Power Priors for Normal Control Outcomes

Why Discounted Normal Controls Need a Clear Information Scale

Historical control outcomes can inform a Bayesian clinical trial, but their relevance to the current trial is rarely certain. A power prior represents this uncertainty by raising the historical likelihood to a discount parameter $a$, where $0 \leq a \leq 1$. For normally distributed outcomes with known variance, the effect of this operation has a direct interpretation: it multiplies the historical contribution to precision by $a$. The outcomes’ measurement scale and the assumed sampling variance remain unchanged.

The distinction matters in practice. A discount of $0.25$ does not mean that each historical outcome is measured with four times the variance. It means that the historical likelihood contributes one quarter of its original information. This tutorial formulates that contribution and shows how to calculate it. The subsequent derivation of a full normal power-prior update builds on these information components.

Definition: Let $Y_1,\ldots,Y_{n_H}$ be historical control outcomes satisfying $Y_i \mid \theta \sim N(\theta,\sigma^2)$, independently, where $\theta$ is the control mean relevant to the current trial and the variance $\sigma^2$ is known. Write the historical data as $D_H$, and let $L_H(\theta;D_H)$ be their likelihood. Given a proper baseline prior density $\pi_0(\theta)$, a fixed-discount power prior is $$ \pi(\theta\mid D_H,a) = \frac{L_H(\theta;D_H)^a\pi_0(\theta)} {C(a;D_H)}, \qquad C(a;D_H) = \int L_H(u;D_H)^a\pi_0(u)\,du. $$ The denominator normalizes the density. At $a=0$, the historical likelihood contributes no information; at $a=1$, it contributes its full likelihood.

The Historical Likelihood and Its Information

This formulation relies on substantive and statistical assumptions. The historical and current control outcomes must represent the same target mean $\theta$, on the same outcome definition and measurement scale. Historical observations are assumed independent conditional on $\theta$, normally distributed, and governed by the stated known variance. Any dependence, material changes in care or eligibility, or unmodeled differences between studies can make the common-mean likelihood inappropriate. Discounting alone does not establish that these assumptions are reasonable.

Under the model, the historical likelihood is $$ L_H(\theta;D_H) = \prod_{i=1}^{n_H} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left\{-\frac{(Y_i-\theta)^2}{2\sigma^2}\right\}. $$ To see how the discount acts, use the identity $\sum_{i=1}^{n_H}(Y_i-\theta)^2 =\sum_{i=1}^{n_H}(Y_i-\bar{Y}_H)^2+n_H(\bar{Y}_H-\theta)^2$, where $\bar{Y}_H$ is the historical sample mean. The first term does not depend on $\theta$. Therefore the part of the powered likelihood that varies with $\theta$ is

L_H(\theta;D_H)^a \ \propto\ \exp\left\{-\frac{a n_H}{2\sigma^2}(\bar{Y}_H-\theta)^2\right\}.

This factorization shows that the historical sample mean carries the information about $\theta$; the centered sum of squares affects the likelihood’s scale but not its shape as a function of $\theta$. For $a>0$, the $\theta$-dependent kernel has the same quadratic form as a normal likelihood for $\bar{Y}_H$ with variance $\sigma^2/(a n_H)$. This is an information-equivalence statement, not a claim that the actual historical observations had a different sampling variance.

Key result: The historical precision contributed by the powered likelihood is $$ I_H(a)=\frac{a n_H}{\sigma^2}. $$ Thus its effective sample size, measured relative to observations with variance $\sigma^2$, is $a n_H$. The discount scales information linearly: $a=0.4$ retains 40% of the historical likelihood’s precision, and $a=1$ retains all of it.

Precision has units equal to the inverse of the outcome variance. If the outcome is measured in millimetres of mercury, for example, precision is expressed per squared millimetre of mercury. The effective sample size is dimensionless. When a proper baseline prior also contributes information, its precision remains separate from $I_H(a)$; the historical contribution should not be reported as though it were the total precision of the resulting prior.

Worked Example: Discounting One Historical Control Study

Suppose a hypothetical historical study records $n_H=40$ independent control outcomes. The known outcome standard deviation is $\sigma=4$ mmHg and the historical mean is $\bar{Y}_H=5.8$ mmHg. Consider a fixed discount $a=0.25$. Assume the outcome definition, population, and control setting make a common target mean plausible for this illustration.

The full historical precision would be $n_H/\sigma^2=40/16=2.5$ per $(\text{mmHg})^2$. The discounted contribution is

I_H(0.25)=\frac{0.25(40)}{4^2} =\frac{10}{16} =0.625\ \text{per }(\text{mmHg})^2.

The effective sample size is $a n_H=0.25(40)=10$ observations. Equivalently, the historical mean’s powered-likelihood kernel has variance $16/10=1.6$ $(\text{mmHg})^2$, and standard deviation $\sqrt{1.6}=1.265$ mmHg. Without discounting, the corresponding standard deviation would be $4/\sqrt{40}=0.6325$ mmHg. The discount therefore reduces the likelihood precision to one quarter and increases the equivalent standard deviation by a factor of two. Neither calculation changes the assumed individual-outcome standard deviation of 4 mmHg.

Several Studies and Their Information Contributions

If historical studies share the same target mean and known variance, their independent likelihoods multiply. Giving study $j$ its own discount $a_j$ leads to a product of powered likelihoods. Each study contributes precision $a_j n_j/\sigma^2$. This makes it possible to check how the total historical information is assembled without treating every study as equally informative.

Worked Example: Adding Study-Specific Discounted Precision

Consider two hypothetical control studies with independent outcomes and a common known standard deviation of 5 outcome units. Study 1 has $n_1=20$, mean 4 units, and discount $a_1=0.50$. Study 2 has $n_2=30$, mean 6 units, and discount $a_2=0.20$. Assume both studies use the same endpoint and population definition, so a shared target mean is defensible for this calculation.

Study 1 contributes $a_1n_1/\sigma^2=0.50(20)/25=0.40$ per squared outcome unit. Study 2 contributes $a_2n_2/\sigma^2=0.20(30)/25=0.24$ per squared outcome unit. Together,

I_H=\frac{0.50(20)+0.20(30)}{25} =\frac{10+6}{25} =0.64\ \text{per squared outcome unit}.

The combined effective sample size is $10+6=16$ observations. Because the variance is common, the historical information is centered at the effective-size-weighted mean $(10(4)+6(6))/16=4.75$ outcome units. This is an information summary, not evidence that the two study means are identical. Their difference should prompt a clinical and design review; a numerical discount does not by itself explain or resolve the discrepancy.

How the Baseline Prior Fits In

A power prior combines historical information with a baseline prior $\pi_0(\theta)$. A proper baseline prior can provide information even when $a=0$; zero discount removes only the historical likelihood. If the baseline prior is $N(m_0,\tau_0^2)$, its precision is $1/\tau_0^2$. The historical likelihood’s precision contribution remains $a n_H/\sigma^2$. Consequently, the quadratic part of the normalized power-prior density has total precision

\frac{1}{\tau_0^2}+\frac{a n_H}{\sigma^2}.

This expression makes the information change explicit without requiring a full update formula. The normalization constant $C(a;D_H)$ is required to make the power prior integrate to one. For a fixed $a$, it does not vary with $\theta$, so it does not change the $\theta$-dependent precision. When the discount itself is assigned a distribution and inferred, the normalization and the model for $a$ matter to that inference; treating them as interchangeable with a fixed discount would be incorrect.

Worked Example: Comparing Historical Information With Baseline Information

Suppose a hypothetical analysis uses a baseline prior $N(0,10^2)$, with outcome units stated on the same scale as the historical data. A compatible historical study has $n_H=25$, known $\sigma=5$ units, and fixed discount $a=0.20$. For this illustration, assume the same common-mean model as above.

The baseline precision is $1/10^2=0.01$ per squared unit. The historical contribution is $0.20(25)/5^2=5/25=0.20$ per squared unit. The total quadratic precision is therefore $0.01+0.20=0.21$ per squared unit. Historical data contribute $0.20/0.21=0.9524$, or about 95.2%, of this total precision. With $a=1$, historical precision would be $25/25=1.00$, giving total precision $1.01$ per squared unit. These calculations compare information contributions; they do not state a posterior mean or replace the full normalized-prior calculation.

Checks, Alternatives, and Limitations

Before using a normal power prior, inspect whether the model and data support its interpretation. Check that all studies report the same endpoint, units, and relevant time point; that the known variance is appropriate; and that observations are independent at the level assumed. Plot study-level means and uncertainty, and examine whether outlying or systematically shifted studies have a clinical explanation. If the study means differ substantially, a shared-mean model may be questionable even if the likelihood can be written down algebraically.

A fixed power prior is one approach among several. Setting $a=0$ excludes historical likelihood information, while $a=1$ uses it fully under the common-parameter model. A fractional value provides a prespecified intermediate contribution. Other approaches, such as a commensurate prior or a hierarchical model, represent historical-current differences differently; they should not be interpreted as simple rescalings of this likelihood. The choice depends on the scientific question and the assumptions the analysis can defend.

Power discounting has important limits. It cannot correct a wrong endpoint mapping, an incorrect known variance, confounding, or a systematic change in control care. It also does not automatically determine a defensible value of $a$: that choice needs scientific justification or a model for uncertainty about borrowing. In particular, a low discount may reduce the impact of a problematic study but should not be used to conceal evidence that the shared-mean assumption fails.

Mistake: treating $a$ as a variance multiplier. The powered likelihood has an equivalent summary variance $\sigma^2/(a n_H)$; this does not change the assumed individual-outcome variance $\sigma^2$.
Mistake: calling $a n_H$ the total sample size. It is the historical effective sample size under this information scale. Any baseline prior and current-trial data are additional information sources.
Mistake: assuming discounting guarantees compatibility. Discounting reduces influence, but it does not establish that historical and current controls share a meaningful target mean.

Key takeaway: For independent normal historical controls with a common mean and known variance, raising the likelihood to $a$ multiplies its precision by $a$. The historical precision contribution is $a n_H/\sigma^2$, and the corresponding effective sample size is $a n_H$. These quantities describe borrowed information, not a change to the outcome model’s known variance.

Check Your Understanding

Answer the questions using the common-mean, known-variance model defined above. Solutions follow each question.

A historical study has $n_H=50$, known $\sigma=3$ outcome units, and $a=0.40$. What are its effective sample size and discounted precision? Solution: The effective sample size is $0.40(50)=20$. The precision is $20/3^2=20/9=2.222$ per squared outcome unit.
Does setting $a=0$ necessarily make the power prior uninformative? Solution: No. It removes the historical likelihood contribution, but a proper baseline prior $\pi_0(\theta)$ can still be informative.
For $n_H=24$, $\sigma=4$, and $a=0.50$, calculate the historical precision. Solution: $a n_H/\sigma^2=0.50(24)/16=12/16=0.75$ per squared outcome unit.
Two studies use the same known variance and target mean. Their effective sample sizes are 8 and 12. What is the combined historical effective sample size? Solution: Their precision contributions add, so the combined effective sample size is $8+12=20$.
Why does a power discount not fix incompatible historical and current outcome definitions? Solution: The discount scales the likelihood contribution under the specified model. It does not make different endpoints measure the same parameter or repair a false common-mean assumption.