Sample Overlap In Two-Sample MR

Why Sample Overlap Matters

In two-sample Mendelian randomization (MR), genetic associations with an exposure are estimated in one sample and associations with an outcome in another. Using separate samples can reduce the weak-instrument bias found in one-sample instrumental-variable analyses. But “two-sample” does not necessarily mean that the participants are entirely separate. If some people contribute to both association estimates, the estimates’ sampling errors can be correlated.

This correlation is a feature of the sampling process: shared participants contribute information about both associations. It does not, by itself, violate the instrumental-variable assumptions. Its practical importance is that when instruments are weak, correlated errors can make the MR estimate behave more like an observational association. The size of the risk depends on the degree of overlap, the residual covariance in the overlapping participants, and instrument strength.

Association Errors With Shared Participants

Consider one genetic instrument $G$, an exposure $X$, and an outcome $Y$. The exposure and outcome associations are estimated by separate linear regressions, each with an intercept. Let $n_X$ and $n_Y$ be their sample sizes, and let $m$ be the number of participants in both samples. Define $\mu_G=E[G]$, $Z=G-\mu_G$, and $v_G=\operatorname{Var}(G)>0$. The use of $Z$ is important: with an intercept, the slope is based on genotype variation around its mean.

Write $\beta_X$ and $\beta_Y$ for the population slopes of the exposure and outcome association regressions. Their residuals are $\varepsilon_X$ and $\varepsilon_Y$, respectively. Assume the linear regressions are correctly specified, $E[\varepsilon_X]=E[\varepsilon_Y]=0$, and $E[Z\varepsilon_X]=E[Z\varepsilon_Y]=0$. For the covariance calculation below, also assume that participants are independent of one another and that, in the target population, genotype is independent of the residual pair $(\varepsilon_X,\varepsilon_Y)$. Define $c=\operatorname{Cov}(\varepsilon_X,\varepsilon_Y)$. These assumptions are simplifying conditions for the displayed formula; in a real study, ancestry, relatedness, or other structure may require additional modelling.

Definition: Sample overlap is the number $m$ of participants contributing to both the exposure-association and outcome-association analyses. Association-error covariance is the covariance between $\hat{\beta}_X-\beta_X$ and $\hat{\beta}_Y-\beta_Y$, where $\hat{\beta}_X$ and $\hat{\beta}_Y$ are the estimated slopes. Under the assumptions stated above, its leading large-sample approximation is $$ \operatorname{Cov}(\hat{\beta}_X-\beta_X,\hat{\beta}_Y-\beta_Y) \approx \frac{m c}{n_X n_Y v_G}. $$

The formula makes the key dependencies explicit. With no shared participants, $m=0$, the covariance is zero if the samples are otherwise independent. With overlap, the sign follows the residual covariance $c$: positive residual covariance induces positive association-error covariance, and negative residual covariance induces negative covariance. The expression is an approximation based on the leading sampling errors, not a universal exact finite-sample identity.

Deriving the Overlap Covariance

For an ordinary least-squares slope with an intercept, the large-sample influence of one observation is proportional to its centered genotype multiplied by its regression residual. Thus, to leading order,

\hat{\beta}_X-\beta_X \approx \frac{1}{n_X v_G}\sum_{i\in S_X} Z_i\varepsilon_{Xi}, \qquad \hat{\beta}_Y-\beta_Y \approx \frac{1}{n_Y v_G}\sum_{i\in S_Y} Z_i\varepsilon_{Yi}, \qquad Z_i=G_i-\mu_G.

Here $S_X$ and $S_Y$ denote the exposure and outcome samples. When we take the covariance of these sums, terms involving different people have zero covariance under independence across participants. Only the $m$ shared participants contribute. For each shared participant, the relevant expectation is $E[Z^2\varepsilon_X\varepsilon_Y]$. Genotype independence from the residual pair gives

E[Z^2\varepsilon_X\varepsilon_Y] =E[Z^2]E[\varepsilon_X\varepsilon_Y] =v_G c.

The second equality uses zero residual means, so $E[\varepsilon_X\varepsilon_Y]=\operatorname{Cov}(\varepsilon_X,\varepsilon_Y)=c$. Combining the $m$ shared contributions with the two slope denominators yields

\operatorname{Cov}(\hat{\beta}_X-\beta_X,\hat{\beta}_Y-\beta_Y) \approx \frac{m}{n_X n_Y v_G^2}E[Z^2\varepsilon_X\varepsilon_Y] = \frac{m c}{n_X n_Y v_G}.

Centering is not optional in this calculation. If $E[G]\ne 0$, the score contribution uses $G-E[G]$, not $G$. For example, if $E[G]=1$ and $v_G=0.5$, then $E[Z^2]=0.5$; replacing $Z$ by $G$ would instead use $E[G^2]=1.5$, which gives the wrong covariance moment.

Worked Example: Quantifying Error Covariance From Overlap

Suppose a hypothetical study estimates the exposure association in $n_X=10{,}000$ people and the outcome association in $n_Y=12{,}000$ people, with $m=3{,}000$ participants shared. Assume $v_G=0.5$, residual standard deviations $\sigma_X=2$ exposure units and $\sigma_Y=1$ outcome unit, and residual correlation $0.4$ among participants. Then $c=0.4(2)(1)=0.8$ exposure-outcome units.

The approximate covariance of the slope errors is

\frac{m c}{n_X n_Y v_G} = \frac{3{,}000(0.8)}{10{,}000(12{,}000)(0.5)} = 0.0000400.

Under the same simplifying assumptions, the approximate standard errors of the two slopes are $\sigma_X/\sqrt{n_Xv_G}$ and $\sigma_Y/\sqrt{n_Yv_G}$. Their product is

\frac{2}{\sqrt{10{,}000(0.5)}}\, \frac{1}{\sqrt{12{,}000(0.5)}} \approx 0.0003651.

Thus the approximate correlation between the association errors is $0.0000400/0.0003651\approx0.110$. The overlap coefficient $m/\sqrt{n_Xn_Y}$ is $3{,}000/\sqrt{120{,}000{,}000}\approx0.274$; multiplying it by the residual correlation $0.4$ also gives about $0.110$. This calculation concerns sampling errors, not the estimated causal effect itself.

Overlap, Instrument Strength, and Bias

In a valid-instrument setting for a single exposure, the population associations satisfy $\beta_Y=\theta\beta_X$, where $\theta$ is the causal effect per unit of exposure. The summary-data Wald estimate is $\hat{\beta}_Y/\hat{\beta}_X$. When $\beta_X$ is well estimated and far from zero, modest association-error covariance may have limited influence. When the instrument is weak, however, the estimated exposure association can be noisy relative to its true value. The denominator of the ratio can then fluctuate substantially, and the estimate can be sensitive to the joint behavior of the two association errors.

In the familiar linear setting, one-sample IV estimates with weak instruments tend to be pulled toward the observational association between exposure and outcome. With independent exposure and outcome samples, the two-sample estimate is instead less subject to that particular pull. Partial overlap can move the behavior between these cases: shared participants correlate the association errors, and greater overlap generally raises the risk that weak-instrument estimates will be drawn toward the observational association. The direction depends on the residual relationships and analysis setting; overlap alone does not guarantee a particular direction or amount of bias.

A useful rule of thumb under balanced, approximately homogeneous conditions is that the bias toward the observational association is roughly scaled by the overlap proportion and inversely by effective instrument strength. This is an approximation for understanding design risk, not a universal correction formula. With one instrument, the first-stage $F$ statistic summarizes strength; with multiple instruments, a relevant measure of aggregate strength is needed. The approximation is most informative when the samples and residual structures are comparable and the instruments are weak enough for weak-instrument bias to matter. It should not replace a design-specific analysis.

Worked Example: Using Overlap and Strength as a Risk Check

Suppose a hypothetical linear MR analysis has an effective first-stage strength of $F_{\mathrm{eff}}=10$, with half of the participants in the exposure sample also included in the outcome sample. Suppose the observational association is $0.30$ outcome units per exposure unit, while the causal effect under consideration is $0.10$ outcome units per exposure unit.

Using the rough design approximation “overlap proportion divided by effective strength,” the fraction of the gap toward which the estimate may be pulled is $0.5/10=0.05$. The observational-causal gap is $0.30-0.10=0.20$, so the illustrative shift is $0.05(0.20)=0.010$. The resulting illustrative estimate is

0.10+0.010=0.110 \quad\text{outcome units per exposure unit}.

This is not a predicted estimate or a bias correction: the rule is only a rough approximation, and the assumptions behind it may not hold. It does show why “50% overlap” cannot be judged in isolation. The same overlap fraction is more concerning when instruments are weak, and less concerning when they are strong.

Worked Example: Comparing Two Overlap Designs

Consider two hypothetical designs with $n_X=8{,}000$, $n_Y=10{,}000$, $v_G=0.4$, and residual covariance $c=0.6$ in exposure-outcome units. Design A has $m=0$ shared participants, while Design B has $m=2{,}000$.

For Design A, the leading covariance is zero, assuming the samples are otherwise independent. For Design B, it is

\frac{2{,}000(0.6)}{8{,}000(10{,}000)(0.4)} = 0.0000375.

The overlap coefficient for Design B is $2{,}000/\sqrt{8{,}000(10{,}000)}\approx0.224$. With residual standard deviations $\sigma_X=1.5$ and $\sigma_Y=1$, the approximate standard errors are $1.5/\sqrt{8{,}000(0.4)}\approx0.02652$ and $1/\sqrt{10{,}000(0.4)}\approx0.01581$. Their product is about $0.0004193$, giving an error correlation of $0.0000375/0.0004193\approx0.0894$. The arithmetic indicates that Design B has positively correlated association errors under the assumptions; it does not establish how much the final causal estimate will shift.

Practical Interpretation and Common Pitfall

Overlap should be considered alongside first-stage strength and the residual relationship between exposure and outcome. It is useful to report the exposure and outcome sample sizes, the number or proportion of participants shared, and whether the association estimates are based on the same underlying measurements or substantially different subsets. If overlap is uncertain, it is better to describe that uncertainty than to treat the samples as definitively independent. The covariance formula also depends on the regression and independence assumptions: when genotype and residuals are not independent, the relevant moment is $E[Z^2\varepsilon_X\varepsilon_Y]$, which need not equal $v_Gc$.

Common mistake: “The analysis is called two-sample MR, so its association errors are independent.” The label does not ensure disjoint participants. Check the sample construction.
Common mistake: “Any overlap invalidates the instrumental-variable assumptions.” Overlap induces sampling-error covariance; it is not by itself evidence that the instrument is associated with confounders or has a direct effect on the outcome.
Common mistake: “The overlap percentage alone determines the bias.” It does not. Instrument strength, residual covariance, and the analysis setting also matter.
Common mistake: Using $G$ rather than $G-E[G]$ in the slope-error derivation when the regression includes an intercept. The score uses centered genotype, and the covariance moment must use $(G-E[G])^2$.

Key takeaway: Shared participants can correlate exposure- and outcome-association errors. Under stated linear-model assumptions, that covariance is approximately $m c/(n_Xn_Yv_G)$. Overlap is most consequential as a weak-instrument concern; it is neither a stand-alone validity test nor a universal bias correction.

Check Your Understanding

Answer these questions using the definitions and assumptions in the tutorial.

Why does the slope-error derivation use $G-E[G]$ rather than $G$ when each association regression includes an intercept?
If there is no sample overlap and the samples are otherwise independent, what does the leading covariance formula imply about association-error covariance?
For $n_X=5{,}000$, $n_Y=8{,}000$, $m=1{,}000$, $v_G=0.25$, and $c=0.5$, calculate the approximate covariance of the association errors.
Does sample overlap itself demonstrate that a genetic instrument violates the instrumental-variable assumptions? Explain the distinction.
Why should the same overlap proportion prompt more concern when instruments are weak than when they are strong?