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Prior Predictive Checks · Tutorial 80 of 696

Predictive Means Under Normal Endpoint Models

Learn to derive and interpret prior predictive distributions for normal-endpoint sample means, including cases with known or uncertain variance and independent treatment-arm priors.

Intermediate 10 min read

What You'll Learn

  • Distinguish a prior distribution for an endpoint mean from the prior predictive distribution of a trial sample mean.
  • Derive a normal prior predictive distribution when the endpoint variance is known.
  • Obtain a Student t predictive distribution under a normal-inverse-gamma prior.
  • Calculate prior predictive distributions for treatment-arm sample means and their difference.
  • Check predictive calculations for assumptions, units, and implausible trial-level outcomes.

Why Predict Trial Means?

A prior for a continuous endpoint mean describes uncertainty about the underlying average response. A trial, however, produces a sample mean that also varies from one set of enrolled participants to another. A prior predictive distribution for the sample mean combines these two sources of uncertainty: uncertainty about the endpoint model parameters and sampling variation among participants.

This distinction matters when judging whether a prior is compatible with plausible trial outcomes. A prior mean may be clinically reasonable while its induced distribution for a trial average is too wide, too narrow, or centered somewhere unexpected. The calculations below use fixed sample sizes and normal endpoint models. As in the earlier material on prior predictive distributions for trial data, the aim is to examine outcomes implied by the model before observing trial data.

Definition: Suppose \(Y_1,\ldots,Y_n\) are endpoint values in a trial arm and \(\bar{Y}=n^{-1}\sum_{i=1}^n Y_i\) is the arm’s sample mean. The prior predictive distribution of \(\bar{Y}\) is its distribution before observing the trial outcomes, found by averaging the sampling distribution of \(\bar{Y}\) over the prior distribution of the model parameters.

Model and Assumptions

For one arm, begin with the model \(Y_i\mid\mu,\sigma^2\sim N(\mu,\sigma^2)\), independently for \(i=1,\ldots,n\). The sample size \(n\) is fixed, \(\mu\) is the arm’s population mean, and \(\sigma^2\) is the within-arm variance. Thus, conditional on the parameters, \(\bar{Y}\mid\mu,\sigma^2\sim N(\mu,\sigma^2/n)\).

These results require independent participants with a common normal outcome distribution within the arm, and a prior on \(\mu\) and, where relevant, \(\sigma^2\). They do not account automatically for clustering, repeated measures, changing variance across participants, or informative missingness. Such features require a model that represents them. All means, standard deviations, and outcomes must be expressed in the endpoint’s units; variances are in squared units.

Known Endpoint Variance: A Normal Predictive Distribution

First suppose \(\sigma^2\) is treated as known and the prior for the arm mean is \(\mu\sim N(m_0,\tau^2)\). Conditional on \(\mu\), the sample mean has mean \(\mu\) and variance \(\sigma^2/n\). Integrating over the normal prior gives:

$$ \bar{Y}\sim N\left(m_0,\ \tau^2+\frac{\sigma^2}{n}\right). $$

The predictive center is \(m_0\), the prior mean. The predictive variance is the sum of the prior variance \(\tau^2\) and the sampling variance \(\sigma^2/n\). Increasing \(n\) reduces the sampling contribution, but it does not remove uncertainty about \(\mu\): as \(n\) grows, the predictive variance approaches \(\tau^2\).

Worked Example: A Control-Arm Mean With Known Variance

Suppose a hypothetical trial measures a continuous endpoint in points. In the control arm, \(n=40\), the prior is \(\mu\sim N(100,5^2)\), and the within-arm standard deviation is treated as known to be \(\sigma=12\) points. Find the prior predictive distribution of the sample mean and the probability that it exceeds 110 points.

The normal model’s assumptions are that the 40 outcomes are independent, normally distributed with common mean and variance, and that the sample size is fixed. Under these assumptions, the predictive variance is:

$$ \tau^2+\frac{\sigma^2}{n} =5^2+\frac{12^2}{40} =25+3.6 =28.6. $$

Therefore, the predictive standard deviation is \(\sqrt{28.6}=5.348\) points, and \(\bar{Y}\sim N(100,28.6)\), where 28.6 is the variance. Standardizing the threshold gives:

$$ P(\bar{Y}>110) =1-\Phi\left(\frac{110-100}{\sqrt{28.6}}\right) =1-\Phi(1.870) \approx 0.0308. $$

Under this prior and model, about 3.08% of hypothetical control-arm sample means would exceed 110 points. This is a prior predictive probability, not a posterior probability about the mean after observing data.

Uncertain Variance: A Student t Predictive Distribution

In many applications, the within-arm variance is uncertain as well as the mean. One conjugate specification is the normal-inverse-gamma prior. Here the inverse-gamma distribution is parameterized so that a variance \(v\) with shape \(a_0\) and scale \(b_0\) has density proportional to \(v^{-a_0-1}\exp(-b_0/v)\).

Definition: A normal-inverse-gamma prior is \(\sigma^2\sim\operatorname{InvGamma}(a_0,b_0)\) and \(\mu\mid\sigma^2\sim N(m_0,\sigma^2/k_0)\), with \(a_0,b_0,k_0>0\). Under the independent normal endpoint model with fixed \(n\), the prior predictive distribution of \(\bar{Y}\) is Student t with \(2a_0\) degrees of freedom, location \(m_0\), and scale \(s_{\mathrm{pred}}=\sqrt{(b_0/a_0)(1/k_0+1/n)}\). The Student t scale is not its standard deviation.

A short derivation shows where this result comes from. Conditional on \(\sigma^2=v\), integrating over \(\mu\) combines the prior uncertainty in \(\mu\) with sampling variation:

$$ \bar{Y}\mid v\sim N\left(m_0,\ v\left(\frac{1}{k_0}+\frac{1}{n}\right)\right). $$

Write \(c=1/k_0+1/n\) and \(x\) for a possible value of \(\bar{Y}\). Integrating this conditional normal density over the inverse-gamma density gives an integral proportional to:

$$ \int_0^\infty v^{-(a_0+1/2)-1} \exp\left\{-\frac{b_0+(x-m_0)^2/(2c)}{v}\right\}\,dv \ \propto\ \left[b_0+\frac{(x-m_0)^2}{2c}\right]^{-(a_0+1/2)}. $$

The proportionality follows by substituting \(u=[b_0+(x-m_0)^2/(2c)]/v\); the remaining integral is a gamma integral. The resulting density has the Student t form with \(2a_0\) degrees of freedom and scale squared \(b_0c/a_0\), which gives the stated scale. The heavier tails relative to a normal distribution reflect uncertainty about the variance as well as the mean.

Worked Example: A Sample Mean With Uncertain Variance

Consider a hypothetical endpoint measured in millimetres. Let \(n=16\), \(m_0=50\) mm, \(k_0=4\), \(a_0=4\), and \(b_0=144\) mm\(^2\). Assume independent normal outcomes within the arm and the normal-inverse-gamma prior just defined. Calculate the predictive distribution and its central 95% interval.

The predictive degrees of freedom are \(2a_0=8\). Its scale is:

$$ s_{\mathrm{pred}} =\sqrt{\frac{144}{4}\left(\frac{1}{4}+\frac{1}{16}\right)} =\sqrt{36(0.3125)} =\sqrt{11.25} =3.354\text{ mm}. $$

Thus, \(\bar{Y}\) has a Student t distribution with 8 degrees of freedom, location 50 mm, and scale 3.354 mm. The 97.5th percentile of a t distribution with 8 degrees of freedom is approximately 2.306. The central 95% prior predictive interval is:

$$ 50\pm 2.306(3.354) =50\pm 7.736 =[42.26,\ 57.74]\text{ mm}. $$

This interval describes the central 95% of sample means generated under the specified prior and sampling model. It is not a 95% posterior interval for \(\mu\). Under this model, variance uncertainty produces a t-shaped predictive distribution rather than a normal one.

Two Arms and a Difference of Sample Means

For two independent trial arms, the same reasoning can be applied to each mean. If the arm-specific priors and outcomes are independent, and each arm’s variance is treated as known, then the difference of sample means is also normal. Define the difference as treatment minus control. Its predictive mean is the difference between the prior means, and its predictive variance is the sum of the two prior variances and the two sampling variances.

$$ \bar{Y}_T-\bar{Y}_C \sim N\left( m_{0T}-m_{0C},\ \tau_T^2+\tau_C^2+\frac{\sigma_T^2}{n_T}+\frac{\sigma_C^2}{n_C} \right). $$

The independence assumptions are important: the formula for the variance adds the two variances because the arm means are independent. If the prior means are dependent, a covariance term is needed; if the outcomes are dependent across arms, the sampling model must represent that dependence. This calculation concerns a difference on the original endpoint scale, not a standardized effect.

Worked Example: Predicting a Difference Between Arm Means

In a hypothetical trial, higher endpoint values are better. The treatment arm has \(n_T=36\), \(\mu_T\sim N(96,5^2)\), and known \(\sigma_T=15\) points. The control arm has \(n_C=36\), \(\mu_C\sim N(100,4^2)\), and known \(\sigma_C=12\) points. Assume independent normal outcomes and independent priors across arms. Find the predictive distribution of treatment minus control and the probability that the treatment sample mean is higher.

The predictive mean difference is \(96-100=-4\) points. Its variance is:

$$ 5^2+4^2+\frac{15^2}{36}+\frac{12^2}{36} =25+16+6.25+4 =51.25\text{ points}^2. $$

The predictive standard deviation is \(\sqrt{51.25}=7.159\) points. Therefore, \(\bar{Y}_T-\bar{Y}_C\sim N(-4,51.25)\), with 51.25 denoting the variance. The probability of a positive difference is:

$$ P(\bar{Y}_T-\bar{Y}_C>0) =1-\Phi\left(\frac{0-(-4)}{7.159}\right) =1-\Phi(0.5587) \approx 0.288. $$

Under these priors and assumptions, about 28.8% of hypothetical trials would have a higher treatment-arm sample mean than control-arm sample mean. This probability describes prior predictive trial outcomes; it is not a probability that the treatment is effective after data have been observed.

Diagnostics, Alternatives, and Limitations

A useful check is to compare the analytical calculation with simulated prior predictive trial means. Draw parameters from the stated prior, generate \(n\) outcomes from the conditional endpoint model, and calculate \(\bar{Y}\) for each replicate. The simulated center and spread should agree with the derived distribution within Monte Carlo error. A mismatch can reveal a coding error, an inconsistent variance parameterization, or a simulation that does not follow the assumed model.

Also check the model’s implications in endpoint units. If the predictive distribution assigns substantial probability to sample means outside the scale’s possible or clinically credible range, investigate the prior, the assumed normal endpoint model, or both. A normal model has unbounded support; that may be an acceptable approximation for some continuous measures but not for every constrained endpoint. The predictive distribution also changes with \(n\): smaller trials have more sampling variation, while larger trials retain uncertainty from the prior on the mean.

A plug-in calculation that fixes \(\mu=m_0\) and \(\sigma^2\) at selected values gives only a conditional sampling distribution. It omits uncertainty in the parameters and therefore is not the prior predictive distribution. The normal and Student t formulas above are convenient when their assumptions and priors apply. For more complicated priors or models, simulation can approximate the prior predictive distribution without requiring a closed-form result, but it does not remove the need to check the model and parameterization.

Common pitfall: “The sample mean is centered at the prior mean, so its predictive variance is just \(\sigma^2/n\).” This omits uncertainty about \(\mu\). With known \(\sigma^2\) and \(\mu\sim N(m_0,\tau^2)\), the correct predictive variance is \(\tau^2+\sigma^2/n\). When \(\sigma^2\) is uncertain, integrating over it also changes the predictive shape, giving heavier Student t tails under the normal-inverse-gamma model.

These formulas depend on fixed sample sizes, normal outcomes, and the stated prior structures. They do not by themselves diagnose every feature of a trial design or validate a prior. In particular, a plausible predictive mean does not ensure plausible individual outcomes, and a plausible marginal distribution for each arm does not establish that a joint model is appropriate when arms or endpoints are dependent.

Check Your Understanding

Work through each question, then compare your reasoning with the solutions below.

  1. Suppose \(\mu\sim N(70,3^2)\), \(\sigma=10\), and \(n=25\). Find the prior predictive mean and standard deviation of \(\bar{Y}\), in endpoint units.
  2. For a normal-inverse-gamma prior with \(m_0=20\), \(k_0=2\), \(a_0=3\), \(b_0=18\), and \(n=9\), find the predictive degrees of freedom and scale.
  3. Why does increasing the sample size not eliminate all prior predictive uncertainty about the sample mean?
  4. A researcher sets \(\mu=m_0\) and computes only \(\sigma^2/n\) as the predictive variance, despite using a normal prior with positive variance for \(\mu\). What is missing?

Solutions. (1) The predictive mean is \(70\). Its variance is \(3^2+10^2/25=9+4=13\), so its standard deviation is \(\sqrt{13}=3.606\) endpoint units. (2) The degrees of freedom are \(2a_0=6\). The scale is \(\sqrt{(18/3)(1/2+1/9)}=\sqrt{6(11/18)}=\sqrt{11/3}=1.915\) endpoint units. (3) As \(n\) increases, the sampling variance \(\sigma^2/n\) decreases, but uncertainty about \(\mu\) remains; with known variance and a normal prior, the predictive variance approaches \(\tau^2\). (4) The calculation omits uncertainty in \(\mu\). The correct variance is \(\tau^2+\sigma^2/n\), not merely \(\sigma^2/n\).

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