Why Average Exposure-Response Predictions?
A PK/PD analysis can support more than one scientifically defensible exposure-response model. For example, candidate models may differ in whether the response approaches a maximum, how the exposure-response curve is parameterized, or whether a delayed response is represented. Even after comparing the models, uncertainty about which structure is most appropriate may remain. A decision based only on the single model with the largest posterior probability can conceal that uncertainty, especially when models make meaningfully different predictions at the exposure of interest.
Bayesian model averaging addresses this by combining predictions, not by requiring the candidate models to share the same parameterization. At a specified exposure and data set, it forms a predictive distribution whose components are the candidate models’ posterior predictive distributions and whose weights are their posterior model probabilities. Those probabilities, calculated from prior model probabilities and the observed data, are treated as given here; see the earlier material on posterior model probabilities with unequal prior odds.
The candidate models should represent a set of plausible alternatives for the stated prediction task. The weights are conditional on that set: they sum to one even if the set does not include every possible scientific model. Therefore, an averaged prediction is not a guarantee that the candidate set contains the true data-generating process. The prediction target and the conditioning information must also be common across models. For instance, all component distributions should refer to the same outcome, time point, exposure history, and patient or population context.
Predictive Means and Variance
Write \(\mu_m=E(Y^*\mid D,x^*,M_m)\) and \(v_m=\operatorname{Var}(Y^*\mid D,x^*,M_m)\). If the component means and variances are finite, the model-averaged predictive mean is the posterior-weighted mean of the component means:
The predictive variance has two parts: average uncertainty within the candidate models, and variability among their predictive means. This follows from conditioning on the model indicator. Adding and subtracting \(\mu_m\) gives \(Y^*-\mu=(Y^*-\mu_m)+(\mu_m-\mu)\). On squaring and taking the conditional expectation under \(M_m\), the cross-term has expectation zero because \(E(Y^*-\mu_m\mid D,x^*,M_m)=0\). Thus
The \(v_m\) terms capture predictive uncertainty within each model; the squared deviations capture disagreement among the candidate-model predictions. Ignoring the second part understates predictive variance when the models’ means differ. The mixture itself is not generally normal, even when every component distribution is normal, so the mean and variance alone do not determine its quantiles or threshold probabilities.
Worked Example: Combining Predictive Means and Variance
At a specified exposure, suppose two hypothetical models have posterior probabilities \(w_1=0.70\) and \(w_2=0.30\). For a future concentration-related response \(Y^*\), measured in mg/L, their posterior predictive distributions are \(N(2.0,0.5^2)\) and \(N(3.0,1.0^2)\), respectively. Assume both distributions target the same future observation under the same conditions.
The model-averaged mean is \(0.70(2.0)+0.30(3.0)=1.4+0.9=2.3\) mg/L. The within-model variance contribution is \(0.70(0.5^2)+0.30(1.0^2)=0.70(0.25)+0.30(1.00)=0.475\;(\text{mg/L})^2\).
The between-model contribution is \(0.70(2.0-2.3)^2+0.30(3.0-2.3)^2 =0.70(0.09)+0.30(0.49)=0.063+0.147=0.210\;(\text{mg/L})^2\). Consequently, the total predictive variance is \(0.475+0.210=0.685\;(\text{mg/L})^2\), and the predictive standard deviation is \(\sqrt{0.685}=0.8276\) mg/L. The extra \(0.210\;(\text{mg/L})^2\) reflects model disagreement; it would be missed by averaging only the component variances.
This mean and standard deviation do not imply that the averaged distribution is \(N(2.3,0.685)\). It remains the weighted mixture of the two specified normal distributions.
Threshold Probabilities and Predictive Intervals
For a decision tied to a response threshold \(c\), calculate the probability in each component and then average those probabilities using the same posterior model weights. For example, the probability of exceeding the threshold is
For continuous outcomes, a predictive quantile \(q_a\) at probability level \(a\) is defined by \(F(q_a)=a\), where \(F(y)=\sum_m w_m F_m(y)\) is the mixture CDF. An equal-tailed \(100(1-\alpha)\%\) predictive interval has endpoints \(q_{\alpha/2}\) and \(q_{1-\alpha/2}\). In general, these endpoints are not weighted averages of the component interval endpoints. Instead, find the quantiles from the mixture CDF, usually using numerical root-finding or simulation.
Worked Example: Calculating a Threshold-Exceedance Probability
Suppose a hypothetical exposure-response prediction has two candidate-model distributions for a future response \(Y^*\), in response units, with weights \(w_1=0.60\) and \(w_2=0.40\). The distributions are \(N(5.00,0.2659^2)\) and \(N(5.0995,0.1500^2)\). Consider the decision threshold \(c=4.54\) response units.
Under model 1, the standardized threshold is \(z_1=(4.54-5.00)/0.2659\approx-1.73\), so \(P(Y^*\leq4.54\mid D,x^*,M_1)=\Phi(-1.73)\approx0.0418\). Under model 2, \(z_2=(4.54-5.0995)/0.1500=-3.73\), so \(P(Y^*\leq4.54\mid D,x^*,M_2)=\Phi(-3.73)\approx9.574\times10^{-5}\).
The model-averaged probability of being at or below the threshold is therefore \(F(4.54)=0.60(0.0418)+0.40(0.00009574) =0.02508+0.000038296 =0.0251183\), or approximately \(0.02512\). Thus the probability of exceeding 4.54 response units is \(1-0.0251183=0.9748817\), approximately \(0.9749\). Under the stated model set and prediction conditions, the model-averaged predictive probability of exceeding the threshold is about 97.49%.
This conclusion describes predictive uncertainty under the specified data, exposure, candidate set, and model probabilities. It is not the probability that a particular model is true, nor does it by itself determine whether an action is appropriate; the decision also depends on the consequences of false positives and false negatives.
Worked Example: Finding an Equal-Tailed Mixture Interval
For a hypothetical response prediction in response units, let the posterior model probabilities be \(0.50\) for each of two models. Suppose the component predictive distributions are \(N(0,1)\) and \(N(2,1)\). Find an approximate equal-tailed 90% predictive interval.
The mixture CDF is \(F(y)=0.50\Phi(y)+0.50\Phi(y-2)\). The lower endpoint \(q_{0.05}\) solves \(0.50\Phi(q_{0.05})+0.50\Phi(q_{0.05}-2)=0.05\). Numerically, at \(q=-1.284\), the two CDF values are approximately \(0.0996\) and \(0.000512\). Their weighted average is \(0.50(0.0996)+0.50(0.000512)=0.050056\), close to \(0.05\); solving more precisely gives a lower endpoint of about \(-1.284\) response units.
Because the equal-weight mixture is symmetric about 1, its upper 0.95 quantile is approximately \(2-(-1.284)=3.284\) response units. The equal-tailed 90% predictive interval is therefore approximately \([-1.284,\,3.284]\) response units. This interval is obtained from the mixture CDF, not from averaging model-specific interval endpoints.
Assumptions, Diagnostics, and Alternatives
The calculations above require well-defined component predictions for a common target and posterior model probabilities that are nonnegative and sum to one. The component predictive distributions must account for uncertainty relevant to the target: for a future patient response, that may include both parameter uncertainty and outcome variability. A prediction for a latent population mean has a different variance and should not be substituted for a future-observation prediction. State which is being predicted before reporting an interval or exceedance probability.
Check that the candidate models produce predictions on the same scale and under the same exposure and covariate conditions, and that the model weights are aligned with those models. Inspect component predictive distributions as well as the mixture summary. A small-weight model can still affect a tail probability if it assigns appreciable probability beyond the decision threshold. Conversely, a high posterior weight does not make a model’s extrapolation scientifically reliable. If predictions diverge sharply near or beyond the observed exposure range, or if conclusions change substantially under plausible prior choices or candidate sets, report that sensitivity rather than presenting a single averaged value as definitive.
Bayesian model averaging differs from selecting only the highest-probability model: selection gives a single component prediction and discards remaining model uncertainty, whereas averaging retains all candidate predictions in proportion to their posterior probabilities. It also differs from averaging parameter estimates. Candidate models can use parameters with different definitions, scales, or scientific interpretations; averaging their predicted outcomes at a shared target avoids treating such parameters as interchangeable. Parameter averaging across models with different interpretations is a separate problem.
Averaging is not a repair for an implausible candidate set, poor component predictions, or a prediction target that is unsupported by the data. It also does not remove uncertainty about the choice of prior model probabilities, nor does it ensure good predictive calibration. Use the averaged distribution as a transparent summary conditional on the stated candidate set and assumptions, and retain model-specific predictions when their differences are relevant to interpretation or decisions.
Check Your Understanding
Answer each question using the definitions and calculations in this tutorial.
- Two models have posterior probabilities 0.8 and 0.2. Their predictive means are 4 and 7 units. What is the model-averaged predictive mean?
- Why can the model-averaged predictive variance exceed the posterior-weighted average of the component variances?
- For a continuous response, how should the lower endpoint of an equal-tailed 90% model-averaged predictive interval be calculated?
- What must be held common across candidate models when averaging predictions for a PK/PD decision?
- Does a model-averaged probability of threshold exceedance represent the probability that the highest-weight model is true? Explain.
Solutions. (1) \(0.8(4)+0.2(7)=4.6\) units. (2) The total variance includes between-model variability, \(\sum_m w_m(\mu_m-\mu)^2\), in addition to the weighted within-model variances. (3) Form \(F(y)=\sum_m w_mF_m(y)\) and solve \(F(q_{0.05})=0.05\); do not average component interval endpoints. (4) The prediction target, outcome scale, exposure or exposure history, and relevant patient or population conditions must match. (5) No. It is the weighted predictive probability of exceeding the threshold, conditional on the data, candidate set, and model probabilities, not a probability that any one model is true.