Deriving Shared-Control Posterior Covariance

Why Shared Controls Create Dependence

When several treatment arms are compared with one control group, each treatment-control contrast contains the same uncertain control mean. Even if the treatment-arm means are independent in the posterior distribution, subtracting the same random control mean from each one makes the contrasts dependent. This matters whenever an analysis uses a joint posterior event, such as the event that two arms both have effects above a specified threshold.

Building on the earlier material on likelihood models with a shared control, this tutorial derives that dependence under a conjugate normal model. The result is specific to the model assumptions below; the covariance is not automatically the same for other outcome distributions or prior structures.

Model and assumptions: There are \(K\) treatment arms indexed by \(i=1,\ldots,K\) and one control arm indexed by \(0\). Conditional on arm means, observations are independent and \(Y_{ij}\sim N(\mu_i,\sigma^2)\) for arm \(i\), where the common sampling variance \(\sigma^2\) is known. The independent priors are \(\mu_i\sim N(a_i,\tau_i^2)\) for \(i=0,\ldots,K\), with \(\tau_i^2>0\). The treatment-control contrast is \(\delta_i=\mu_i-\mu_0\), so positive values favor treatment when larger outcomes are better.

For arm \(i\), let \(n_i\) be its sample size and \(\bar{x}_i\) its observed sample mean. Conditional on the data, the posterior for each arm mean is normal. Define its posterior mean \(m_i\) and posterior variance \(v_i\) by

v_i=\left(\frac{1}{\tau_i^2}+\frac{n_i}{\sigma^2}\right)^{-1}, \qquad m_i=v_i\left(\frac{a_i}{\tau_i^2}+\frac{n_i\bar{x}_i}{\sigma^2}\right).

The independent priors and separate arm-specific likelihood terms imply that the posterior means \(\mu_0,\ldots,\mu_K\) are mutually independent under this model. The contrasts are not independent, because each uses \(\mu_0\).

Deriving the Contrast Covariance

For any treatment arm \(i\), the posterior mean and variance of its contrast follow from linearity and independence of \(\mu_i\) and \(\mu_0\):

E(\delta_i\mid\text{data})=m_i-m_0, \qquad \operatorname{Var}(\delta_i\mid\text{data})=v_i+v_0.

For two distinct treatment arms \(i\) and \(j\), expand the covariance. Posterior independence makes the covariance between their treatment means zero, and also makes each treatment mean independent of the control mean. The remaining term is the variance of the shared control mean:

\begin{aligned} \operatorname{Cov}(\delta_i,\delta_j\mid\text{data}) &=\operatorname{Cov}(\mu_i-\mu_0,\mu_j-\mu_0\mid\text{data})\\ &=\operatorname{Cov}(\mu_i,\mu_j\mid\text{data}) -\operatorname{Cov}(\mu_i,\mu_0\mid\text{data})\\ &\quad-\operatorname{Cov}(\mu_0,\mu_j\mid\text{data}) +\operatorname{Var}(\mu_0\mid\text{data})\\ &=v_0,\qquad i\ne j. \end{aligned}

This establishes the key result: under the stated independent-prior normal model, every pair of distinct treatment-control contrasts has posterior covariance \(v_0\). It is positive because posterior uncertainty about the control mean is positive. If the control mean were known exactly, \(v_0=0\), and this source of dependence would disappear.

The posterior correlation between two contrasts is therefore

\operatorname{Corr}(\delta_i,\delta_j\mid\text{data}) =\frac{v_0}{\sqrt{(v_i+v_0)(v_j+v_0)}}.

For all \(K\) contrasts together, their posterior covariance matrix is \(\operatorname{diag}(v_1,\ldots,v_K)+v_0\mathbf{1}\mathbf{1}^{\mathsf T}\), where \(\mathbf{1}\) is a \(K\)-vector of ones. The diagonal entries are \(v_i+v_0\); every off-diagonal entry is \(v_0\). Since the contrasts are linear combinations of jointly normal posterior arm means, their joint posterior distribution is multivariate normal with this covariance matrix and means \(m_i-m_0\).

Worked Example: Two Contrasts With a Shared Control

Suppose a hypothetical continuous-outcome trial has known within-arm variance \(\sigma^2=4\) outcome-units squared. Give the control and both treatment means independent \(N(5,4)\) priors. Each arm has \(n_i=4\); the observed means are \(\bar{x}_0=5.5\), \(\bar{x}_1=6.5\), and \(\bar{x}_2=6.0\) outcome units. These assumptions match the model: the variance is common and known, and the priors are independent.

For each arm, the posterior precision is \(1/4+4/4=1.25\), so \(v_i=1/1.25=0.8\) outcome-units squared. The control posterior mean is

m_0=\frac{1}{1.25}\left(\frac{5}{4}+\frac{4(5.5)}{4}\right) =\frac{6.75}{1.25}=5.4\text{ outcome units}.

The treatment posterior means are

m_1=\frac{1}{1.25}\left(\frac{5}{4}+\frac{4(6.5)}{4}\right) =\frac{7.75}{1.25}=6.2, \qquad m_2=\frac{1}{1.25}\left(\frac{5}{4}+\frac{4(6.0)}{4}\right) =\frac{7.25}{1.25}=5.8.

Thus the posterior contrast means are \(m_1-m_0=0.8\) and \(m_2-m_0=0.4\) outcome units. Each contrast has posterior variance \(0.8+0.8=1.6\) outcome-units squared, and their covariance is \(v_0=0.8\) outcome-units squared. Their posterior correlation is \(0.8/\sqrt{1.6(1.6)}=0.5\). The positive correlation is not evidence that the treatments act similarly; in this model it arises because both contrasts subtract the same uncertain control mean.

How Dependence Changes a Joint Posterior Probability

A joint probability generally cannot be obtained by multiplying marginal probabilities unless the events are independent. Here, positive contrast covariance makes independence an unjustified assumption. For example, the probability that both contrasts exceed zero depends on their correlation as well as on their separate posterior means and variances.

Worked Example: Joint Probability That Both Contrasts Are Positive

Use the same prior and sampling variance as in the previous example, but suppose all three hypothetical sample means are \(5.5\) outcome units, with four observations in each arm. The model assumptions still hold. The posterior variance for each arm remains \(0.8\), and each posterior mean is

m_i=\frac{1}{1.25}\left(\frac{5}{4}+5.5\right) =\frac{6.75}{1.25}=5.4\text{ outcome units}.

Both contrast posterior means are therefore zero, both contrast variances are \(1.6\), and their correlation is \(0.5\). Standardizing each contrast gives a bivariate standard normal pair \(Z_1,Z_2\) with correlation \(\rho=0.5\). For a zero-mean bivariate standard normal pair,

P(Z_1>0,Z_2>0)=\frac{1}{4}+\frac{\arcsin(\rho)}{2\pi}.

One justification is to differentiate the bivariate normal probability over the positive quadrant with respect to \(\rho\); its derivative is \(1/(2\pi\sqrt{1-\rho^2})\). At \(\rho=0\), independence gives probability \(1/4\). Integrating that derivative from \(0\) to \(\rho\) gives the displayed expression. Substituting \(\rho=0.5\), for which \(\arcsin(0.5)=\pi/6\), yields

P(\delta_1>0,\delta_2>0\mid\text{data}) =\frac{1}{4}+\frac{\pi/6}{2\pi} =\frac{1}{3}\approx0.3333.

If the contrasts were incorrectly treated as independent, multiplying their marginal probabilities of \(0.5\) would give \(0.25\), not \(0.3333\). In this example the shared control uncertainty makes both contrasts more likely to be on the same side of zero. The calculation is a model-based posterior probability for the stated joint event, not a probability that a hypothesis is true.

Control Information and the Strength of Dependence

The shared-control covariance is the control posterior variance \(v_0\), not the raw sampling variance of the control observations. More precise control information decreases \(v_0\), reducing the covariance. The correlation also depends on treatment-arm posterior variances, so it is useful to calculate the full expression rather than assume that every increase in information changes correlation in exactly the same way.

Worked Example: Adding Control Observations

Suppose a hypothetical study has known \(\sigma^2=1\), independent \(N(0,1)\) priors for all arm means, five observations in treatment 1, and twenty in treatment 2. Initially there are ten control observations. The posterior variances are

v_0=\frac{1}{1+10}=\frac{1}{11}\approx0.09091,\quad v_1=\frac{1}{1+5}=\frac{1}{6}\approx0.1667,\quad v_2=\frac{1}{1+20}=\frac{1}{21}\approx0.04762.

The contrast covariance is \(v_0\approx0.09091\). Their variances are \(v_1+v_0\approx0.2576\) and \(v_2+v_0\approx0.1385\), so their correlation is

\frac{1/11}{\sqrt{(1/6+1/11)(1/21+1/11)}}\approx0.481.

Now increase the control sample size to 30 while leaving both treatment arms unchanged. Then \(v_0=1/31\approx0.03226\), the contrast variances are approximately \(0.1989\) and \(0.07988\), and the correlation is

\frac{1/31}{\sqrt{(1/6+1/31)(1/21+1/31)}}\approx0.256.

The extra control observations reduce both the shared covariance and the correlation in this example. All variances are in outcome-units squared; the correlations are dimensionless. The calculation illustrates why control allocation can affect joint inference, even when treatment-arm sample sizes are held fixed.

Diagnostics, Alternatives, and Limits

Before using this covariance result, check whether its assumptions are credible for the analysis. In particular, inspect whether a common known sampling variance is a reasonable approximation, whether the normal outcome model is suitable, and whether the prior really factorizes across arm means. Also verify which observations are being treated as the shared control data. If arms use different control subsets, the covariance depends on the overlap and cannot be represented by this single \(v_0\) result without further derivation.

A direct alternative is to model each treatment-control contrast separately. That can provide marginal summaries, but it does not by itself specify their joint posterior distribution. Treating those marginal contrasts as independent discards the shared-control covariance and can produce incorrect probabilities for events involving multiple arms. A joint multivariate posterior calculation retains the covariance and is the appropriate approach for such events under this model.

Prior dependence: if arm means have correlated priors, their posterior covariances may contribute additional terms. The covariance need not equal \(v_0\) alone.
Different outcome models: for binary outcomes, unknown variances, or nonnormal likelihoods, the normal conjugate formulas above do not directly apply. The dependence must be derived or computed under the chosen model.
Other sources of dependence: shared patients, clustering, or model parameters common across arms may create dependence beyond the shared control mean.

A common mistake is to calculate two marginal posterior variances correctly and then multiply marginal probabilities as if the contrasts were independent. The correction is to derive their covariance from their shared random components, then evaluate the joint event under the resulting joint posterior distribution. A second pitfall is to confuse covariance with correlation: covariance has squared outcome units, while correlation is unitless and is scaled by both contrast variances.

Key takeaway: With independent normal priors and a shared control mean, distinct treatment-control contrasts have posterior covariance equal to the posterior variance of that control mean. Joint posterior probabilities must account for this dependence.

Check Your Understanding

Answer each question using the stated independent-prior normal model. Solutions are included below.

If \(v_0=0.20\), \(v_1=0.30\), and \(v_2=0.50\), find the covariance and correlation of the two contrasts.
Why are treatment-arm posterior means independent while their control contrasts are dependent?
In the zero-mean bivariate example, what joint probability would an independence calculation give, and why is it inappropriate?
With the treatment posterior variances fixed, what happens to the contrast covariance as control information increases and \(v_0\) decreases?

Solutions. (1) The covariance is \(v_0=0.20\). The correlation is \(0.20/\sqrt{(0.30+0.20)(0.50+0.20)}=0.20/\sqrt{0.35}\approx0.338\). (2) The independent priors and arm-specific likelihoods give a factorized posterior for the arm means. Subtracting the same uncertain \(\mu_0\) from both treatment means introduces covariance \(v_0\). (3) It gives \(0.5\times0.5=0.25\), but this assumes independence; with correlation \(0.5\), the joint probability is \(1/3\). (4) The covariance decreases because it equals \(v_0\). The correlation formula also includes the contrast variances, so its exact change should be checked using that formula.