Deriving the Normal Power Prior Update

Why Derive the Update?

When a trial has few concurrent controls, historical control observations may contribute information about the control mean. A fixed power prior makes that contribution explicit: the historical likelihood is raised to a chosen power $a$, while the current observations contribute their ordinary likelihood. For a normal outcome with known variance, the posterior remains normal, and its precision and mean have a direct weighted-sum form.

The formula is useful only when its assumptions match the analysis. In particular, it treats the historical and current observations as measurements of the same control mean, with a common known variance. The introductory material on Power Priors for Normal Control Outcomes covers the model setting; here the update is derived and examined in detail.

Definition: Let $\mu$ be the unknown control mean, and suppose the baseline prior is $\mu\sim N(\mu_0,\tau_0^2)$, where $\mu_0$ and $\tau_0^2>0$ are specified. Historical observations $Y_{0i}$, $i=1,\ldots,n_0$, and current control observations $Y_i$, $i=1,\ldots,n$, are conditionally independent given $\mu$, with $Y_{0i}\mid\mu\sim N(\mu,\sigma^2)$ and $Y_i\mid\mu\sim N(\mu,\sigma^2)$. The variance $\sigma^2$ is known and common to both data sources. For a fixed discount $a\in[0,1]$, the posterior density is proportional to the baseline prior times the historical likelihood raised to $a$ times the current likelihood.

Write $\bar{y}_0$ for the historical sample mean and $\bar{y}$ for the current control sample mean. With known common variance, each data source contributes information about $\mu$ through its sample size, sample mean, and variance. The residual variation around a sample mean affects the likelihood’s overall scale, but not its dependence on $\mu$. The historical source contributes $a n_0/\sigma^2$ units of precision, whereas the current source contributes $n/\sigma^2$.

Deriving the Posterior

The prior kernel is proportional to $\exp\{-\frac{1}{2\tau_0^2}(\mu-\mu_0)^2\}$. For either data source, the sum of squared deviations can be decomposed into a part that depends on the sample mean and a residual part that does not:

\sum_{i=1}^{n_0}(Y_{0i}-\mu)^2 = \sum_{i=1}^{n_0}(Y_{0i}-\bar{Y}_0)^2 +n_0(\bar{Y}_0-\mu)^2.

The first term on the right is constant with respect to $\mu$. Applying this decomposition to the powered historical likelihood and to the current likelihood gives the following posterior kernel:

\begin{aligned} \pi(\mu\mid \text{data},a) &\propto \exp\left\{-\frac{1}{2\tau_0^2}(\mu-\mu_0)^2 -\frac{a n_0}{2\sigma^2}(\mu-\bar{y}_0)^2 -\frac{n}{2\sigma^2}(\mu-\bar{y})^2\right\}. \end{aligned}

To identify the posterior, collect the coefficients of $\mu^2$ and $\mu$. Define the total posterior precision $P$ and the precision-weighted numerator $Q$ as follows:

P=\frac{1}{\tau_0^2}+\frac{a n_0}{\sigma^2}+\frac{n}{\sigma^2}, \qquad Q=\frac{\mu_0}{\tau_0^2} +\frac{a n_0\bar{y}_0}{\sigma^2} +\frac{n\bar{y}}{\sigma^2}.

The exponent’s terms that involve $\mu$ are $-\frac{1}{2}(P\mu^2-2Q\mu)$. Completing the square gives $-\frac{P}{2}(\mu-Q/P)^2$, apart from a term constant in $\mu$. This is the kernel of a normal distribution with mean $Q/P$ and variance $1/P$.

Theorem: Under the model and fixed-discount assumptions above, the posterior is normal with precision $P$, variance $1/P$, and mean $Q/P$. Equivalently, $$ \mu\mid \text{data},a\sim N\left( \frac{\mu_0/\tau_0^2+a n_0\bar{y}_0/\sigma^2+n\bar{y}/\sigma^2} {1/\tau_0^2+a n_0/\sigma^2+n/\sigma^2}, \frac{1}{1/\tau_0^2+a n_0/\sigma^2+n/\sigma^2} \right). $$

The posterior mean is a precision-weighted average: the prior mean, historical mean, and current mean receive weights $1/\tau_0^2$, $a n_0/\sigma^2$, and $n/\sigma^2$, respectively. The variance of the posterior is smaller when any source adds positive precision. It is important not to interpret that as proof that the sources are compatible; the formula assumes a shared mean whether or not the data support that assumption.

Worked Updates

Worked Example: Combining Historical and Current Controls

Suppose, hypothetically, a control outcome is measured in points. The prior is $N(8,25)$, the known outcome variance is $\sigma^2=16$ points squared, and a historical control sample has $n_0=20$ with $\bar{y}_0=10$ points. The current control sample has $n=10$ with $\bar{y}=12$ points. Use fixed discount $a=0.5$.

The prior precision is $1/25=0.04$ per point squared. The discounted historical precision is $0.5(20)/16=0.625$ per point squared, and the current precision is $10/16=0.625$ per point squared. Thus $P=0.04+0.625+0.625=1.29$ per point squared.

The weighted numerator is $Q=8/25+0.5(20)(10)/16+10(12)/16=0.32+6.25+7.50=14.07$ points per point squared. Therefore, the posterior mean is $14.07/1.29=10.91$ points, and the posterior variance is $1/1.29=0.7752$ points squared. The posterior standard deviation is $\sqrt{0.7752}=0.8804$ points.

A central 95% posterior credible interval, using the normal quantile 1.96, is $10.91\pm1.96(0.8804)$, or approximately 9.18 to 12.63 points. Conditional on this model and the observed data, the posterior probability that $\mu$ lies in this interval is 0.95. The interval is not a statement that the fixed parameter has a 95% probability of lying in it under repeated sampling.

Worked Example: Changing the Discount

Consider hypothetical control outcomes in points with prior $N(0,100)$, known variance $\sigma^2=25$ points squared, historical sample size $n_0=12$ and mean $\bar{y}_0=14$ points, and current sample size $n=8$ and mean $\bar{y}=10$ points. Compare $a=0$, $a=0.5$, and $a=1$. The common denominator and numerator are $P=0.01+0.32+0.48a$ and $Q=3.2+6.72a$.

At $a=0$, $P=0.33$, $Q=3.2$, and the posterior mean is $3.2/0.33=9.697$ points. The historical likelihood contributes no precision.

At $a=0.5$, $P=0.57$, $Q=6.56$, and the posterior mean is $6.56/0.57=11.51$ points. At $a=1$, $P=0.81$, $Q=9.92$, and the posterior mean is $9.92/0.81=12.25$ points. Increasing $a$ moves the posterior mean toward the higher historical mean and increases posterior precision. These calculations show the consequence of choosing different discounts; they do not determine which discount is appropriate.

Worked Example: Updating From Individual Observations

In a hypothetical study, historical outcomes in points are $4,6,8,10,12,14$, and current control outcomes are $7,9,11,13$. Let the prior be $N(8,16)$, the known common variance be $\sigma^2=9$ points squared, and the fixed discount be $a=0.25$. The historical mean is $\bar{y}_0=9$ points and $n_0=6$; the current mean is $\bar{y}=10$ points and $n=4$.

The posterior precision is $P=1/16+0.25(6)/9+4/9=9/144+24/144+64/144=97/144=0.6736$ per point squared. The numerator is $Q=8/16+0.25(6)(9)/9+4(10)/9=0.5+1.5+4.444=6.444$ points per point squared. Thus the posterior mean is $Q/P=928/97=9.567$ points, and the posterior variance is $144/97=1.485$ points squared.

The individual historical values have residual sum of squares $70$ points squared around their mean, and the current values have residual sum of squares $20$ points squared around their mean. With known variance, these residual terms multiply the likelihood by factors that do not depend on $\mu$, so they do not change this posterior update. This simplification would not apply unchanged if the outcome variance were also unknown.

Assumptions, Diagnostics, and Alternatives

The derivation relies on several substantive and mathematical conditions. The prior variance must be positive; the discount is fixed in $[0,1]$; the observations are conditionally independent given $\mu$; and the outcome variance is known and common across the historical and current samples. Most importantly, both samples must be reasonably represented as observations from the same control-mean model. Differences in eligibility, endpoint definition, follow-up, or care may make a shared mean implausible even when the outcomes have the same name.

Useful checks include comparing the historical and current sample means in context, inspecting the individual observations for outliers or unusual spread, and verifying that the same outcome units and measurement rules apply. A large mean difference is a warning that borrowing may pull the posterior toward a potentially irrelevant historical level. A posterior variance that looks reassuringly small does not diagnose or repair that mismatch. Sensitivity calculations across plausible fixed values of $a$ reveal how strongly conclusions depend on the borrowing choice.

For a fixed $a$, the normalizing constant of the power prior is constant with respect to $\mu$, so it does not change the posterior kernel derived here. If $a$ is itself assigned a probability model and inferred, however, that normalizing constant can depend on $a$ and must be handled in the joint analysis. The fixed-discount formula should not be used to silently treat a data-dependent discount as fixed.

Approach	Historical contribution	Key implication
No borrowing	Set $a=0$	Posterior uses the prior and current controls only.
Power prior	Use a fixed $0<a<1$	Historical precision is reduced by the chosen discount.
Full pooling	Set $a=1$	Historical and current observations contribute as equally modeled data.

The normal update is compact and transparent, but it is not a safeguard against model misspecification. It can be misleading when the historical and current control populations differ, when variances differ, or when a few observations dominate the means. In particular, treating the variance as known can understate uncertainty if that value is poorly specified. Those concerns call for a model suited to the data and design; they are not resolved merely by changing the arithmetic in this known-variance update.

Common Mistake

Wrong reasoning: “The historical sample has many observations, so it must dominate the update.”
Correction: Its precision is $a n_0/\sigma^2$, not simply $n_0$. A small discount or large outcome variance can substantially reduce its contribution. Compare the actual precision terms, and assess whether borrowing is scientifically justified.

Key takeaway: For a known common variance and fixed discount, add prior, discounted historical, and current precisions. The posterior mean is their corresponding precision-weighted mean. The calculation is exact under the model, but the shared-mean and known-variance assumptions remain substantive.

Check Your Understanding

Use the fixed-discount normal update above. Each question uses hypothetical data, and outcomes are measured in points.

With prior $N(5,4)$, $\sigma^2=4$, $n_0=4$, $\bar{y}_0=7$, $n=4$, $\bar{y}=9$, and $a=0.5$, find the posterior mean and variance.
What happens to the historical contribution when $a=0$?
For $a=1$, does the formula establish that the historical and current control populations are exchangeable? Explain.
If the historical and current means differ substantially, name one diagnostic response and explain why a narrow posterior interval alone is not reassuring.

Solutions. (1) The precision is $P=1/4+0.5(4)/4+4/4=0.25+0.5+1=1.75$. The numerator is $Q=5/4+0.5(4)(7)/4+4(9)/4=1.25+3.5+9=13.75$. The posterior mean is $13.75/1.75=7.857$ points and the variance is $1/1.75=0.5714$ points squared. (2) The historical precision and numerator terms are both multiplied by zero, so the historical data do not affect this posterior. (3) No. Setting $a=1$ gives the historical likelihood its full weight under the specified model; it does not establish that the shared-mean assumption is valid. (4) Compare the means and inspect the study populations and outcome definitions, then conduct sensitivity analyses over plausible discounts. The interval reflects uncertainty conditional on the assumed shared-mean model and can be narrow even if that model is inappropriate.

Approach	Historical contribution	Key implication
No borrowing	Set \(a=0\)	Posterior uses the prior and current controls only.
Power prior	Use a fixed \(0<a<1\)	Historical precision is reduced by the chosen discount.
Full pooling	Set \(a=1\)	Historical and current observations contribute as equally modeled data.