association direction, while the square measures variance explained in this comparison.
Wrong reasoning: “The Steiger test favors \(X\), so \(X\) must cause \(Y\).”
Correction: It is directionality evidence conditional on measurement, sampling, and causal assumptions; it is not proof of causation or instrument validity.

Key takeaway: A Steiger directionality test compares \(\rho_X^2\) with \(\rho_Y^2\), not \(\rho_X\) with \(\rho_Y\). For independent samples and nonzero correlations, the Fisher-transformation statistic compares \(\operatorname{atanh}(|r_X|)\) and \(\operatorname{atanh}(|r_Y|)\); its conclusion must be interpreted in light of measurement reliability and phenotype scale.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A variant has \(r_X=0.25\) and \(r_Y=-0.35\). Which phenotype has the larger estimated variance explained?
What is the one-sided alternative for a test that \(X\) explains more variance than \(Y\)?
Why is the independent-samples Fisher statistic not automatically appropriate when both correlations come from the same participants?
In a single-predictor linear regression, calculate \(\widehat{R^2}\) if \(t=4\) and \(n=1{,}000\).
Give one reason a Steiger comparison could favor the phenotype with lower underlying genetic correlation.

Solutions. 1. \(r_X^2=0.25^2=0.0625\), while \(r_Y^2=(-0.35)^2=0.1225\); the estimate favors \(Y\). 2. \(H_a:|\rho_X|>|\rho_Y|\), equivalently \(\rho_X^2>\rho_Y^2\). 3. The estimates are dependent, so adding their independent-sample variances does not give the correct uncertainty. 4. \(\widehat{R^2}=4^2/(4^2+1{,}000-2)=16/1{,}014=0.0158\), or about \(1.58\%\). 5. Unequal measurement reliability can attenuate one observed correlation more than the other and reverse the comparison.

What You'll Learn