Diagnosing Instrument Strength With F Statistics

Why Diagnose Instrument Strength?

In Mendelian randomization (MR), a genetic variant is relevant as an instrument when it is associated with the exposure. The strength of that association matters: weak instruments can make causal estimates unstable and, under some designs, more susceptible to bias. An F statistic summarizes evidence about exposure prediction in a form that can be compared across variants or for a set of variants.

An F statistic is a strength diagnostic, not a test of all the instrumental-variable assumptions. A large value does not establish that variants are independent of confounders or affect the outcome only through the exposure. Those assumptions remain separate requirements for a causal interpretation.

Definition: For a single variant with estimated association $\hat{\beta}_{Xj}$ with the exposure and standard error $SE(\hat{\beta}_{Xj})$, the variant-level F statistic is $$ F_j=\left(\frac{\hat{\beta}_{Xj}}{SE(\hat{\beta}_{Xj})}\right)^2. $$ It is the square of the one-degree-of-freedom association $t$ statistic. The estimate and standard error must refer to the same exposure, variant, effect allele, and analysis.

For example, if $\hat{\beta}_{Xj}$ is measured in exposure standard deviations per effect allele, its standard error has the same units, so their ratio—and hence $F_j$—is dimensionless. The formula can be used with other consistent scales as well.

Where the F Statistic Comes From

Consider a linear regression of an exposure on one genetic variant and adjustment covariates. Under the usual linear-model assumptions, the $t$ statistic tests whether the variant's regression coefficient is zero. The corresponding $F$ test for this single coefficient has one numerator degree of freedom.

Justification: Let $RSS_0$ be the residual sum of squares from a model without the variant, and $RSS_1$ the residual sum of squares from the model with it. If the full model has residual degrees of freedom $df$, then the one-coefficient partial $F$ statistic is $$ F=\frac{(RSS_0-RSS_1)/1}{RSS_1/df}. $$ The estimated coefficient's standard error is based on the same residual variance estimate $RSS_1/df$. For a single coefficient, the squared $t$ statistic equals the ratio of its explained sum of squares to that variance estimate. Therefore $F=t^2=(\hat{\beta}_{Xj}/SE(\hat{\beta}_{Xj}))^2$.

In a conventional linear regression, the exact $F$ reference distribution relies on a correctly specified model with independent errors of constant variance and normally distributed errors. In large genome-wide association studies, association statistics may instead use robust or other standard errors. Squaring the reported coefficient-to-standard-error ratio still gives a useful Wald strength statistic, but its exact small-sample $F$ interpretation need not apply.

Worked Example: F for One Variant

Suppose a hypothetical exposure GWAS estimates a variant's effect as $0.04$ exposure standard deviations per effect allele, with $SE=0.01$ standard deviations per allele. The estimates are aligned to the same effect allele and come from the same association analysis.

Substitute the estimate and standard error: $$ F_j=\left(\frac{0.04}{0.01}\right)^2=4^2=16. $$

The variant-level F statistic is 16. This indicates a comparatively strong exposure association by the commonly used rule of thumb discussed below; it does not show that the variant meets the other MR assumptions. Because the estimate and its standard error use the same units, the ratio is unitless.

Strength for a Set of Variants

With multiple instruments, “aggregate F” can refer to different summaries. They answer related but distinct questions, so the calculation and interpretation should be stated explicitly.

One summary is the mean variant-level F statistic, the arithmetic average of the individual statistics. It describes the average strength across the selected variants, but can hide individual weak instruments. It is not generally the same quantity as the joint first-stage F statistic.

A joint first-stage F statistic tests whether a set of variants jointly contributes to predicting the exposure in a regression. Suppose $N$ individuals have exposure data, $K$ variants are included, and the model adjusts for $c$ covariates in addition to an intercept. Let $RSS_0$ denote the residual sum of squares from the covariate-only model and $RSS_1$ that from the model including the variants. The joint statistic is

F_{\mathrm{joint}}= \frac{(RSS_0-RSS_1)/K}{RSS_1/(N-c-K-1)}.

If $R^2_{\mathrm{partial}}=(RSS_0-RSS_1)/RSS_0$ is the proportion of variation left after covariate adjustment that is explained by the variants, the same statistic can be written

F_{\mathrm{joint}}= \frac{R^2_{\mathrm{partial}}/K}{(1-R^2_{\mathrm{partial}})/(N-c-K-1)}.

This expression assumes the specified regression and degrees of freedom are appropriate. It is not safe to substitute a measure of total model $R^2$ without checking how that $R^2$ was defined. In summary-data settings, an aggregate approximation may be available from reported explained variance and sample size, but its validity depends on the data and assumptions used to obtain those quantities. With correlated variants, a joint analysis must account for their correlation; simply treating their associations as independent can misstate joint strength.

Worked Example: Joint F from Partial Explained Variance

Suppose a hypothetical exposure study has $N=10{,}000$ participants, includes $K=3$ variants, and adjusts for $c=5$ covariates besides the intercept. The variants jointly explain $R^2_{\mathrm{partial}}=0.006$ of the exposure variation remaining after covariate adjustment. Assume the regression and the stated degrees of freedom are appropriate.

The residual degrees of freedom are $$ df=10{,}000-5-3-1=9{,}991. $$

Substitution gives $$ F_{\mathrm{joint}}= \frac{0.006/3}{(1-0.006)/9{,}991} =\frac{0.002}{0.994/9{,}991} \approx 20.10. $$

The joint first-stage F statistic is about 20.10. It summarizes the joint contribution of these three variants under the regression model; it does not mean each variant has an F statistic of 20.10. For contrast, if their individual F statistics were 25, 9, and 4, their mean variant-level F would be $(25+9+4)/3=12.67$. The mean and joint statistics are different summaries and need not agree.

Interpreting Thresholds and Other Summaries

An F statistic is a measure of association strength, not a probability that an instrument is valid. A small p-value for the exposure association is also not a substitute for an F statistic: p-values depend on sample size as well as effect precision, and a very large sample can produce a small p-value for a modest association. Conversely, $R^2$ directly describes explained variation but does not incorporate sample size in the same way as an F statistic.

A conventional guideline is that an F statistic greater than 10 suggests that weak-instrument concerns may be less severe in some settings. Treat this as a heuristic, not a universal cutoff. Its usefulness depends on the design, how strength was calculated, the number and dependence of instruments, and the precision and sample structure of the exposure associations. State whether the threshold is being applied to each variant, a mean F, or a joint first-stage F; these are not interchangeable.

In two-sample MR, variant-level F statistics are often calculated from exposure-association estimates and standard errors. A mean across variants can be useful as a summary, but it should not obscure a weak variant. In one-sample settings, a joint first-stage F from the exposure regression is a direct measure of the set's contribution in that sample. The appropriate summary depends on the design and available data.

Worked Example: Strong F Does Not Guarantee an Unbiased Estimate

Consider hypothetical exposure associations for two variants: $\hat{\beta}_{X1}=0.10$ and $\hat{\beta}_{X2}=0.08$, each with standard error $0.02$, in exposure standard deviations per effect allele. Both estimates are assumed to be correctly aligned and reported with their corresponding standard errors.

Their strength statistics are $$ F_1=(0.10/0.02)^2=25,\qquad F_2=(0.08/0.02)^2=16. $$ Both exceed 10. Now suppose the true causal effect were $0.50$ outcome units per exposure standard deviation, but the variants also had direct effects on the outcome of $0.02$ and $-0.01$ outcome units per allele, respectively. Their outcome associations would then be $$ \hat{\beta}_{Y1}=0.10(0.50)+0.02=0.07,\qquad \hat{\beta}_{Y2}=0.08(0.50)-0.01=0.03. $$

The corresponding Wald ratios are $0.07/0.10=0.70$ and $0.03/0.08=0.375$, rather than the causal effect of $0.50$. The large exposure F statistics do not remove these direct effects. This example illustrates why strength is only one part of instrument assessment: F statistics cannot verify the exclusion restriction or guarantee an unbiased causal estimate.

Diagnostics, Failure Modes, and Limits

Before interpreting a reported F statistic, check that the exposure effect and standard error refer to the same variant, effect allele, exposure scale, and sample. A mismatched allele or incorrectly paired standard error can produce a meaningless value. Also check how the statistic was calculated: a squared association $t$, a mean of variant-level statistics, and a joint first-stage test should not be described as though they were the same diagnostic.

For a set of instruments, inspect the individual F statistics as well as an aggregate summary. A high mean can coexist with one or more weak variants. If variants are correlated, use a method that incorporates their dependence for joint assessment. If the necessary individual-level data or correlation information is unavailable, describe the limitations of the available summary rather than treating an independence-based calculation as exact.

Most importantly, a high F statistic is not a certificate of unbiasedness. It speaks to the relevance of the variants for the exposure, not to genetic independence, the exclusion restriction, or the accuracy of every other modeling choice. In addition, the strength estimate may be imprecise, and its interpretation depends on the sample and analysis that produced it. A diagnostic should inform the design and interpretation of an MR analysis, not replace them.

Common pitfall: “Every variant has $F>10$, so the MR estimate must be unbiased.” This reasoning confuses exposure relevance with the full set of assumptions needed for causal interpretation. The correction is to report how strength was measured, inspect both variant-level and appropriate aggregate diagnostics, and assess the other instrumental-variable assumptions separately.

Check Your Understanding

Try each question, then compare your work with the solutions below.

A variant has an exposure association estimate of $0.06$ and standard error $0.015$, in the same units. Calculate its F statistic.
Explain why the mean of individual variant-level F statistics is not necessarily the joint first-stage F statistic.
A study has $N=2{,}000$, $K=2$, $c=3$, and partial $R^2=0.004$. Calculate the joint F statistic using the stated formula.
What does the conventional $F>10$ guideline suggest, and what does it fail to establish?

Solutions. (1) $F=(0.06/0.015)^2=4^2=16$. (2) The mean averages separate one-variant statistics, whereas the joint F tests the variants together in a regression and depends on their combined explanatory contribution and degrees of freedom. (3) The residual degrees of freedom are $2{,}000-3-2-1=1{,}994$. Thus $F=(0.004/2)/[(1-0.004)/1{,}994]=0.002/(0.996/1{,}994)\approx4.004$. (4) It is a heuristic suggesting that weak-instrument concerns may be less severe under some conditions. It does not prove instrument validity or guarantee an unbiased causal estimate.