How Discounting Changes the Posterior Mean

Why Discounting Changes the Posterior Mean

In a trial with a binary control outcome, historical control data can contribute information about the control event rate. A power-prior discount controls how much that historical likelihood contributes. Increasing the discount does not necessarily increase the posterior mean: historical successes tend to raise it, while historical failures tend to lower it. The direction depends on how the historical event rate compares with the posterior mean based on the prior and current controls.

This tutorial derives that relationship for a fixed discount in a beta-binomial model. The calculation assumes a common event probability for current and historical controls. It therefore describes the model’s borrowing behavior, not a guarantee that the historical and current control populations are clinically exchangeable.

Model and Assumptions

Definition: Let $p$ be the event probability in the control population. Assign it a proper beta prior, $p\sim\operatorname{Beta}(a,b)$, with $a>0$ and $b>0$. Let $y$ successes among $n$ current control patients follow a binomial model, and let $y_H$ successes among $n_H$ historical control patients follow the same binomial model conditional on $p$. For a fixed discount $\delta\in[0,1]$, raise the historical likelihood to the power $\delta$. Under these assumptions, the posterior distribution for $p$ is $\operatorname{Beta}(a+y+\delta y_H,\ b+n-y+\delta(n_H-y_H))$.

The common-$p$ assumption means that the model treats the current and historical controls as sharing one event probability, conditional on the model. It also assumes binomial sampling within each data set and uses a fixed, externally specified $\delta$. The discount is not an estimated event rate; it is a weight applied to the historical likelihood. At $\delta=0$, historical outcomes contribute nothing. At $\delta=1$, their likelihood is included without discounting.

Write $y_H$ for historical successes and $n_H-y_H$ for historical failures. The likelihoods, up to factors that do not depend on $p$, are $p^y(1-p)^{n-y}$ for current data and $p^{y_H}(1-p)^{n_H-y_H}$ for historical data. The powered historical likelihood is therefore proportional to $p^{\delta y_H}(1-p)^{\delta(n_H-y_H)}$. Multiplying it by the current likelihood and the beta prior gives the posterior kernel.

Deriving the Posterior Mean

\begin{aligned} \pi(p\mid \text{data},\delta) &\propto p^{a-1}(1-p)^{b-1} p^y(1-p)^{n-y} \left[p^{y_H}(1-p)^{n_H-y_H}\right]^\delta\\ &=p^{a+y+\delta y_H-1} (1-p)^{b+n-y+\delta(n_H-y_H)-1}. \end{aligned}

This is the kernel of a beta distribution. A $\operatorname{Beta}(A,B)$ distribution has mean $A/(A+B)$, so the posterior mean is

E(p\mid \text{data},\delta) = \frac{a+y+\delta y_H} {a+b+n+\delta n_H}.

The expression makes the separate roles of historical successes and failures visible. Each historical success contributes $\delta$ to the first shape parameter, and each historical failure contributes $\delta$ to the second. Thus, increasing the discount adds weighted successes and failures; it does not simply add successes or move the mean in one predetermined direction.

Comparing the Mean With and Without Historical Data

Define the posterior mean from the prior and current controls alone as $m_0$, and the historical event rate as $r_H$, when $n_H>0$. Also let $C=a+b+n$, the sum of the two beta shape parameters before historical data are added. Then $m_0=(a+y)/C$, and algebra gives a useful weighted-average form:

\begin{aligned} m(\delta) &=E(p\mid \text{data},\delta)\\ &=\frac{C m_0+\delta n_H r_H}{C+\delta n_H}. \end{aligned}

The posterior mean is a weighted average of $m_0$ and $r_H$, with weights $C$ and $\delta n_H$. In particular, it lies between those two values. If the historical rate exceeds $m_0$, borrowing moves the mean upward; if it is below $m_0$, borrowing moves the mean downward. If they are equal, the mean does not change with the discount.

Subtracting the no-historical-data mean shows the size and direction of the change directly:

\begin{aligned} m(\delta)-m_0 &=\frac{\delta(y_H C-n_H(a+y))} {C(C+\delta n_H)}\\ &=\frac{\delta n_H(r_H-m_0)} {C+\delta n_H}. \end{aligned}

For $n_H>0$, the denominator is positive. Consequently, for any $\delta>0$, the sign of the change is the sign of $r_H-m_0$. The change is zero at $\delta=0$, and its magnitude increases as the historical data receive more weight, unless the historical rate equals $m_0$.

Justification: Differentiating the posterior mean with respect to the fixed discount gives $$ \frac{d m(\delta)}{d\delta} =\frac{y_H(a+b+n)-n_H(a+y)} {(a+b+n+\delta n_H)^2} =\frac{n_H C(r_H-m_0)} {(C+\delta n_H)^2}. $$ The denominator is positive, so the derivative has the sign of $r_H-m_0$. Thus the mean changes monotonically with $\delta$: it increases if $r_H>m_0$, decreases if $r_H<m_0$, and is constant if the two rates are equal. This derivative concerns the posterior mean only; it does not state how posterior precision changes.

Worked Example: Historical Data Raise the Mean

Suppose the prior is $\operatorname{Beta}(2,2)$, and current controls have 18 successes among 40 patients. Historical controls have 30 successes among 50 patients. Use a fixed discount of $\delta=0.4$.

The current-data posterior without historical information has shape parameters $2+18=20$ and $2+(40-18)=24$, so its mean is $m_0=20/(20+24)=20/44=0.4545$. The historical event rate is $r_H=30/50=0.6000$, which exceeds $m_0$. The model therefore predicts an upward movement in the posterior mean when historical data receive positive weight.

The discounted posterior shapes are $2+18+0.4(30)=32$ and $2+22+0.4(20)=32$. Hence

m(0.4)=\frac{32}{32+32}=\frac{32}{64}=0.5000.

The weighted-average calculation gives the same result: $(44(0.4545)+20(0.6000))/(44+20)=32/64=0.5000$, using unrounded values in the arithmetic. The historical data raise the mean from approximately 0.455 to 0.500, or about 0.0455 in event probability.

Worked Example: Historical Failures Lower the Mean

Suppose the prior is $\operatorname{Beta}(1,1)$, current controls have 12 successes among 30 patients, and historical controls have 8 successes among 40 patients. Let $\delta=0.5$.

With only the prior and current controls, the posterior shapes are $1+12=13$ and $1+18=19$. Thus $m_0=13/32=0.4063$. The historical event rate is $8/40=0.2000$, below the current-only posterior mean. The historical failures also contribute: there are $40-8=32$, of which the discount gives weight $0.5(32)=16$.

The discounted posterior shapes are $1+12+0.5(8)=17$ and $1+18+0.5(32)=35$. Therefore,

m(0.5)=\frac{17}{17+35}=\frac{17}{52}=0.3269.

Equivalently, the historical contribution has total weight $0.5(40)=20$, so the weighted average is $(32(0.40625)+20(0.2000))/(32+20)=17/52=0.3269$. Borrowing lowers the posterior mean by approximately $0.0793$. It would be incorrect to describe this as the discount “adding only eight successes”: it also adds weighted historical failures to the denominator and the second shape parameter.

Worked Example: Matching Rates Leave the Mean Unchanged

Suppose the prior is $\operatorname{Beta}(2,2)$, with 14 successes among 28 current controls and 20 successes among 40 historical controls. Consider $\delta=0.25$.

The current-only posterior is $\operatorname{Beta}(16,16)$, because the current data contain 14 successes and 14 failures. Its mean is $m_0=16/32=0.5000$. The historical rate is also $r_H=20/40=0.5000$. The discounted posterior shapes are $2+14+0.25(20)=21$ and $2+14+0.25(20)=21$, giving

m(0.25)=\frac{21}{21+21}=\frac{21}{42}=0.5000.

The mean is unchanged, even though historical data have been added. This equality of means does not mean the historical data have no effect on the posterior distribution; it means only that their event rate matches the current-only posterior mean.

Limits, Comparisons, and Practical Checks

At $\delta=0$, the formula reduces to $m(0)=(a+y)/(a+b+n)=m_0$: the analysis uses the prior and current controls only. At $\delta=1$, it becomes $(a+y+y_H)/(a+b+n+n_H)$, the ordinary beta-binomial update using both data sets at full weight. Values between zero and one give fractional likelihood contributions. These are algebraic weights, not literal fractional patients.

The no-borrowing analysis is the main comparison when assessing the effect of historical information: it isolates how far the mean moves from the current-only baseline. Full pooling is another comparison, corresponding to $\delta=1$, but it may be inappropriate if the historical and current control settings differ. A fixed discount reduces the historical contribution; it does not correct a wrong common-rate assumption or ensure that the selected discount is clinically justified.

Useful checks are to verify that the posterior shapes include both historical successes and historical failures, that the denominator equals the sum of those shapes, and that the result is between $m_0$ and $r_H$ whenever $\delta>0$ and $n_H>0$. For $\delta=0$, the result must equal $m_0$; for $\delta=1$, it must equal the full-data update. If an intermediate result lies outside the two component means, recheck the arithmetic or the model specification.

A central limitation is that the direction-of-change result is conditional on the assumed common event probability and the chosen data. If outcome definitions, patient characteristics, calendar time, or care differ in ways relevant to the event rate, $r_H$ may not be an appropriate target for borrowing. A lower mean is not automatically a better estimate, nor is an upward shift evidence that borrowing is valid. Clinical comparability and data quality must be considered when selecting and interpreting historical information.

A common mistake is to reason that “more historical successes always raise the posterior mean.” The correction is to compare the full historical rate $y_H/n_H$ with the current-only posterior mean $(a+y)/(a+b+n)$. The discount moves the mean toward the historical rate, so historical data can raise or lower it. Another mistake is to compare the historical rate only with the raw current proportion $y/n$; the relevant reference here includes the prior, so it is $m_0$, not necessarily $y/n$.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

With $a=2$, $b=3$, $y=10$, $n=20$, $y_H=18$, $n_H=30$, and $\delta=0.5$, calculate the posterior mean.
For the data in Question 1, does borrowing raise or lower the current-only posterior mean? Show the two rates used to determine the direction.
State the posterior mean at $\delta=0$ for $a=1$, $b=1$, $y=6$, and $n=10$, regardless of the historical data.
Suppose the historical rate equals the current-only posterior mean. What does the derivative of $m(\delta)$ imply about the posterior mean as the fixed discount varies?

Solutions. (1) The posterior shapes are $2+10+0.5(18)=21$ and $3+10+0.5(12)=19$. The mean is $21/(21+19)=21/40=0.5250$. (2) The current-only mean is $(2+10)/(2+3+20)=12/25=0.4800$, while the historical rate is $18/30=0.6000$. Since the historical rate is higher, borrowing raises the mean; the computed value $0.5250$ is above $0.4800$. (3) At $\delta=0$, the mean is $(1+6)/(1+1+10)=7/12=0.5833$; historical data have zero weight. (4) If $r_H=m_0$, then $dm(\delta)/d\delta=0$, so the posterior mean remains equal to that shared value for every fixed discount, although other features of the posterior distribution can still change.