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Mendelian Randomization Foundations · Tutorial 841 of 989

Instrument Strength And The F Statistic

Calculate explained exposure variance and first-stage F statistics, and interpret what these strength measures do—and do not—establish about a genetic instrument.

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What You'll Learn

  • Define marginal and conditional explained variance for genetic instruments
  • Calculate a single-variant instrument’s explained exposure variance from allele frequency and effect size
  • Derive the first-stage F statistic from the partial R-squared
  • Assess a combined instrument with a joint first-stage test
  • Recover approximate strength measures from summary-data exposure estimates
  • Interpret the F statistic without treating a rule of thumb as a guarantee

Why Instrument Strength Matters

In Mendelian randomization (MR), an instrument must predict the exposure for an instrumental-variable estimate to be informative. This requirement is called instrument relevance. A genetic variant can be associated with an exposure but explain only a very small fraction of its variation, especially in a large sample. Measuring that explained fraction and the corresponding first-stage F statistic helps describe the strength of the exposure association.

Strength is a property of the instrument’s association with the exposure in the study data. It is not, by itself, evidence that the instrument satisfies every condition needed for a causal interpretation. Here the focus is on quantifying relevance, for a single variant and for variants tested together.

Definition: In a first-stage regression, the exposure is the response and the genetic instrument or instruments are predictors, usually alongside any covariates included in the analysis. Instrument relevance concerns whether the instrument coefficients predict the exposure. The partial \(R^2\) is the fraction of exposure variation left after accounting for covariates that is additionally explained by the instrument set.

Explained Exposure Variance

For a single variant, the explained exposure variance is often reported as an \(R^2\). In a simple regression with an intercept and no additional covariates, this is the proportion of sample variation in the exposure explained by the variant. With covariates, it is important to distinguish the variant’s contribution beyond those covariates from the total \(R^2\) of the full model.

Let \(R^2_{\text{cov}}\) denote the \(R^2\) from a regression of the exposure on covariates alone, and let \(R^2_{\text{full}}\) denote the \(R^2\) after adding the \(k\) genetic predictors. The partial \(R^2\) is the increase in explained variation divided by the variation not already explained by the covariates:

$$ R^2_{\text{partial}} = \frac{R^2_{\text{full}}-R^2_{\text{cov}}}{1-R^2_{\text{cov}}}. $$

This is a conditional measure: it describes the fraction of the remaining exposure variation explained by the instruments after adjustment for the covariates. If there are no covariates, \(R^2_{\text{cov}}=0\), so partial \(R^2\) equals the ordinary \(R^2\).

For an additive dosage \(G\) with effect \(\beta_{XG}\) on a continuous exposure \(X\), an approximate population calculation for one variant is

$$ R^2 \approx \frac{\operatorname{Var}(G)\beta_{XG}^2}{\operatorname{Var}(X)} = \frac{2f(1-f)\beta_{XG}^2}{\operatorname{Var}(X)}. $$

Here \(f\) is the frequency of the coded effect allele. The second equality uses the approximation \(\operatorname{Var}(G)=2f(1-f)\), which holds under Hardy–Weinberg equilibrium for an allele dosage coded 0, 1, or 2. The effect estimate and variance must use compatible units. For example, if \(X\) is standardized to variance 1, \(\beta_{XG}\) is measured in standard deviations of exposure per effect allele.

Worked Example: Explained Variance for One Variant

Suppose a hypothetical study has \(N=5{,}000\) participants. A variant has effect-allele frequency \(f=0.30\) and an estimated additive effect of \(0.08\) exposure standard deviations per allele. Assume the exposure is standardized, the genotype dosage follows the stated Hardy–Weinberg approximation, and there are no covariates in this first-stage calculation.

The dosage variance is \(2f(1-f)=2(0.30)(0.70)=0.42\). Since the exposure variance is 1, the estimated explained variance is

$$ R^2 \approx \frac{0.42(0.08)^2}{1} = 0.42(0.0064) = 0.002688. $$

Thus the variant explains about \(0.2688\%\) of the exposure variance in this approximation. The percentage is \(100(0.002688)=0.2688\%\), not \(2.688\%\). The assumptions used here are explicit: the effect is per allele, the exposure variance is 1, and the dosage variance is approximated from the allele frequency under Hardy–Weinberg equilibrium.

From Partial \(R^2\) to the First-Stage F Statistic

The first-stage F statistic tests the joint null hypothesis that all \(k\) instrument coefficients in the exposure regression are zero, conditional on the covariates. Under the usual linear regression model with independent, normally distributed errors of constant variance, the ordinary partial F test has an exact F distribution under that null. Let \(c\) be the number of covariate columns, excluding the intercept, and assume the design matrix has full column rank. The residual degrees of freedom in the full model are \(N-c-k-1\).

The relationship between this F statistic and partial \(R^2\) is

$$ F = \frac{R^2_{\text{partial}}/k} {(1-R^2_{\text{partial}})/(N-c-k-1)} = \frac{N-c-k-1}{k} \frac{R^2_{\text{partial}}}{1-R^2_{\text{partial}}}. $$

The numerator measures the average fraction of residual exposure variation explained per tested instrument coefficient. The denominator measures the remaining unexplained variation per residual degree of freedom. A larger partial \(R^2\), a larger sample size, or fewer tested coefficients generally yields a larger F statistic, all else equal.

Why the F Formula Uses Partial \(R^2\)

The relationship follows directly from the usual partial F test. Write \(\mathrm{SSE}_{\text{cov}}\) for the residual sum of squares from the covariate-only regression and \(\mathrm{SSE}_{\text{full}}\) for that from the full regression. Adding \(k\) predictors reduces the residual sum of squares by \(\mathrm{SSE}_{\text{cov}}-\mathrm{SSE}_{\text{full}}\). The partial F statistic is this reduction per added coefficient divided by the full-model residual mean square:

$$ F = \frac{(\mathrm{SSE}_{\text{cov}}-\mathrm{SSE}_{\text{full}})/k} {\mathrm{SSE}_{\text{full}}/(N-c-k-1)}. $$

The partial \(R^2\) can also be written as

$$ R^2_{\text{partial}} = \frac{\mathrm{SSE}_{\text{cov}}-\mathrm{SSE}_{\text{full}}} {\mathrm{SSE}_{\text{cov}}}, \qquad 1-R^2_{\text{partial}} = \frac{\mathrm{SSE}_{\text{full}}}{\mathrm{SSE}_{\text{cov}}}. $$

Substituting these two ratios into the F expression gives the stated formula: the sum-of-squares reduction is \(R^2_{\text{partial}}\mathrm{SSE}_{\text{cov}}\), while the full-model error sum of squares is \((1-R^2_{\text{partial}})\mathrm{SSE}_{\text{cov}}\). The common factor \(\mathrm{SSE}_{\text{cov}}\) cancels.

For one instrument and no covariates, \(k=1\), \(c=0\), and the residual degrees of freedom are \(N-2\). Therefore,

$$ F = (N-2)\frac{R^2}{1-R^2}. $$

Worked Example: F Statistic for One Variant

Use the hypothetical single-variant study above, with \(N=5{,}000\), \(R^2=0.002688\), one tested variant, and no covariates. The ordinary first-stage F statistic has \(1\) numerator degree of freedom and \(4{,}998\) denominator degrees of freedom.

$$ F = \frac{4{,}998(0.002688)}{1-0.002688} = \frac{13.434624}{0.997312} \approx 13.47. $$

The numerator is \(4{,}998(0.002688)=13.434624\), and the denominator is \(0.997312\). The resulting F statistic is approximately \(13.47\). Under the conventional linear-model test, this is the observed first-stage test statistic for the null of no exposure association. It is not a measure of the variant’s effect on the outcome.

Assessing a Combined Instrument

When several variants are used together, the relevant strength measure is generally the joint first-stage test of their coefficients, conditional on the covariates. The corresponding partial \(R^2\) measures how much additional exposure variation the set explains. Variants in linkage disequilibrium can carry overlapping information, so adding their separate marginal \(R^2\) values does not in general give the set’s conditional \(R^2\). The joint regression accounts for their relationships through its design matrix.

Worked Example: Joint Strength of Two Variants

In a hypothetical sample of \(N=10{,}000\), suppose a covariate-only model explains \(R^2_{\text{cov}}=0.200\) of exposure variation, and a full model adding two genetic variants has \(R^2_{\text{full}}=0.208\). There are \(c=3\) covariate columns, excluding the intercept, and the full model has full column rank. The instruments’ partial \(R^2\) is

$$ R^2_{\text{partial}} = \frac{0.208-0.200}{1-0.200} = \frac{0.008}{0.800} = 0.010. $$

Thus the two variants jointly explain \(1.00\%\) of the exposure variation remaining after the covariates. The full-model residual degrees of freedom are \(10{,}000-3-2-1=9{,}994\). The joint F statistic is

$$ F = \frac{9{,}994}{2} \frac{0.010}{1-0.010} = 4{,}997\frac{0.010}{0.990} \approx 50.47. $$

The test has \(2\) numerator and \(9{,}994\) denominator degrees of freedom. It tests whether both variant coefficients are zero in the exposure regression, given the covariates. It does not imply that each variant individually has an F statistic of \(50.47\), nor can the joint strength be inferred by simply adding two marginal \(R^2\) values.

Using Summary-Data Associations

For a single variant, a commonly available summary-data approximation is \(F\approx(\hat{\beta}_{XG}/SE)^2\), where \(\hat{\beta}_{XG}\) is the estimated variant–exposure association and \(SE\) is its standard error. This is the squared Wald statistic for testing a zero coefficient. In ordinary linear regression with one tested coefficient and conventional standard errors, the corresponding partial F statistic equals that squared statistic. With summary data, differing standard-error methods, or approximations in the reported estimates, treat the equality as approximate.

Worked Example: Strength From a Summary Association

Suppose a hypothetical exposure GWAS reports a per-allele association of \(\hat{\beta}_{XG}=0.045\) exposure units, with \(SE=0.012\) exposure units, for one variant. The squared association statistic is

$$ F \approx \left(\frac{0.045}{0.012}\right)^2 = (3.75)^2 = 14.0625 \approx 14.06. $$

If this is a conventional one-variant first-stage test with \(7{,}994\) residual degrees of freedom, the corresponding partial-\(R^2\) relation can be rearranged as \(R^2_{\text{partial}}=F/(F+df)\). Hence

$$ R^2_{\text{partial}} \approx \frac{14.0625}{14.0625+7{,}994} = \frac{14.0625}{8{,}008.0625} \approx 0.001756. $$

The estimated partial explained fraction is about \(0.1756\%\). This conversion relies on the stated residual degrees of freedom and the one-coefficient F relationship; it is not a general conversion for a multi-variant joint test from a single variant’s summary association.

Interpreting Strength Measures Carefully

The F statistic combines the amount of exposure variation explained with the information available in the sample. Consequently, the same \(R^2\) can produce different F statistics in samples of different sizes. Conversely, a large F statistic in a very large sample can coexist with a small explained fraction. Reporting both a variance measure and an F statistic gives a more complete description than reporting either alone.

An F statistic above 10 is often used as a rough screening rule in discussions of weak-instrument concerns. It is a heuristic, not a universal boundary: it does not guarantee that an instrument is strong enough for every analysis, and falling below it does not itself quantify the bias or uncertainty in a particular estimate. Interpretation depends on such features as the study design, number of instruments, sample size, and how the statistic was calculated. A strength statistic also cannot establish the other assumptions required for causal interpretation.

Common Mistakes

  • Using total \(R^2\) when covariates are present. The instrument-set F statistic uses the partial \(R^2\) for the added genetic predictors, not the full model’s total \(R^2\).
  • Adding marginal \(R^2\) values for correlated variants. Their explained variation can overlap. Use the joint regression’s partial \(R^2\) and joint F statistic.
  • Calling a large F statistic proof of a valid instrument. It quantifies exposure prediction; by itself it does not establish the assumptions needed for a causal interpretation.
  • Treating 10 as a guarantee. The F>10 rule is a rough convention, not a universal pass/fail test or a promise of negligible bias.
  • Ignoring degrees of freedom. For a joint test, use the number of tested instrument coefficients in the numerator and the full-model residual degrees of freedom in the denominator.
Key takeaway: Quantify genetic instrument relevance using the exposure variation explained beyond covariates and the corresponding first-stage F statistic. For a joint instrument set, calculate both from the joint first-stage regression; neither measure alone establishes causal validity.

Check Your Understanding

Work through these questions using the definitions and assumptions above.

  1. With no covariates, what are the numerator and denominator degrees of freedom for a first-stage F test of one variant in a sample of \(N\) individuals?
  2. A covariate-only model has \(R^2=0.25\), and the full model with instruments has \(R^2=0.28\). Calculate the instruments’ partial \(R^2\).
  3. Why is it generally incorrect to add the marginal \(R^2\) values of two variants in linkage disequilibrium to obtain their joint explained variance?
  4. A one-variant summary association has estimate \(0.06\) and standard error \(0.02\). Calculate the approximate F statistic and state what association it tests.
  5. In one or two sentences, explain why an F statistic above 10 is not proof that an instrument satisfies all requirements for causal interpretation.
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