Why a Normal Posterior Update Is Useful
A continuous trial endpoint may be summarized by an arm-specific mean, such as average change in a symptom score or average reduction in blood pressure. When the sampling variance is treated as known and both the sampling model and prior are normal, the posterior distribution for that mean can be calculated exactly. The update combines the prior information and the trial estimate, giving more weight to whichever is more precise.
This is a model-based calculation, not a rule for averaging any two numbers. The sampling variance, the prior variance, and the data must match the assumptions below. In particular, a variance estimated from the same small trial is not literally known just because it has been inserted into the formula.
From Individual Outcomes to the Trial Estimate
One common model starts with individual outcomes \(X_1,\ldots,X_n\) in one trial arm. Conditional on the arm mean \(\theta\), assume that the observations are independent and identically distributed as \(X_i\mid\theta\sim N(\theta,\sigma^2)\), where the sampling variance \(\sigma^2\) is known. The sample mean \(\bar{x}\) then satisfies \(\bar{X}\mid\theta\sim N(\theta,\sigma^2/n)\). Thus the update uses \(y=\bar{x}\) and \(v=\sigma^2/n\).
The normal model is exact under these assumptions. The variance \(\sigma^2\) describes the variation of individual outcomes around the arm mean; the variance of their mean is smaller by a factor of \(n\). Alternatively, a trial may report an estimate \(y\) and a known standard error \(SE\). In that case, use \(v=SE^2\), provided the estimate is modeled as unbiased and normally distributed around \(\theta\) with that variance.
The assumptions matter: observations must be independent under the specified sampling model, the mean must be the parameter being estimated, and the variance used in the likelihood must be known and appropriate. If participants are clustered, repeated measurements are dependent, or the variance changes in a way the model ignores, then \(\sigma^2/n\) may not be the correct variance for \(\bar{x}\).
Deriving the Precision-Weighted Update
The posterior density is proportional to the likelihood times the prior density. For an estimate \(y\), the parts of those densities that depend on \(\theta\) are
Write \(\lambda_0=1/\tau_0^2\) and \(\lambda_y=1/v\). Expanding the expression inside the brackets gives a quadratic in \(\theta\):
To complete the square, set \(m=(\lambda_0\mu_0+\lambda_y y)/(\lambda_0+\lambda_y)\). The \(\theta\)-dependent terms then equal \((\lambda_0+\lambda_y)(\theta-m)^2\), plus a constant that does not depend on \(\theta\). Consequently the posterior density is proportional to a normal density with mean \(m\) and variance \(1/(\lambda_0+\lambda_y)\). This establishes both the posterior mean and variance in the definition.
The variance decreases because the posterior precision is the sum of prior and data precision. The posterior mean is a weighted average: its weights are the two precisions divided by their sum. Since both precisions are positive, the mean must lie between \(\mu_0\) and \(y\). A more precise source pulls it closer.
Worked Example: Updating a Prior With a Trial-Arm Mean
Suppose a hypothetical trial measures change in a clinical score, in points, for 16 participants in one arm. The sample mean change is 6 points. Assume individual outcomes are independent normal observations with known standard deviation 8 points. Before seeing these data, the arm mean has prior distribution \(N(10,16)\), so the prior mean is 10 points and the prior standard deviation is 4 points.
Check the likelihood variance. The individual-outcome variance is \(8^2=64\) points squared. Therefore the sampling variance of the mean is \(v=64/16=4\) points squared; its standard error is 2 points. This calculation uses the stated independent, common-variance normal sampling model.
The prior precision is \(1/16=0.0625\) per point squared, and the data precision is \(1/4=0.25\) per point squared. The posterior variance is
The posterior mean is
Thus \(\theta\mid\bar{x}=6\sim N(6.8,3.2)\), with posterior standard deviation \(\sqrt{3.2}=1.789\) points. A central 95% posterior interval is approximately \(6.8\pm1.96(1.789)\), or [3.29, 10.31] points. Under the stated model and prior, 95% of the posterior probability for the arm mean lies in this interval. It is not a statement that the parameter has a 95% probability of being in the interval under repeated sampling.
Using a Reported Estimate and Updating in Stages
The same formula applies when the trial provides an estimate and its known standard error rather than individual observations. It also permits sequential updating: after one data set has been used, its posterior can serve as the prior for new, conditionally independent data. The new likelihood must contain only the new data. Reusing an estimate that already includes earlier observations would count those observations twice.
Worked Example: Updating From an Aggregate Trial Estimate
Suppose a hypothetical trial reports an arm mean estimate of 2.4 points with known standard error 1.2 points. Use a prior \(N(0,9)\), meaning a prior mean of 0 points and prior variance of 9 points squared. Treat the reported estimate as unbiased and normally distributed with the stated known standard error.
The data variance is \(v=1.2^2=1.44\) points squared. The prior precision is \(1/9=0.1111\), and the data precision is \(1/1.44=0.6944\), both in reciprocal points squared. Therefore
and
The posterior standard deviation is approximately \(\sqrt{1.241}=1.114\) points. A central 95% posterior interval is \(2.069\pm1.96(1.114)\), or approximately [-0.114, 4.252] points. The posterior mean is closer to the estimate than to the prior mean because the data precision, \(0.6944\), is greater than the prior precision, \(0.1111\).
Worked Example: Updating Sequentially Without Double-Counting
Consider a hypothetical arm mean \(\theta\) in points with initial prior \(N(0,4)\). At the first look, an estimate \(y_1=1\) has known sampling variance \(v_1=4\) points squared. The first posterior precision is \(1/4+1/4=0.5\), so its variance is 2 points squared. Its mean is \((0(1/4)+1(1/4))/0.5=0.5\) points.
At a later look, suppose a new, independent estimate is \(y_2=2\) points with known sampling variance \(v_2=2\) points squared. Updating the first posterior with this new likelihood gives total precision \(1/2+1/2=1\) per point squared. The updated posterior is
The same result follows by combining the original prior and both independent estimates directly: precision \(1/4+1/4+1/2=1\), and precision-weighted mean \([0(1/4)+1(1/4)+2(1/2)]/1=1.25\) points. This agreement checks the arithmetic and shows why the second update must use only the new estimate.
Checking the Model and Choosing an Approach
The update is exact only for the specified normal likelihood and normal prior. Before relying on it, check whether the endpoint and trial design make a normal mean model reasonable, whether the reported standard error reflects the actual design, and whether the prior mean and spread are scientifically defensible. Earlier material on prior predictive checks is useful for examining the implications of prior choices in terms of possible outcomes; it does not make a misspecified sampling model correct.
Potential warning signs include strong skewness or heavy tails in individual outcomes, influential extreme observations, outliers inconsistent with the normal sampling model, dependence between participants, and substantial differences in variability across groups that the model does not represent. For the arm mean, also verify that the variance used is a variance of the estimate, not the individual-outcome variance mistakenly used without dividing by \(n\). When only an aggregate estimate is available, the normal known-variance assumption and the appropriateness of its reported standard error must be justified from the study design and analysis.
A conjugate normal update is attractive when its assumptions are credible: it gives an exact posterior with a simple calculation. If the outcome distribution or sampling process is materially nonnormal, a likelihood appropriate to that endpoint may be needed, and computation may no longer have this closed form. If the variance is not known but must be estimated jointly with the mean, this fixed-variance update is not the appropriate full model; that setting requires a different posterior calculation. These alternatives address different modeling assumptions rather than being interchangeable computational shortcuts.
A normal prior also has support over the entire real line. That can be reasonable for an unconstrained mean, but may assign material prior probability to impossible values when the endpoint has hard bounds. Check the prior’s implications in context. A posterior calculation can be numerically correct and still be unhelpful if the prior or likelihood represents implausible values or an unsuitable data-generating process.
Check Your Understanding
Work through each question, then compare your reasoning with the solutions below.
- A hypothetical estimate is \(y=4\) units with known sampling variance \(v=9\) units squared. The prior is \(N(1,3)\). Find the posterior mean and variance.
- In a normal individual-outcome model, there are \(n=25\) independent participants and the known individual standard deviation is 10 units. What is the sampling variance of the sample mean?
- With prior precision 1 and data precision 3, the prior mean is 10 units and the estimate is 6 units. Is 9 the posterior mean? Explain.
- Name two checks that could reveal that the known-variance normal update is unsuitable for a trial.
Solutions. (1) The prior variance is 3, so prior precision is \(1/3\); data precision is \(1/9\). The posterior variance is \(1/(1/3+1/9)=9/4=2.25\) units squared. The posterior mean is \([1(1/3)+4(1/9)]/(1/3+1/9)=(1/3+4/9)/(4/9)=1.75\) units. (2) The individual variance is \(10^2=100\) units squared, so the mean’s sampling variance is \(100/25=4\) units squared. (3) No. The specified precisions give \([10(1)+6(3)]/(1+3)=7\) units; 9 is between the two input means but uses different weights. (4) Examples include dependent observations that make \(\sigma^2/n\) incorrect, pronounced skewness or influential outliers inconsistent with a normal sampling model, or a variance estimated from the data but treated as known. The checks should be tied to the trial design and endpoint, not applied as a mechanical list.