Why Unequal Allocation Changes Predictive Success
In a two-arm trial, the number of future participants need not be the same in the treatment and control arms. A planned 2:1 allocation, for example, gives different amounts of future information about each arm. Predictive success must therefore average over the possible outcomes using the treatment-arm and control-arm sample sizes separately. Replacing those two sizes with one pooled number can give the wrong predictive distribution and the wrong probability of meeting the final decision rule.
Predictive success is the probability, given the information currently available and the specified model, that the trial’s final data will satisfy a prespecified success criterion. It is not the probability that the treatment works, nor is it the probability that the current evidence is correct. The criterion might be a threshold on a final posterior probability, a posterior mean contrast, or another explicitly defined rule.
The allocation is part of the prediction. Write \(m_T\) and \(m_C\) for the numbers of future participants assigned to treatment and control. In general, \(m_T\ne m_C\), and the future outcomes in the two arms have different predictive distributions because their sample sizes differ. The analysis below assumes the future allocation counts are fixed. If enrollment, dropout, or the allocation mechanism makes the final counts random, their uncertainty must also be represented.
Binary Outcomes: Separate Beta-Binomial Predictions
Suppose the binary outcome is a response, with response probabilities \(p_T\) and \(p_C\). Assume independent beta priors for the two arm-specific response probabilities and independent participant outcomes within each arm conditional on those probabilities. After the observed data, write the posterior distributions as \(p_T\mid D\sim\operatorname{Beta}(a_T,b_T)\) and \(p_C\mid D\sim\operatorname{Beta}(a_C,b_C)\). Here each pair of parameters incorporates the prior and the observed arm-specific responses and nonresponses.
Let \(X_T\) and \(X_C\) be the future numbers of responses. Conditional on the respective response probabilities, \(X_T\sim\operatorname{Binomial}(m_T,p_T)\) and \(X_C\sim\operatorname{Binomial}(m_C,p_C)\). Integrating over each beta posterior gives separate beta-binomial predictive distributions:
Under independent arm posteriors and independent future outcomes, the joint predictive probability for a pair of counts is the product of the two arm-specific probabilities. To calculate predictive success, sum those joint probabilities over all pairs \((x_T,x_C)\) that make the final data satisfy the chosen rule. Unequal allocation changes both the range of possible counts and their predictive probabilities.
The formula follows directly by integrating the binomial likelihood over the beta posterior. The beta function identity \(B(u,v)=\int_0^1p^{u-1}(1-p)^{v-1}\,dp\) gives:
This integration matters: plugging a posterior mean into a binomial model ignores uncertainty about \(p_j\). For example, the predictive mean is \(E(X_j\mid D)=m_j a_j/(a_j+b_j)\), while its variance is \(m_j a_jb_j(a_j+b_j+m_j)/\{(a_j+b_j)^2(a_j+b_j+1)\}\). The extra variability relative to a binomial distribution with fixed probability reflects posterior uncertainty.
Worked Example: Exact Binary Predictive Success With Unequal Future Counts
Consider a hypothetical binary-response trial with independent current posteriors \(p_T\mid D\sim\operatorname{Beta}(1,1)\) and \(p_C\mid D\sim\operatorname{Beta}(1,1)\). Suppose two future participants are assigned to treatment and one to control. For this example, define final success as a higher treatment than control posterior mean response probability. This is a deliberately simple, fully specified decision rule, not a universal choice of trial criterion.
If \(x_T\) and \(x_C\) are the future response counts, the final posterior means are \((1+x_T)/(2+2)\) and \((1+x_C)/(2+1)\), respectively. A \(\operatorname{Beta}(1,1)\) prior predictive distribution assigns equal probability to every possible number of successes: the treatment count \(X_T=0,1,2\) has probability \(1/3\) for each value, and the control count \(X_C=0,1\) has probability \(1/2\) for each value.
Enumerate the six pairs. If \(x_C=0\), the control posterior mean is \(1/3\); treatment succeeds for \(x_T=1\) or 2, with joint probability \((2/3)(1/2)=1/3\). If \(x_C=1\), the control posterior mean is \(2/3\); treatment succeeds only for \(x_T=2\), with joint probability \((1/3)(1/2)=1/6\). Therefore $$ \operatorname{PS}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}=0.500. $$
The calculation uses two treatment future outcomes and one control future outcome; it does not treat the future data as three interchangeable observations. With other posterior parameters or a different final rule, the same method applies, but the set of successful count pairs and their probabilities must be recalculated.
Continuous Outcomes: Predicting the Future Contrast
For a continuous endpoint, consider a normal model in which the future treatment-minus-control sample mean contrast \(Z\) estimates an effect \(\delta\), with known sampling variance determined by the two future sample sizes. Assume the current posterior is \(\delta\mid D\sim N(m,v)\), and future sampling satisfies \(Z\mid\delta\sim N(\delta,w)\), where $$ w=\sigma^2\left(\frac{1}{m_T}+\frac{1}{m_C}\right). $$ Here \(\sigma^2\) is the common known within-arm variance, and the future arm outcomes are independent conditional on their means. This contrast formulation assumes the likelihood information relevant to the decision is represented by \(Z\); it is not a substitute for a model with an important nuisance parameter or a more complex endpoint structure.
A final rule can require the posterior probability of a positive effect to reach \(q\), where \(0<q<1\). Conditional on future contrast \(Z\), normal updating gives posterior variance \(v_f=(1/v+1/w)^{-1}\) and posterior mean \(m_f=v_f(m/v+Z/w)\). The criterion \(P(\delta>0\mid D,Z)\ge q\) is equivalent to \(m_f\ge z_q\sqrt{v_f}\), where \(z_q\) is the \(q\) quantile of the standard normal distribution. The predictive distribution before observing \(Z\) is \(Z\mid D\sim N(m,v+w)\).
To see the cutoff, substitute the posterior mean into the final rule: \(v_f(m/v+Z/w)\ge z_q\sqrt{v_f}\). Rearranging for \(Z\) gives \(Z\ge \frac{v+w}{v}z_q\sqrt{v_f}-\frac{w}{v}m\), which is \(Z\ge z_{\mathrm{cut}}\). Because \(Z\mid D\) is normal with mean \(m\) and variance \(v+w\), its upper-tail probability is the expression above. The allocation enters through \(w\): a smaller future count in either arm increases the contrast variance.
Worked Example: Continuous Endpoint With a 4:1 Future Allocation
Suppose, hypothetically, that the current posterior for the treatment-minus-control effect is \(N(0.500,0.250)\), where the second value is the variance. Assume a known within-arm variance \(\sigma^2=1\), four future treatment participants and one future control participant. The final success rule is \(P(\delta>0\mid\text{final data})\ge0.95\).
The predictive contrast variance is \(w=1(1/4+1/1)=1.25\). The final posterior variance is \(v_f=(1/0.250+1/1.25)^{-1}=(4+0.8)^{-1}=0.2083\), with posterior standard deviation \(\sqrt{0.2083}=0.4564\). Using \(z_{0.95}=1.6449\), the posterior mean must reach \(1.6449(0.4564)=0.7507\) for the final rule to pass.
The final posterior mean as a function of future contrast \(Z\) is $$ m_f=\frac{1.25(0.500)+0.250Z}{1.25+0.250} =0.4167+0.1667Z. $$ Thus the criterion passes when \(0.4167+0.1667Z\ge0.7507\), or \(Z\ge2.005\) (using unrounded values). Before seeing future data, \(Z\mid D\sim N(0.500,1.500)\), so its standard deviation is \(\sqrt{1.500}=1.225\). Consequently $$ \operatorname{PS} =1-\Phi\left(\frac{2.005-0.500}{1.225}\right) \approx1-\Phi(1.229) \approx0.110. $$
Under this model and the specified allocation, the probability of meeting the final 0.95 posterior-probability rule is about 11.0%. It is a predictive probability conditional on the current posterior and design assumptions, not a probability that the effect is positive.
Allocation Comparisons, Diagnostics, and Limitations
The allocation ratio affects precision through the sum of the two reciprocal future sample sizes. With a fixed total, a more balanced allocation generally gives a more precise unadjusted contrast when the arm variances and per-participant costs are equal. It does not follow that every decision rule’s predictive success must increase under every reallocation: the current posterior mean, variances, criterion, and allocation direction all matter. Allocation should be evaluated by recomputing the predictive probability, not by relying on a general slogan.
Worked Example: Comparing Two Allocations for the Same Total
Use the hypothetical normal-model quantities from the preceding example, but compare allocations of four treatment and one control participant with three treatment and two control participants. Both allocate five participants in total. Under the 3:2 allocation, \(w=1(1/3+1/2)=0.8333\), and \(v_f=(4+1.2)^{-1}=0.1923\). The required final posterior mean is \(1.6449\sqrt{0.1923}=0.7213\).
For this allocation, the final posterior mean is $$ m_f=\frac{0.8333(0.500)+0.250Z}{0.8333+0.250} \approx0.3846+0.2308Z. $$
The posterior rule therefore passes for \(Z\ge1.4590\) (using unrounded values). The predictive contrast has mean \(0.500\) and variance \(0.250+0.8333=1.0833\), so $$ \operatorname{PS}\approx 1-\Phi\left(\frac{1.4590-0.500}{\sqrt{1.0833}}\right) =1-\Phi(0.9214) \approx0.178. $$
In this specific comparison, the balanced allocation gives predictive success of about 17.8%, compared with about 11.0% under the 4:1 allocation. The result reflects greater contrast precision for this equal-variance model and the positive effect direction in the current posterior. It is not evidence that 3:2 is universally optimal.
Useful checks help reveal mistakes. For binary outcomes, verify that each arm’s predictive probabilities sum to 1 and that the treatment count ranges from 0 to \(m_T\), while the control count ranges from 0 to \(m_C\). For continuous outcomes, check that increasing either future sample size lowers \(w\), and check that the predictive variance is \(v+w\), not \(w\) alone. If a calculation produces a probability outside 0 to 1, or a success event inconsistent with the stated final rule, revisit the threshold and tail direction.
An alternative to exact summation or a normal-tail calculation is simulation: draw arm-specific parameters from their current posteriors, generate future data using each arm’s own future sample size, update the posterior, and record whether the final rule passes. Simulation is useful when the model or rule makes exact calculation impractical, but it introduces Monte Carlo error. A plug-in calculation that fixes each response probability at its posterior mean is simpler but generally understates predictive variability; it does not integrate over posterior uncertainty.
These calculations depend on the model. Independence between arm parameters may be inappropriate under a shared hierarchical model; overdispersed binary outcomes or correlated continuous outcomes require a suitable likelihood; and uncertainty about the control response model is not resolved simply by using unequal counts. Also, a planned allocation is not the same as the realized final allocation if participants are lost or enrollment stops early. Model those mechanisms when material rather than silently substituting planned counts.
- Pooling the future counts. The arms have distinct outcomes and distinct predictive distributions. Keep \(m_T\) and \(m_C\) separate.
- Using a binomial plug-in prediction. A posterior mean response probability does not represent posterior uncertainty. Integrate over the posterior or simulate from it.
- Confusing a final posterior threshold with predictive success. The former is evaluated after the final data; the latter is the probability, before those data arrive, that the rule will be met.
- Assuming the most balanced allocation always wins. That conclusion requires conditions such as equal variances and costs; assess the actual model and decision rule.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- For independent beta posteriors in a binary trial, what determines the range of future treatment response counts?
- Why does a beta-binomial predictive distribution generally have more variance than a binomial distribution using the posterior mean response probability?
- In the normal contrast model, suppose \(m_T\) increases while \(m_C\) is unchanged. What happens to the future contrast sampling variance \(w\)?
- Explain the difference between a final posterior success rule and predictive success.
- Name one reason a fixed planned allocation might not be adequate for predicting the final trial result.
Solutions. (1) The treatment future count ranges from 0 through \(m_T\); the control range is separately 0 through \(m_C\). (2) The beta-binomial includes uncertainty in the response probability, which adds predictive variation beyond conditional binomial variation. (3) Since \(w=\sigma^2(1/m_T+1/m_C)\), increasing \(m_T\) decreases \(w\). (4) A final posterior rule is applied to the completed data; predictive success is the current probability that those future data will make the rule pass. (5) Dropout, early stopping, or variable enrollment can make realized final arm counts differ from planned counts.