Why the Number and Choice of Controls Matter
In a case-control study, recruiting more controls can improve precision, but the improvement is not unlimited: each additional control generally contributes less than the one before. Moreover, the number alone does not tell the whole story. Controls who add useful variation in the covariates of interest may provide more information than the same number drawn from a narrow part of the covariate distribution.
A design decision therefore has two parts: how many controls to recruit relative to cases, and how to sample those controls. The calculations below use a binary exposure and a log odds ratio to make the trade-offs explicit. The principles extend to more complex models, but the numerical formulas depend on the model, covariate distribution, and sampling design.
A Basic Precision Formula for a Binary Exposure
Suppose cases and controls are sampled independently, with fixed sample sizes \(n_1\) and \(n_0\). Let \(q_1\) be the probability that a sampled case is exposed and \(q_0\) the probability that a sampled control is exposed. Assume \(0<q_1,q_0<1\). The sample exposure proportions are \(\hat q_1\) and \(\hat q_0\). The log odds ratio can be estimated by the difference of their logits; if controls were sampled with exposure-dependent probabilities, the known sampling-odds correction must also be included.
To justify the formula, the case exposure count is binomial with parameters \(n_1,q_1\), so \(\hat q_1\) has variance \(q_1(1-q_1)/n_1\); similarly, \(\hat q_0\) has variance \(q_0(1-q_0)/n_0\). The derivative of \(g(q)=\log(q/(1-q))\) is \(g'(q)=1/[q(1-q)]\). The delta method therefore gives variance \(1/[n_1q_1(1-q_1)]\) for \(g(\hat q_1)\), and \(1/[n_0q_0(1-q_0)]\) for \(g(\hat q_0)\). The two samples are independent, so the variance of their difference is the sum of these variances. A known sampling-odds correction is a constant and adds no sampling variance. \(\square\)
The formula shows why a large control count does not automatically guarantee a small standard error. The control contribution depends both on \(n_0\) and on \(q_0(1-q_0)\), which is largest when exposed and unexposed controls are equally common. If nearly all controls have the same exposure status, the sample provides little information about the exposure odds.
Worked Example: Comparing One and Four Controls per Case
Suppose \(n_1=200\), \(q_1=0.40\), and controls are sampled so that \(q_0=0.20\). Compare the approximate standard errors with 200 controls and 800 controls.
The case contribution is $$ \frac{1}{200(0.40)(0.60)} =\frac{1}{48} \approx 0.0208333. $$ The control contribution per control is governed by $$ \frac{1}{q_0(1-q_0)} =\frac{1}{(0.20)(0.80)} =6.25. $$ With 200 controls, the variance is $$ \frac{1}{48}+\frac{6.25}{200} =0.0208333+0.03125 =0.0520833, $$ so the standard error is \(\sqrt{0.0520833}\approx 0.2282\).
With 800 controls, the variance is $$ \frac{1}{48}+\frac{6.25}{800} =0.0208333+0.0078125 =0.0286458, $$ and the standard error is \(\sqrt{0.0286458}\approx 0.1693\). Quadrupling the controls reduces the standard error, but it does not divide it by two: the unchanged case contribution remains. The calculation is an approximation, not a guarantee of the realized standard error in a particular data set.
Choosing an Allocation Under Recruitment Costs
A useful planning approximation is $$ V(n_1,n_0)=\frac{A}{n_1}+\frac{B}{n_0}, $$ where \(V\) approximates the variance of the target estimator and \(A,B>0\) summarize the information contributed by cases and controls under the proposed design. For the binary-exposure log odds ratio above, \(A=1/[q_1(1-q_1)]\) and \(B=1/[q_0(1-q_0)]\). Other models can produce different constants or more complicated information expressions.
Suppose each case costs \(c_1>0\) to recruit and each control costs \(c_0>0\), and the total recruitment budget is fixed at \(C\). Under this approximation, the continuous allocation that minimizes variance has a simple ratio.
For a proof, the variance decreases as either sample size increases, so the budget constraint is tight at the optimum. Substitute \(n_0=(C-c_1n_1)/c_0\), for \(0<n_1<C/c_1\), into \(V\). The resulting function is $$ V(n_1)=\frac{A}{n_1}+\frac{B c_0}{C-c_1n_1}. $$ Its first derivative is $$ V'(n_1)=-\frac{A}{n_1^2}+\frac{B c_0c_1}{(C-c_1n_1)^2}, $$ and its second derivative is $$ V''(n_1)=\frac{2A}{n_1^3}+ \frac{2B c_0c_1^2}{(C-c_1n_1)^3}>0. $$ Thus the objective is strictly convex, and its unique minimum occurs where \(V'(n_1)=0\). Since both terms in $$ \frac{A}{n_1^2}=\frac{B c_0c_1}{(C-c_1n_1)^2} $$ are positive, taking positive square roots and using \(C-c_1n_1=c_0n_0\) gives \(n_0/n_1=\sqrt{Bc_1/(Ac_0)}\). \(\square\)
Worked Example: A Cost-Aware Control-to-Case Ratio
Use the preceding binary-exposure setting with \(q_1=0.40\) and \(q_0=0.20\). Then \(A=1/(0.40\cdot0.60)=25/6\) and \(B=1/(0.20\cdot0.80)=25/4\). Suppose recruiting a case costs \(c_1=\$1{,}200\) and recruiting a control costs \(c_0=\$200\).
The approximate cost-optimal ratio is $$ \frac{n_0}{n_1} = \sqrt{\frac{(25/4)(1200)}{(25/6)(200)}} = \sqrt{\frac{7500}{5000/3}} = \sqrt{4.5} \approx 2.12. $$ Thus the formula suggests about 2.12 controls per case under these assumptions, not three. The ratio is greater than one because controls cost less and, in this example, their variance coefficient \(B\) is larger than the cases' coefficient \(A\). For integer recruitment plans, one would compare nearby feasible allocations under the actual budget. The result is not universal: changing the relative costs, exposure distributions, or target parameter changes the ratio.
When the number of cases is fixed and only controls can be added, the control term \(B/n_0\) declines with diminishing returns: its derivative is \(-B/n_0^2\). One can also express the marginal variance reduction from one more control approximately as $$ \frac{B}{n_0}-\frac{B}{n_0+1} =\frac{B}{n_0(n_0+1)}. $$ This quantity shrinks as \(n_0\) grows. A design team should therefore compare the expected gain from additional controls with their recruitment and measurement costs, rather than choosing a ratio by convention alone.
Choosing Which Controls to Recruit
Sampling controls to improve covariate coverage can be useful when the exposure or other important covariates are unevenly distributed. For a binary exposure, the control information term is largest when the sampled controls have \(q_0=0.5\), because \(q_0(1-q_0)\leq 0.25\). But a design that deliberately changes the exposure composition of controls must account for that selection. If exposed controls are sampled at a different rate from unexposed controls, the known sampling odds must be incorporated into the analysis; treating the sampled controls as a simple random sample can shift the estimated odds ratio.
Worked Example: Random Versus Exposure-Stratified Controls
Keep \(n_1=200\) and \(q_1=0.40\). Compare 400 controls sampled with exposure proportion \(q_0=0.20\) to 400 controls sampled within exposure strata to achieve \(q_0=0.40\). Assume the stratum-specific sampling probabilities are known and the analysis includes the corresponding correction.
With random controls at \(q_0=0.20\), the variance approximation is $$ \frac{1}{200(0.40)(0.60)} +\frac{1}{400(0.20)(0.80)} = \frac{1}{48}+\frac{1}{64} = \frac{7}{192} \approx 0.0364583. $$ The standard error is approximately \(\sqrt{7/192}=0.1909\).
With the stratified sample at \(q_0=0.40\), the corrected estimator has control contribution $$ \frac{1}{400(0.40)(0.60)} =\frac{1}{96}. $$ The total variance approximation is $$ \frac{1}{48}+\frac{1}{96} =\frac{1}{32} =0.03125, $$ so the standard error is approximately \(0.1768\). In this illustration the stratified strategy yields more information for the log odds ratio with the same number of controls. The improvement relies on correctly accounting for the known selection probabilities; without that correction, the changed control exposure distribution would alter the estimated contrast.
For multiple covariates, the best control strategy depends on which coefficient or contrast is the precision target. In a generalized linear model, the expected information is built from contributions involving the model weights and covariate vectors. Controls are especially useful when they provide observations in covariate regions that inform the target contrast, rather than repeatedly adding nearly redundant observations. Stratification or targeted sampling can help, but it may also require known selection probabilities, weighting, or a design-appropriate likelihood. The information and variance methods for the specific sampling design must be used.
Meeting a Standard-Error Target
A practical planning calculation starts with a fixed case contribution and solves for the minimum number of controls that brings the approximate standard error below a target. Because the control count is an integer, the final candidate must be checked directly in the original inequality.
Worked Example: Finding the Minimum Control Count
Suppose the case contribution to variance is \(A=1/48\), the control coefficient is \(B=6.25\), and the target standard error is \(0.18\). Under the approximation \(V=1/48+6.25/m\), where \(m\) is the number of controls, require $$ \sqrt{\frac{1}{48}+\frac{6.25}{m}}\leq 0.18. $$ Both sides are nonnegative, so squaring preserves the inequality. Rearranging gives $$ \frac{6.25}{m}\leq 0.18^2-\frac{1}{48} =0.0324-0.0208333333 =0.0115666667, $$ and hence $$ m\geq \frac{6.25}{0.0115666667} \approx 540.3458. $$ The smallest eligible integer is therefore 541. Checking the neighboring integers confirms the rounding: at \(m=540\), the variance is \(1/48+6.25/540\approx0.0324074\), which exceeds \(0.18^2=0.0324\); at \(m=541\), it is \(1/48+6.25/541\approx0.0323860\), which is below \(0.0324\). Thus 541 controls meet the approximate target, whereas 540 do not.
Common Pitfall
Wrong reasoning: “Four times as many controls must cut the standard error in half.” This treats the entire variance as if it were the control contribution. In the formula \(A/n_1+B/n_0\), increasing controls reduces only \(B/n_0\); the case contribution remains unchanged when \(n_1\) is fixed. The correction is to calculate the total variance at each proposed allocation and then take its square root. Also, a targeted control sample should not be analyzed as if it were randomly sampled when the selection probabilities depend on exposure or other covariates.
The practical lesson is to plan around the estimand, the expected covariate distribution, and recruitment costs. More controls are most valuable when they contribute information that is otherwise scarce, and less valuable once the control component is small relative to the case component. A control ratio is therefore a design choice to justify, not a universal rule.
Check Your Understanding
Use the assumptions in each question and show the calculation where relevant.
- For \(q_0=0.10\), compare the control information per person with that for \(q_0=0.50\). Which exposure distribution contributes more information to the log odds ratio?
- If cases are fixed and the control contribution to variance is \(B/n_0\), explain why adding 100 controls has a smaller effect when \(n_0=900\) than when \(n_0=100\).
- In the cost-aware allocation formula, what happens to the recommended control-to-case ratio if controls become more expensive while all other quantities remain fixed?
- Why must a known correction be included when controls are sampled at different rates across exposure strata?
- With case variance contribution \(0.02\), control coefficient \(B=4\), and 100 controls, calculate the approximate total variance and standard error.