Why the Boundary Cases Need a Proof
A power prior discounts a historical-data likelihood by raising it to a value between zero and one. The boundary values have intuitive names: zero discount means no borrowing, and a discount of one means full borrowing. But those labels can be misleading if the posterior construction is not written out. In particular, full borrowing does not mean that the historical posterior replaces the baseline prior. The baseline prior remains part of the analysis at both boundaries.
The key is to distinguish the historical likelihood contribution from the complete prior distribution used in the current analysis. At discount zero, the historical likelihood contributes no parameter information. At discount one, it contributes the full historical likelihood. In both cases, the baseline prior and the current-data likelihood are still present.
The Two Boundary Identities
For \(a_0=0\), the power-prior convention is that the historical likelihood factor is omitted: \(L_0(\theta\mid D_0)^0\) contributes the constant one. Thus the posterior is based on the current data and the baseline prior:
For \(a_0=1\), the power is one, so the full historical likelihood enters alongside the current likelihood and the baseline prior:
These are statements about posterior kernels: two densities proportional to the same function of \(\theta\) are equal after division by their respective normalizing integrals, provided those integrals are finite and positive. The data likelihoods are multiplied because, under the assumed model, the current and historical data contribute their likelihood information about \(\theta\). The baseline prior appears in each expression because it is part of the specified power-prior construction, not part of the discountable historical data.
Worked Example: Beta-Binomial Boundary Cases
Suppose a hypothetical control response probability is \(\theta\). Historical data contain \(x_0=12\) responses among \(n_0=20\) participants, and current control data contain \(x=8\) responses among \(n=15\) participants. Use a baseline prior \(\theta\sim\operatorname{Beta}(2,3)\), with density proportional to \(\theta^{2-1}(1-\theta)^{3-1}\). Assume the two binomial likelihoods are conditionally independent given \(\theta\), and both studies use the same response definition and binomial model.
Ignoring binomial coefficients that do not depend on \(\theta\), the historical likelihood is \(\theta^{12}(1-\theta)^8\), and the current likelihood is \(\theta^8(1-\theta)^7\). The posterior kernel is therefore $$ \theta^{8+12a_0+2-1}(1-\theta)^{7+8a_0+3-1}, $$ so its normalized form is a beta density with parameters \(10+12a_0\) and \(10+8a_0\).
At \(a_0=0\), the historical exponents contribute zero. The posterior is \(\operatorname{Beta}(10,10)\): its first parameter is \(2+8=10\), and its second is \(3+(15-8)=10\). Historical data have not been used, but the baseline parameters \(2\) and \(3\) remain.
At \(a_0=1\), the historical data enter at full likelihood strength. The posterior is \(\operatorname{Beta}(22,18)\), since \(2+8+12=22\) and \(3+7+8=18\). The baseline prior still contributes \(2\) and \(3\); it has not been replaced by a historical-only posterior.
Derivation and Conditions
The boundary result follows directly from the definition, but the conditions matter. For a fixed discount, the posterior density is obtained by multiplying the baseline density by each likelihood contribution and normalizing. Write the normalizing constant as \(Z(a_0)\):
For the posterior to be a probability density, \(Z(a_0)\) must be finite and strictly positive. At zero, the historical factor is one by the no-borrowing boundary convention, so
At one, substitution gives
This proves the boundary identities, as well as the continued presence of \(\pi_0(\theta)\). The argument does not require conjugacy. It does require a coherent likelihood and prior model and a finite, positive normalizing integral at the boundary in question. If a likelihood is zero at some parameter values, the zero-discount case is defined by omitting the historical factor, rather than by treating a pointwise expression \(0^0\) as an ordinary real-number operation. In common continuous models with parameter values restricted to the interior, the likelihood is positive and this technical distinction does not arise.
Keep the current likelihood, historical likelihood raised to \(a_0\), and baseline prior visible in one posterior kernel.
At zero, omit the historical likelihood; at one, include it to the first power.
Verify that the resulting kernel has a finite, positive integral before interpreting it as a posterior density.
A Second Model: Normal Historical Data
Worked Example: Normal Likelihood at Zero and One
Consider a hypothetical continuous endpoint with known variance \(\sigma^2=16\) in both studies. Historical observations have sample size \(n_0=10\) and sample mean \(\bar{x}_0=5\); current observations have sample size \(n=6\) and sample mean \(\bar{x}=7\). Assume independent normal observations with common mean \(\theta\), the stated common known variance, and a proper baseline prior \(\theta\sim N(0,100)\). The normal likelihood kernel for observations with sample size \(m\) and mean \(\bar{x}\) is proportional to \(\exp[-m(\theta-\bar{x})^2/(2\sigma^2)]\).
The baseline prior has precision \(1/100\), the current likelihood has precision \(6/16\), and the historical likelihood raised to \(a_0\) has precision \(10a_0/16\). Multiplying the kernels gives posterior precision and precision-weighted mean numerator $$ \frac{1}{100}+\frac{6}{16}+\frac{10a_0}{16}, \qquad \frac{0}{100}+\frac{6(7)}{16}+\frac{10a_0(5)}{16}, $$ respectively. At \(a_0=0\), these are \(0.385\) and \(2.625\), so the posterior is \(N(2.625/0.385,\ 1/0.385)=N(6.818,\ 2.597)\). The historical observations contribute nothing; the baseline prior remains, with precision \(0.01\).
At \(a_0=1\), the precision is \(0.01+0.375+0.625=1.010\), and the numerator is \(2.625+3.125=5.750\). The posterior is \(N(5.750/1.010,\ 1/1.010)=N(5.693,\ 0.9901)\). Both data sets contribute at their full likelihood strengths, but the baseline prior still contributes precision \(0.01\) and weighted numerator zero. The results are conditional on the stated common-mean, common-variance model; if the historical and current populations do not support that model, full borrowing may not be scientifically justified.
Likelihood Constants and Normalization
Likelihoods are often written either as full probability functions or as kernels that retain only factors depending on \(\theta\). For a fixed \(a_0\), a factor depending only on observed data is constant with respect to \(\theta\) and cancels when the posterior is normalized. For example, the binomial coefficient in a binomial likelihood can be omitted when deriving the posterior for \(\theta\). Raising that coefficient to \(a_0\) changes the unnormalized density by a data-only constant, not the posterior distribution.
A normalized power prior may divide the discounted historical likelihood by a factor such as \(\int L_0(\theta\mid D_0)^{a_0}\pi_0(\theta)\,d\theta\). For a fixed \(a_0\), this factor does not depend on \(\theta\), so it also cancels from the posterior density for \(\theta\), assuming it is finite and positive. This cancellation does not mean that the baseline prior can be dropped: the baseline density inside the integral and posterior kernel is still essential. If the discount itself is assigned a probability distribution and inferred, terms depending on \(a_0\) cannot generally be discarded from the joint model.
Worked Example: Checking the Normalizing Constant at a Boundary
Return to the hypothetical beta-binomial example, but consider only the historical contribution to the power prior. The historical likelihood kernel is \(\theta^{12}(1-\theta)^8\), and the baseline prior is \(\operatorname{Beta}(2,3)\). At \(a_0=0\), the historical factor is one, so its baseline-prior integral is $$ \int_0^1 1\cdot\pi_0(\theta)\,d\theta=1. $$
At \(a_0=1\), using the full binomial likelihood kernel, the integral is $$ \int_0^1\theta^{12}(1-\theta)^8 \frac{\theta^{2-1}(1-\theta)^{3-1}}{B(2,3)}\,d\theta = \frac{B(14,11)}{B(2,3)}. $$ Here \(B(r,s)=\int_0^1\theta^{r-1}(1-\theta)^{s-1}\,d\theta\) is the beta function. The integral is finite and positive. Dividing the powered likelihood by this value normalizes the historical power-prior density, but for this fixed boundary value the divisor is constant in \(\theta\) and does not alter the posterior kernel after normalization. The baseline prior remains in the integrand and in the resulting posterior.
Interpretation, Diagnostics, and Alternatives
The two boundary analyses are useful checks on an implementation. At \(a_0=0\), changing historical outcomes while keeping the current data and baseline prior fixed must not change the posterior. At \(a_0=1\), the posterior kernel must include the historical likelihood once, not zero times and not twice. In a conjugate calculation, these checks can be made by inspecting the sufficient-statistic updates; in a general model, compare the implemented log-posterior kernel at the two boundary values with the expressions derived above.
A common mistake is to say, “At full borrowing, use the historical posterior as the prior and then multiply by the baseline prior again.” That generally counts the baseline prior twice. The power-prior definition instead gives \(L_0(\theta\mid D_0)\pi_0(\theta)\) as the historical-data-informed prior kernel at \(a_0=1\); multiplying by the current likelihood yields the posterior. A different construction can deliberately use an historical posterior as a prior, but it must be specified as a different model, and its baseline-prior contribution must be tracked to avoid duplication.
Another mistake is to equate zero borrowing with a flat prior. Zero discount removes only the historical likelihood. It does not erase \(\pi_0(\theta)\). If the baseline prior is informative, it continues to influence the posterior; if it is improper, posterior propriety must be checked rather than assumed. Similarly, one means full historical-likelihood contribution under the specified model, not proof that the historical and current populations are exchangeable or that borrowing is scientifically appropriate.
| Boundary or approach | Posterior kernel for \(\theta\) | Role of baseline prior |
|---|---|---|
| Power prior, \(a_0=0\) | \(L(\theta\mid D)\pi_0(\theta)\) | Retained; historical likelihood omitted |
| Power prior, \(a_0=1\) | \(L(\theta\mid D)L_0(\theta\mid D_0)\pi_0(\theta)\) | Retained; historical likelihood included fully |
| Current-data analysis with the baseline prior | \(L(\theta\mid D)\pi_0(\theta)\) | Same kernel as the zero-discount boundary |
| Historical-only analysis with the baseline prior | \(L_0(\theta\mid D_0)\pi_0(\theta)\) | Not the full power-prior posterior when current data are present |
The boundary identities are algebraic and do not resolve model mismatch, endpoint incompatibility, calendar-time changes, or data-quality concerns. The power-prior formulation assumes that both likelihoods describe the same parameter, unless a richer model explicitly connects different study parameters. Thus \(a_0=1\) is a transparent mathematical endpoint, not an automatic recommendation to borrow all available historical information. A practical boundary diagnostic is to verify both algebra and modeling assumptions before interpreting the result.
Check Your Understanding
Answer each question using the posterior-kernel definition above. Brief solutions follow the questions.
- Write the posterior kernel at \(a_0=0\), and state what happens to the baseline prior.
- For a binomial historical sample with \(x_0=4\) responses among \(n_0=9\), write the historical likelihood contribution at \(a_0=1\), omitting factors independent of \(\theta\).
- With a \(\operatorname{Beta}(1,1)\) baseline and current data of \(x=3\) responses among \(n=5\), what posterior distribution results at \(a_0=0\)?
- True or false: at \(a_0=1\), the historical posterior based on \(\pi_0\) replaces \(\pi_0\), so the baseline prior should not appear in the final posterior kernel. Explain.
Solutions. (1) The kernel is \(L(\theta\mid D)\pi_0(\theta)\); the baseline prior remains unchanged. (2) The historical contribution is \(\theta^4(1-\theta)^5\), since there are \(9-4=5\) nonresponses. (3) The posterior is \(\operatorname{Beta}(1+3,1+(5-3))=\operatorname{Beta}(4,3)\); the historical data do not contribute at zero discount. (4) False. The full posterior kernel is \(L(\theta\mid D)L_0(\theta\mid D_0)\pi_0(\theta)\). Treating the historical posterior as a replacement and multiplying the baseline prior again would generally double-count that prior.