Power Priors for Time-to-Event Controls

Borrowing Survival Information Through a Model

A historical control group can inform the control event-time distribution in a new trial, but only through assumptions linking the two studies. A power prior expresses this borrowing by raising the historical likelihood to a discount power. For survival data, the form of that likelihood depends on the chosen event-time model, and the parameters that receive information are exactly those appearing in the powered likelihood.

To keep the survival contribution clear, this tutorial first considers independent, fully observed event times under a specified parametric model. Thus every historical participant contributes an observed event time, and no participant is censored. The likelihood contribution from censoring is a distinct issue. The essential assumptions here are that the model is appropriate for the event-time distribution, observations are conditionally independent under it, and the shared parameters have the same interpretation in the historical and current control populations. These assumptions do not follow merely from using a power prior.

Definition: Let $D_H$ denote historical control event-time data, $L_H(\theta\mid D_H)$ their likelihood under parameter $\theta$, and $\pi_0(\theta)$ a proper baseline prior. For a fixed discount $a_0\in[0,1]$, the power-prior kernel is $L_H(\theta\mid D_H)^{a_0}\pi_0(\theta)$. It defines a conditional prior density only if $$ C(a_0)=\int L_H(\theta\mid D_H)^{a_0}\pi_0(\theta)\,d\theta $$ is finite and nonzero. When this condition holds, the density is $L_H(\theta\mid D_H)^{a_0}\pi_0(\theta)/C(a_0)$.

At $a_0=0$, the historical likelihood contributes no information and, for a proper $\pi_0$, $C(0)=1$. At $a_0=1$, the full historical likelihood is used. Intermediate values discount the likelihood; they do not change the definition of the event-time model. The value of $a_0$ is treated as fixed here. A proper baseline prior alone does not ensure that $C(a_0)$ is finite when $a_0>0$.

Exponential Event Times: The Discounted Rate Contribution

Suppose the event time $T$ follows an exponential distribution with rate $\lambda>0$. Its density is $f(t\mid\lambda)=\lambda\exp(-\lambda t)$ for $t>0$. For $n_H$ independent, fully observed historical event times $t_1,\ldots,t_{n_H}$, the likelihood is

L_H(\lambda\mid D_H) =\prod_{i=1}^{n_H}\lambda\exp(-\lambda t_i) =\lambda^{n_H}\exp(-\lambda S_H), \qquad S_H=\sum_{i=1}^{n_H}t_i.

Raising this likelihood to $a_0$ gives $\lambda^{a_0n_H}\exp(-a_0\lambda S_H)$. Thus the discount scales both the event-count term and the total-time term. Under this model, historical information is about the rate $\lambda$; it does not directly borrow a separate shape parameter because the exponential model has none.

For a conjugate illustration, let the baseline prior be $\lambda\sim\operatorname{Gamma}(\alpha_0,\beta_0)$, using the shape-rate convention. Its density is proportional to $\lambda^{\alpha_0-1}\exp(-\beta_0\lambda)$. Multiplication by the powered likelihood gives the kernel

\lambda^{\alpha_0+a_0n_H-1} \exp\{-(\beta_0+a_0S_H)\lambda\}.

This is a gamma kernel with shape $\alpha_0+a_0n_H$ and rate $\beta_0+a_0S_H$, both positive when $\alpha_0,\beta_0>0$. In this case the normalizing integral is finite. This convenient result depends on the exponential likelihood and gamma baseline prior; it is not a guarantee for arbitrary survival models and priors.

Worked Example: Discounting Historical Exponential Controls

Suppose a hypothetical historical control group has $n_H=8$ fully observed event times, with total follow-up to event $S_H=40$ months. Assume an exponential model and a baseline prior $\lambda\sim\operatorname{Gamma}(2,0.1)$, where the rate parameter $0.1$ has units of months. Take $a_0=0.25$.

The historical likelihood and its powered form are

L_H(\lambda)=\lambda^8\exp(-40\lambda), \qquad L_H(\lambda)^{0.25} =\lambda^2\exp(-10\lambda).

The baseline prior kernel is $\lambda^{2-1}\exp(-0.1\lambda)$. Multiplying kernels gives

\lambda^{1}\exp(-0.1\lambda)\, \lambda^2\exp(-10\lambda) = \lambda^3\exp(-10.1\lambda).

Therefore the normalized conditional power prior is $\operatorname{Gamma}(4,10.1)$. Its shape is $2+0.25(8)=4$, and its rate is $0.1+0.25(40)=10.1$ months. The gamma integral is finite because the shape and rate are positive.

Now suppose a hypothetical current control group has 6 fully observed events with total event time 18 months. Its likelihood is $\lambda^6\exp(-18\lambda)$. Multiplying this unpowered current likelihood by the conditional power prior gives a posterior kernel proportional to

\lambda^{4-1}\exp(-10.1\lambda)\, \lambda^6\exp(-18\lambda) = \lambda^9\exp(-28.1\lambda).

Thus the posterior is $\operatorname{Gamma}(10,28.1)$, with posterior mean $10/28.1=0.3559$ events per month. The historical data contribute through the shared exponential rate, at the specified discount; the result does not establish that historical and current controls truly have identical rates.

Weibull Event Times: Identifying Shared Parameters

A Weibull model allows the event hazard to vary over time. Use the parameterization with shape $k>0$ and rate-like parameter $\lambda>0$:

f(t\mid k,\lambda) =k\lambda t^{k-1}\exp(-\lambda t^k),\qquad t>0.

Here $\lambda$ has units of time to the power $-k$. For independent, fully observed historical event times, the likelihood is

L_H(k,\lambda\mid D_H) =k^{n_H}\lambda^{n_H} \left(\prod_{i=1}^{n_H}t_i^{k-1}\right) \exp\left(-\lambda\sum_{i=1}^{n_H}t_i^k\right).

Its power $a_0$ discounts the likelihood contribution involving both $k$ and $\lambda$. If both parameters are shared between historical and current controls, both can receive historical information. If $k$ is fixed in advance and only $\lambda$ is unknown, the historical contribution informs $\lambda$ conditional on that fixed shape. A parameter is not borrowed simply because it has a familiar name: it must be unknown in the model and linked across the data being combined.

Worked Example: A Weibull Model With Fixed Shape

Consider hypothetical historical event times of 2, 4, and 5 days. Assume a Weibull model with shape fixed at $k=2$, and take $a_0=0.4$. Let the baseline prior for $\lambda$ be $\operatorname{Gamma}(1.5,0.02)$, with rate parameter measured in days squared because $\lambda$ has units of days to the power $-2$.

For these data, $n_H=3$, $\prod_i t_i=2(4)(5)=40$, and $\sum_i t_i^2=4+16+25=45$ days squared. Substituting into the likelihood with $k=2$ gives

L_H(\lambda) =2^3\lambda^3(40)\exp(-45\lambda) =320\lambda^3\exp(-45\lambda).

Raising this expression to $0.4$ gives $320^{0.4}\lambda^{1.2}\exp(-18\lambda)$. The constant $320^{0.4}$ does not depend on $\lambda$, so it cancels when the conditional prior is normalized. Multiplying the remaining kernel by the baseline gamma kernel yields

\lambda^{1.5-1}\exp(-0.02\lambda)\, \lambda^{1.2}\exp(-18\lambda) \ \propto\ \lambda^{2.7-1}\exp(-18.02\lambda).

The conditional power prior is $\operatorname{Gamma}(2.7,18.02)$. Its normalizing integral is finite because its gamma shape and rate are positive. Historical information updates $\lambda$ conditional on $k=2$; this calculation does not estimate or borrow the shape. If $k$ were unknown, the powered likelihood would also involve $k$, and this gamma calculation would no longer normalize the joint prior.

Why the Normalizing Integral Must Be Checked

A power-prior kernel is not automatically a probability density. The following hypothetical example shows the issue even when the baseline prior is proper and the event-time model is standard. Let there be one fully observed event at $t=1$ under the Weibull model, with both $k$ and $\lambda$ unknown, and set $a_0=0.5$. The likelihood is $L_H(k,\lambda)=\lambda k\exp(-\lambda)$.

Worked Example: A Proper Baseline Prior With an Infinite Normalizer

Take independent baseline priors $\lambda\sim\operatorname{Gamma}(1,1)$ and a density for $k>1$ proportional to $k^{-1.25}$. Both are proper: the gamma prior is exponential, and $\int_1^\infty k^{-1.25}\,dk=4$. The powered likelihood is

L_H(k,\lambda)^{0.5} =\lambda^{0.5}k^{0.5}\exp(-0.5\lambda).

After multiplication by the baseline kernels, the joint kernel is proportional to $\lambda^{0.5}\exp(-1.5\lambda)k^{-0.75}$ on $\lambda>0,\ k>1$. Its integral factors into a finite integral over $\lambda$ and an infinite integral over $k$:

\left(\int_0^\infty \lambda^{0.5}\exp(-1.5\lambda)\,d\lambda\right) \left(\int_1^\infty k^{-0.75}\,dk\right) =\text{finite}\times\infty =\infty.

Consequently $C(0.5)=\infty$, so no finite normalizing constant turns this kernel into a conditional prior density. The remedy is not to divide by an infinite quantity: the model or baseline prior must instead be changed, or additional conditions must be imposed so the integral is finite.

This failure is a diagnostic, not a technicality. When parameters are weakly identified by a small historical data set, the powered likelihood may not control the tails of the baseline prior sufficiently. Before using a power prior, determine whether the conditional normalizing integral exists for the chosen model, data, baseline prior, and discount. Conjugate algebra can make this check straightforward in special cases; otherwise analytical arguments or reliable numerical integration may be needed.

Assumptions, Alternatives, and Practical Limits

The calculations above rely on complete event times, conditional independence, and a correctly specified parametric density. They also rely on a meaningful parameter link: the historical and current controls must use the same endpoint definition and the shared parameter must represent the same feature of the event-time distribution. Differences in population, care, or follow-up can make that link questionable. Discounting reduces the historical likelihood's influence, but does not repair a misspecified model or make noncomparable data exchangeable.

The main alternatives answer different questions. A current-data-only analysis sets $a_0=0$, so historical controls do not contribute. Full pooling corresponds to $a_0=1$, using the entire historical likelihood under the shared model. An intermediate fixed discount uses a fraction of that likelihood. These choices alter the weight placed on historical evidence, not the event-time model itself. The earlier material on power-prior discounts distinguishes the discount's interpretation from the choice of likelihood.

A practical check is to state explicitly which parameters are shared before writing the power prior. For example, if Weibull shape differs across eras but the analysis fixes one common shape, the model assumes equality rather than learning it from the data. If shape is shared but weakly identified, the normalizer may be especially sensitive to the baseline prior. Power priors also do not by themselves account for unrecorded differences between studies. Model and data comparability must be assessed separately from the discount.

Choice	Historical contribution	Interpretation
$a_0=0$	None	Baseline prior and current data determine inference
$0<a_0<1$	Discounted likelihood	Historical evidence informs shared parameters with reduced likelihood weight
$a_0=1$	Full likelihood	Historical data enter at full weight under the shared model

Key takeaway: Write the event-time likelihood first, raise that likelihood to the chosen discount, and identify the parameters that actually appear in it. Then verify that the integral against the baseline prior is finite and nonzero before calling the result a conditional prior density.

Check Your Understanding

Answer each question by naming the model assumptions and showing the relevant calculation where appropriate.

Under an exponential model, what historical summaries appear in the powered likelihood for fully observed event times?
For Weibull historical data with fixed shape $k$, which parameter is updated in the fixed-shape calculation? What changes if $k$ is also unknown?
Suppose $n_H=5$, $S_H=12$ months, $a_0=0.2$, and the baseline prior is $\operatorname{Gamma}(3,0.5)$. Give the conditional power-prior parameters.
Does a proper baseline prior guarantee a proper conditional power prior? Explain using the normalizing integral.
What does setting $a_0=0$ do to the historical likelihood contribution?

Solutions. (1) The powered likelihood is $\lambda^{a_0n_H}\exp(-a_0\lambda S_H)$, so it uses the historical event count $n_H$ and total event time $S_H$. (2) With fixed $k$, only $\lambda$ is updated; with unknown $k$, the historical likelihood also depends on $k$, so it can inform both parameters if they are shared, and the joint normalizer must be checked. (3) The shape is $3+0.2(5)=4$, and the rate is $0.5+0.2(12)=2.9$ months, giving $\operatorname{Gamma}(4,2.9)$; its integral is finite for these positive parameters. (4) No. Properness requires $0<C(a_0)<\infty$; a proper baseline prior can still yield an infinite integral after multiplication by the powered likelihood. (5) For a proper baseline prior, $L_H^0=1$, so $C(0)=1$ and the historical data contribute nothing.

Choice	Historical contribution	Interpretation
\(a_0=0\)	None	Baseline prior and current data determine inference
\(0<a_0<1\)	Discounted likelihood	Historical evidence informs shared parameters with reduced likelihood weight
\(a_0=1\)	Full likelihood	Historical data enter at full weight under the shared model