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Bayesian Multi-Arm Approaches · Tutorial 391 of 499

Likelihood Models With a Shared Control

Learn to write a shared-control normal likelihood correctly, interpret treatment-control contrasts, and recognize the dependence created by reusing the control observations.

Advanced 12 min read

What You'll Learn

  • Specify a normal-outcome model with one control arm and several treatment arms.
  • Write the likelihood using individual observations and arm-level summaries.
  • Express each treatment effect as a contrast with the common control mean.
  • Identify how the shared control observations contribute to every contrast without duplicating their likelihood contribution.
  • Calculate the sampling covariance between treatment-control contrasts.
  • Diagnose the consequences of treating shared-control comparisons as independent.

Why the Shared-Control Likelihood Needs Care

When several treatment arms are compared with one control arm, each treatment-control comparison uses the same control data. That reuse is part of the design, not a reason to copy the control observations into separate independent data sets. The likelihood should describe each observed outcome once, while the treatment effects are defined relative to the same control mean.

This tutorial uses a normal-outcome model with independent observations and a common variance. The model makes the role of the shared control explicit and gives a direct way to see why the observed treatment-control contrasts are dependent. The focus is the likelihood and its sampling structure, not the posterior calculations that may follow when the likelihood is combined with prior distributions.

Definition: Let arm \(0\) be the control and arms \(j=1,\ldots,K\) be the treatment arms. Write \(Y_{ij}\) for the outcome of participant \(i\) in arm \(j\), and let \(n_j\) be the sample size in that arm. Under the common-variance normal model, observations are independent conditional on their arm means and variance, with \(Y_{ij}\sim N(\mu_j,\sigma^2)\). Parameterize the means as \(\mu_j=\mu_0+\delta_j\), where \(\mu_0\) is the control mean and \(\delta_j\) is the treatment-minus-control mean difference for arm \(j\).

The model assumes that each participant contributes one outcome, that participants are independent, and that outcomes in every arm follow normal distributions with a common variance \(\sigma^2\). It also assumes the arm labels and data correspond to the comparisons being modeled: the control group is the one shared by all \(K\) treatment comparisons. For example, if the analysis is intended to compare concurrent randomized groups, the model should use the relevant concurrent control observations rather than automatically combining observations from different calendar periods.

The variance may be treated as known for a simple likelihood expression or as an unknown parameter to estimate. Equal variance is a modeling assumption, not a consequence of having a shared control. If variance differs materially among arms, this likelihood may not represent the data adequately.

Writing the Likelihood Once

For known \(\sigma^2\), the likelihood of all the individual observations is the product of their normal densities. With \(\mu_j=\mu_0+\delta_j\), it is

$$ L(\mu_0,\delta_1,\ldots,\delta_K) =(2\pi\sigma^2)^{-N/2} \exp\left\{-\frac{1}{2\sigma^2} \left[ \sum_{i=1}^{n_0}(Y_{i0}-\mu_0)^2 +\sum_{j=1}^{K}\sum_{i=1}^{n_j}(Y_{ij}-\mu_0-\delta_j)^2 \right]\right\}, \qquad N=\sum_{j=0}^{K}n_j. $$

The control observations appear in the first sum exactly once. Each treatment arm contributes its own residual terms, with its mean written as the shared control mean plus that arm's effect. The likelihood does not contain a separate copy of the control residual sum for each treatment arm.

If \(\sigma^2\) is unknown, it is included as a parameter and the same expression applies with the variance left unrestricted:

$$ L(\mu_0,\delta_1,\ldots,\delta_K,\sigma^2) =(2\pi\sigma^2)^{-N/2}\exp\left\{-\frac{\mathrm{SSE}(\mu_0,\delta_1,\ldots,\delta_K)}{2\sigma^2}\right\}, \qquad \sigma^2>0, $$

where \(\mathrm{SSE}\) is the bracketed sum of squared residuals in the preceding expression. Substituting an estimated variance into the known-variance formula is not the same as specifying this likelihood with unknown variance.

Deriving the Arm-Summary Form

For each arm, define the sample mean \(\bar{Y}_j=n_j^{-1}\sum_iY_{ij}\). The within-arm sum of squares decomposes into a part that depends on the proposed arm mean and a part that does not:

$$ \sum_{i=1}^{n_j}(Y_{ij}-\mu_j)^2 = \sum_{i=1}^{n_j}(Y_{ij}-\bar{Y}_j)^2 +n_j(\bar{Y}_j-\mu_j)^2. $$

To verify the identity, write \(Y_{ij}-\mu_j=(Y_{ij}-\bar{Y}_j)+(\bar{Y}_j-\mu_j)\), square, and sum. The cross-product is zero because \(\sum_i(Y_{ij}-\bar{Y}_j)=0\). The first term is fixed by the observed data, so for likelihood comparisons about the arm means it is a constant. Thus the arm means have the likelihood kernel

$$ L(\mu_0,\delta_1,\ldots,\delta_K) \propto \exp\left\{-\frac{1}{2\sigma^2} \left[n_0(\bar{Y}_0-\mu_0)^2+ \sum_{j=1}^{K}n_j(\bar{Y}_j-\mu_0-\delta_j)^2\right]\right\}. $$

The sample means summarize the part of the likelihood that depends on the arm means when the common variance is known. If the variance is unknown, within-arm variation also carries information about it; the sample means alone do not preserve that information.

Worked Example: Building the Likelihood From Individual Outcomes

Suppose a hypothetical experiment has one control arm and two treatment arms. The control outcomes are 8, 10, and 12; treatment 1 outcomes are 11, 13, and 12; and treatment 2 outcomes are 9, 10, and 11. Assume independent normal outcomes with known common standard deviation \(\sigma=2\). Evaluate the likelihood kernel at \(\mu_0=10,\delta_1=2,\delta_2=0\).

The proposed means are \(\mu_0=10\), \(\mu_1=\mu_0+\delta_1=12\), and \(\mu_2=\mu_0+\delta_2=10\). The control residual sum of squares is \((8-10)^2+(10-10)^2+(12-10)^2=8\). Treatment 1 contributes \((11-12)^2+(13-12)^2+(12-12)^2=2\), and treatment 2 contributes \((9-10)^2+(10-10)^2+(11-10)^2=2\). Therefore, the total \(\mathrm{SSE}\) is \(8+2+2=12\).

There are \(N=9\) observations, so the likelihood at these parameter values is

$$ L=(2\pi\cdot 2^2)^{-9/2}\exp\left(-\frac{12}{2\cdot 2^2}\right) =(8\pi)^{-9/2}\exp(-1.5). $$

The control residual contribution is 8, not 16: those three control observations enter the data likelihood once. They are relevant to both treatment-control contrasts through the estimate of \(\mu_0\), but that does not create additional control observations.

Contrasts and the Shared-Control Contribution

The observed treatment-control contrast for arm \(j\) is \(D_j=\bar{Y}_j-\bar{Y}_0\), which estimates \(\delta_j\). Under the model, each contrast has expectation \(\delta_j\). Because the treatment and control samples are independent, its variance is

$$ \operatorname{Var}(D_j)=\frac{\sigma^2}{n_j}+\frac{\sigma^2}{n_0}. $$

The term \(\sigma^2/n_0\) is present in every comparison because every contrast subtracts the same \(\bar{Y}_0\). For two different treatment arms \(j\) and \(k\), the treatment sample means are independent of each other and of the control mean. Consequently,

$$ \operatorname{Cov}(D_j,D_k) =\operatorname{Cov}(\bar{Y}_j-\bar{Y}_0,\bar{Y}_k-\bar{Y}_0) =\operatorname{Var}(\bar{Y}_0) =\frac{\sigma^2}{n_0}, \qquad j\ne k. $$

This is a sampling covariance between estimated contrasts. It arises from reusing the same random control mean; it does not mean the participant-level outcomes in different arms are dependent. The vector of contrasts is jointly normal under the model, with means \((\delta_1,\ldots,\delta_K)\), diagonal variances \(\sigma^2/n_j+\sigma^2/n_0\), and off-diagonal covariances \(\sigma^2/n_0\).

Worked Example: Contrasts From Arm Summaries

Consider hypothetical data with \(n_0=40,\bar{Y}_0=50\), \(n_1=30,\bar{Y}_1=56\), and \(n_2=25,\bar{Y}_2=54\). Assume a known common standard deviation of 10 outcome units and independent normal observations.

The observed contrasts are \(D_1=56-50=6\) outcome units and \(D_2=54-50=4\) outcome units. Their standard errors are

$$ SE(D_1)=\sqrt{10^2\left(\frac{1}{30}+\frac{1}{40}\right)} =\sqrt{5.833}=2.415\ \text{outcome units}, $$
$$ SE(D_2)=\sqrt{10^2\left(\frac{1}{25}+\frac{1}{40}\right)} =\sqrt{6.500}=2.550\ \text{outcome units}. $$

For example, the contrast likelihood contribution for arm 1 is proportional to \(\exp\{-(6-\delta_1)^2/[2(5.833)]\}\). The corresponding contribution for arm 2 is proportional to \(\exp\{-(4-\delta_2)^2/[2(6.500)]\}\). These two marginal expressions describe each contrast separately; multiplying them as if independent would omit their shared-control dependence. The control mean of 50 and sample size of 40 contribute to both standard errors.

Worked Example: Quantifying Dependence Between Two Comparisons

Suppose a hypothetical three-arm experiment has \(n_0=n_1=n_2=20\), known common standard deviation \(\sigma=4\), and observed means \(\bar{Y}_0=100,\bar{Y}_1=103,\bar{Y}_2=101\). The two observed contrasts are \(D_1=3\) and \(D_2=1\) outcome units.

For either contrast, the variance is \(16/20+16/20=1.6\) square outcome units. Their covariance is \(16/20=0.8\) square outcome units, because the same control mean appears in both. Their correlation is \(0.8/\sqrt{1.6\cdot1.6}=0.5\). Thus the sampling covariance matrix is

$$ \begin{pmatrix} 1.6 & 0.8\\ 0.8 & 1.6 \end{pmatrix}. $$

The two estimated effects are positively correlated even though the treatment-arm observations are independent. If the shared-control contribution were incorrectly omitted, the covariance would be set to zero and the correlation would be reported as zero. That would describe a different sampling model.

Diagnostics, Alternatives, and Limits

Before using this likelihood, check whether its assumptions are plausible for the outcome and design. Useful checks include plots of outcomes by arm for marked skewness or outliers, comparisons of arm-specific spreads for major variance differences, and review of the data collection and randomization records for dependence or mismatched control populations. A small sample cannot establish normality or equal variance, so these checks are diagnostic rather than proofs of the assumptions.

The full arm-level likelihood is usually the clearest representation when the goal is to model all arm means together. Pairwise analyses can be useful for reporting individual contrasts, but their uncertainty calculations must account for shared control data. Treating each pair as if it had an independent control group is not an equivalent alternative. If there truly are separate control groups with independently observed participants, their likelihood contributions and contrast covariance structure differ.

The common-variance model can be inappropriate when outcome variability differs substantially by arm. A heteroscedastic normal model can instead assign a variance \(\sigma_j^2\) to each arm, changing both the likelihood and the contrast variances. Non-normal outcomes may call for a likelihood suited to their measurement scale and distribution. Moreover, the simple model does not automatically handle missing outcomes, clustering, repeated measurements, or calendar-time changes in the control response; these require additional modeling assumptions.

Common pitfall: Writing one likelihood for each treatment-control comparison and multiplying them as independent likelihoods duplicates the shared control information. The correction is to write the participant-level likelihood with the control observations included once, or to use a joint likelihood for the contrasts that includes their shared-control covariance.
Key takeaway: In a shared-control normal model, each treatment mean is \(\mu_0+\delta_j\), while the control observations provide one common contribution to the likelihood. Every treatment-control contrast contains the same control estimate, so the contrasts are correlated even when the arm-specific observations are independent.

Check Your Understanding

Answer each question using the shared-control normal model described above. Solutions are included after the questions.

  1. With \(K\) treatment arms and one control, how many times should the control residual sum of squares appear in the individual-outcome likelihood?
  2. If \(n_0=25\), \(n_1=25\), and \(\sigma=5\), calculate the variance and standard error of \(D_1=\bar{Y}_1-\bar{Y}_0\).
  3. For two treatment arms with \(n_0=25\) and \(\sigma=5\), calculate the covariance of their estimated treatment-control contrasts. Briefly explain its source.
  4. Why is multiplying separate marginal contrast likelihoods as though contrasts were independent generally incorrect?

Solutions. (1) Once: each control observation is one observed outcome and contributes one residual term. (2) The variance is \(5^2(1/25+1/25)=2\) square outcome units, so the standard error is \(\sqrt{2}=1.414\) outcome units. (3) The covariance is \(\sigma^2/n_0=25/25=1\) square outcome unit; both contrasts subtract the same random control mean. (4) The marginal contrasts share that control estimate and therefore have nonzero covariance. Multiplying their marginal likelihoods as independent would omit this dependence and misrepresent the joint sampling model.

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