Power Trade-Offs Between Families and Unrelated Participants

Why the Number Sequenced Is Not the Whole Comparison

A sequencing budget can buy many unrelated participants or fewer pedigrees with several relatives each. Counting sequenced people alone does not tell us which design has more power. Relatives can share rare alleles, so their genotype information is correlated; yet comparing relatives within the same family can also remove some between-family variation. The value of that within-family comparison depends on the covariance it actually creates, not simply on how many relatives were enrolled.

A fair comparison starts with the same scientific target and a clearly defined budget. For illustration, this tutorial uses a binary gene-level carrier indicator and compares its prevalence in cases and controls. In an actual sequencing study, the indicator might encode a prespecified burden definition. The power calculations below are approximate and do not account for every feature of a particular burden test.

Definition: Let $X=1$ indicate that a participant meets a prespecified gene-level carrier definition, and $X=0$ otherwise. Let $p_1=P(X=1\mid\text{case})$ and $p_0=P(X=1\mid\text{control})$. The target contrast is $\Delta=p_1-p_0$. A design comparison is meaningful only when both designs target the same contrast, use comparable variant definitions and sequencing quality, and handle ascertainment and confounding appropriately.

The contrast is an association target, not a claim that carrier status causes disease. Family recruitment can depend strongly on the phenotype that led to a proband being selected. The case and control probabilities used in a calculation must therefore correspond to the intended analysis after the relevant ascertainment handling, rather than being treated as unbiased population probabilities automatically.

Independent Participants and Clustered Relatives

For unrelated participants sampled independently within each status group, the estimated carrier proportions $\hat p_1$ and $\hat p_0$ have approximate variances $p_1(1-p_1)/n_1$ and $p_0(1-p_0)/n_0$. If the two groups are independent, the variance of their difference is the sum of those terms. Relatives are different: their carrier indicators can be correlated because alleles are inherited within pedigrees.

Definition: For equal-sized clusters of $m$ participants with a common within-cluster intraclass correlation $\rho$, the usual design effect for a group mean or proportion is $D=1+(m-1)\rho$. Its approximate effective sample size is $n_{\mathrm{eff}}=n/D$, where $n$ is the number of participants in that group. This approximation assumes comparable cluster sizes, exchangeable within-cluster correlation, and independent clusters; it is not a universal substitute for a pedigree-specific variance calculation.

When $\rho>0$, $D>1$, so the effective sample size is smaller than the headcount. The approximation reflects that adding another relative to a pedigree usually contributes less independent information than adding an unrelated person. If cases and controls are sampled from the same families, however, the two estimated proportions may themselves be correlated. A design effect for each group alone does not determine the variance of their difference.

\operatorname{Var}(\hat p_1-\hat p_0) =\operatorname{Var}(\hat p_1)+\operatorname{Var}(\hat p_0) -2\operatorname{Cov}(\hat p_1,\hat p_0).

This covariance term is important. Shared rare alleles can make relatives’ carrier indicators positively associated. For a paired case-control comparison within each pedigree, that shared component can cancel in the difference, reducing its variance. But the covariance must be estimated from suitable data, justified by a model, or varied in a sensitivity analysis. It cannot be inferred from the two marginal design effects alone.

Worked Example: Effective Sample Size With Three Relatives per Cluster

Suppose a hypothetical design enrolls 60 cases from 20 distinct pedigrees, with three cases per pedigree. Assume the pedigrees are independent and that the carrier indicator has a within-pedigree intraclass correlation of $\rho=0.20$. Use the equal-cluster approximation to find the effective sample size for the case group.

There are $n=60$ participants and $m=3$ per cluster, so $D=1+(m-1)\rho=1+(3-1)(0.20)=1.40$.

n_{\mathrm{eff}}=\frac{60}{1.40}=42.86.

Thus, for estimating this group proportion under the stated approximation, these 60 related cases provide about the precision of 42.86 independent participants. The control group needs its own calculation: its cluster size and correlation may differ. This effective sample size is not a count of independent pedigrees, nor does it capture a possible covariance between case and control estimates drawn from the same pedigrees.

Approximate Power for a Case-Control Contrast

For independent case and control groups, or for groups whose covariance is explicitly assumed to be zero, a large-sample standard error is obtained by substituting the group variances. With clustered groups, one may use effective sample sizes under the design-effect assumptions. For a paired family design, instead use the variance of the family-level contrast. These are different variance calculations; they should not be combined as if they were interchangeable.

SE(\hat p_1-\hat p_0) \approx \sqrt{\frac{p_1(1-p_1)}{n_{\mathrm{eff},1}} +\frac{p_0(1-p_0)}{n_{\mathrm{eff},0}}} \quad\text{when the group estimates have zero covariance.}

For a two-sided normal-approximation test at significance level $\alpha$, define $\lambda=\Delta/SE$ and let $z_{1-\alpha/2}$ be the corresponding standard normal quantile. The approximate power is the probability that a normal test statistic with mean $\lambda$ and variance 1 falls outside the null rejection limits.

\operatorname{Power} \approx P\!\left(Z>z_{1-\alpha/2}-\lambda\right) + P\!\left(Z<-z_{1-\alpha/2}-\lambda\right), \qquad Z\sim N(0,1).

This approximation presumes a valid test calibration, a prespecified analysis and threshold, and enough independent information for a normal approximation to be reasonable. In a real gene-level analysis, the appropriate $\alpha$ may be much smaller than 0.05 because of the testing plan. Sparse carrier counts, uncertain variant effects, or complex pedigree structure may make simulation or a model-specific calculation preferable.

Worked Example: Approximate Power With Unrelated Participants

Consider a hypothetical study with 60 unrelated cases and 60 unrelated controls. Suppose the assumed carrier probabilities are $p_1=0.04$ and $p_0=0.01$. Test the two-sided contrast at $\alpha=0.05$, assuming independent sampling, correct classification, and a normal approximation.

The target difference is $\Delta=0.04-0.01=0.030$. Under the stated independence assumption, the variance estimate under the alternative is $0.04(0.96)/60+0.01(0.99)/60=0.000640+0.000165=0.000805$.

SE=\sqrt{0.000805}\approx0.02837, \qquad \lambda=\frac{0.030}{0.0283725}\approx1.0574.

For a two-sided 5% test, $z_{0.975}\approx1.96$. Therefore, $z_{0.975}-\lambda\approx0.9026$ and $-z_{0.975}-\lambda\approx-3.0174$. The approximate power is $P(Z>0.9026)+P(Z<-3.0174)\approx0.1833+0.0013=0.1846$. Under these hypothetical assumptions, power is about 18.5%. This low value reflects the small sample and the rarity of the carrier definition; it is not evidence that a particular real study has this power.

Paired Families: Use the Family-Level Contrast

For $F$ independent pedigrees each contributing one case and one control, define the within-family difference $D_j=X_{1j}-X_{0j}$, where $j$ indexes pedigrees. The estimate of the contrast is the mean of these $F$ differences. Its variance depends on the covariance between the case and control indicators within a pedigree. This is the direct way to represent shared family information in a paired design.

\operatorname{Var}(D_j) =p_1(1-p_1)+p_0(1-p_0) -2\operatorname{Cov}(X_{1j},X_{0j}), \qquad \operatorname{Var}(\bar D)=\frac{\operatorname{Var}(D_j)}{F}.

The second equality assumes independent pedigrees with a common variance. Positive within-family covariance lowers the variance of the difference relative to independent case and control indicators with the same marginal probabilities. That can make a paired design efficient when family-level shared factors affect both relatives similarly. It does not guarantee that a family design is more powerful: fewer pedigrees, different costs, ascertainment, and the actual covariance all matter.

Worked Example: A Paired Family Variance

Suppose a hypothetical study has 40 independent pedigrees, each contributing one case and one control. Assume $p_1=0.04$, $p_0=0.01$, and a within-pedigree covariance of $0.0085$ between their binary carrier indicators. This covariance is an explicit hypothetical assumption, not a value implied by the marginal probabilities.

The case-indicator variance is $0.04(0.96)=0.0384$, and the control-indicator variance is $0.01(0.99)=0.0099$. Thus, the variance of one paired difference is $0.0384+0.0099-2(0.0085)=0.0313$.

\operatorname{Var}(\bar D)=\frac{0.0313}{40}=0.0007825, \qquad SE(\bar D)=\sqrt{0.0007825}\approx0.02797.

The covariance assumption is feasible for these margins: the implied probability that both relatives are carriers is $p_1p_0+0.0085=0.0089$, which is between 0 and the smaller marginal probability, 0.01. If covariance were instead zero, the paired contrast variance would be $0.0483/40=0.0012075$, with standard error about 0.03475. The difference illustrates why the family covariance must be specified; it is not determined by the number of relatives or by the marginal probabilities alone.

Putting Recruitment Costs and Information Together

A fixed-budget comparison should include the costs that affect how many usable observations can be collected: recruitment, sequencing, family tracing, consent, follow-up, and any additional quality-control or analysis requirements. It should then calculate precision for the same estimand under each design. A lower sequencing cost per participant is not enough if the participants contribute highly redundant information; conversely, a family design can gain efficiency if a paired contrast cancels shared variation.

Worked Example: Which Design Uses a Hypothetical Budget Better?

Suppose a hypothetical budget is $120,000. An unrelated participant costs $1,000, so the study can enroll 60 cases and 60 controls. A family costs $3,000 and yields one sequenced case and one sequenced control, so the budget funds 40 paired pedigrees. Assume both designs target $p_1=0.04$ versus $p_0=0.01$, both use the same valid carrier definition and sequencing quality, and pedigrees are independent. For this illustration only, assume the paired within-family covariance is 0.0085.

For the unrelated design, the standard error was calculated as $\sqrt{0.000805}\approx0.02837$. For the paired family design, the family-level calculation gives $\sqrt{0.0313/40}\approx0.02797$. Under these particular assumptions, the family design has a slightly smaller standard error despite having only 40 cases and 40 controls rather than 60 of each. The calculation does not establish a general family advantage: it depends on the specified covariance and costs.

The covariance needed for the family design to match the unrelated design can be found directly. Set its variance equal to the unrelated variance $0.000805$: $(0.0483-2c)/40=0.000805$, where $c$ is the within-pair covariance. Solving gives $0.0483-2c=0.0322$, so $c=0.00805$. Thus, with these assumed margins and costs, the paired family design has lower variance only if its covariance exceeds approximately 0.00805. If that covariance is smaller, the unrelated design has lower variance. In practice, uncertainty in this quantity should be examined rather than hidden behind a single optimistic assumption.

The number of pedigrees is often a crucial constraint. Additional relatives from the same pedigree may add less information than recruiting another pedigree, particularly when the relevant carrier indicators are highly correlated. But one should not translate this observation into a universal rule to recruit unrelated participants: a well-chosen within-family comparison may control shared background and reduce variance. A useful comparison therefore reports the budget, the number of independent pedigrees, the number and roles of sequenced relatives, the assumed or estimated correlations, and the resulting variance or power under each design.

Derivation and Interpretation of the Design Effect

The equal-cluster design effect follows from the variance of a cluster sum. Let $X_{ij}$ be a binary indicator for participant $i$ in cluster $j$, with variance $v=p(1-p)$. If every pair in a cluster has correlation $\rho$, their covariance is $\rho v$. The variance of the mean across $n$ participants, arranged in clusters of size $m$, is the sum of $n$ individual variances and the within-cluster pair covariances, divided by $n^2$.

\operatorname{Var}(\bar X) =\frac{nv+n(m-1)\rho v}{n^2} =\frac{v}{n}\{1+(m-1)\rho\} =\frac{vD}{n}.

The term $n(m-1)$ counts ordered within-cluster pair contributions across the $n$ participants; equivalently, each participant has $m-1$ within-cluster partners in the covariance sum. Equating this variance with the independent-sample form $v/n_{\mathrm{eff}}$ gives $n_{\mathrm{eff}}=n/D$. This justification applies to a group mean or proportion under the stated equal-size, common-correlation structure. A paired case-control contrast has an additional covariance component and should instead be evaluated at the family level, as above.

Common Mistakes and Practical Checks

Equating headcount with information. Treating 120 relatives as 120 independent observations ignores shared alleles. Use a justified cluster model or a pedigree-level variance calculation.
Adding two design effects and stopping. Separate case and control design effects do not specify $\operatorname{Cov}(\hat p_1,\hat p_0)$. State an independence assumption only when it is justified; otherwise include the covariance or analyze paired family contrasts.
Assuming relatives always help or always hurt. Correlation reduces independent information for a group proportion, but shared variation can cancel in a within-family difference. Which effect matters depends on the estimand and design.
Comparing designs with different targets or quality. Different carrier definitions, callability, ascertainment, or phenotype definitions can change the probabilities being compared. Align these features before interpreting a power difference.
Reporting a precise power from uncertain inputs. Power depends on effect size, carrier prevalence, covariance, costs, and the testing threshold. Show assumptions and examine plausible alternatives, especially for rare variants and modest numbers of pedigrees.

Key takeaway: Under a fixed budget, compare designs through the variance of the same case-control estimand, not through the number of sequenced people alone. Account for clustering in each group and for case-control covariance when relatives are paired; the family design’s advantage, if any, depends on the information and costs actually available.

Check Your Understanding

Use the stated assumptions in each question and distinguish group-level clustering from paired-family covariance.

With cluster size $m=4$ and within-cluster correlation $\rho=0.10$, what is the design effect? If 80 participants are enrolled, what is the approximate effective sample size?
Why does knowing the case and control design effects separately not determine the variance of $\hat p_1-\hat p_0$ when both groups come from the same pedigrees?
For 30 independent case-control pairs, suppose the variance of one family-level difference is 0.036. Calculate the standard error of the mean paired difference.
Under what circumstances can positive within-family covariance reduce the variance of a paired contrast, and why does that not guarantee greater power for every family study?
Name at least three costs or design features that should be aligned or included in a fixed-budget comparison of family and unrelated-participant sequencing.