A Bayesian Estimand as a Posterior Distribution

From an Estimand to a Posterior Distribution

An estimand defines the clinical quantity of interest: the population, outcome, treatment conditions, and strategy for intercurrent events determine what is being targeted. A Bayesian model represents uncertainty about parameters relevant to that quantity. The link between the two is a function: for each possible value of the model parameters, evaluate the estimand. Applying that function to the posterior distribution gives a posterior distribution for the estimand.

This is more than reporting a posterior mean for a model coefficient. A clinical estimand may be a probability, a difference between population summaries, or another function of several parameters. Its posterior can be asymmetric, bounded, or otherwise shaped differently from the parameter posteriors. The essential task is to preserve the estimand’s definition while carrying parameter uncertainty through the required calculation.

Definition: Let $D$ denote the observed trial data, let $\theta$ be the vector of model parameters, and let $p(\theta\mid D)$ be their posterior distribution under a specified likelihood and prior. If the clinical estimand is the function $\Psi=g(\theta)$, its posterior distribution is the distribution induced by applying $g$ to $\theta\sim p(\theta\mid D)$. Equivalently, for any set $A$, $$ P(\Psi\in A\mid D)=\int I\{g(\theta)\in A\}\,p(\theta\mid D)\,d\theta. $$

The estimand is fixed once its clinical definition and target are specified. The posterior distribution describes uncertainty about its value given the data, model, and prior. It does not mean that the clinical question changes from draw to draw. Rather, each parameter draw represents one plausible parameter configuration under the posterior, and evaluating $g$ at that configuration gives the corresponding plausible value of the estimand.

Assumptions and the Transformation Principle

The calculation is meaningful only when the model and the estimand refer to compatible quantities. The model must describe the outcome under the treatment conditions relevant to the estimand, and the parameters must identify the population summary or contrast that the estimand requires. Any population averaging, intercurrent-event strategy, or outcome definition in the estimand must be represented in the mapping from parameters to $\Psi$. If the model instead targets a different population or outcome, transforming its posterior does not repair that mismatch.

For a posterior distribution to be used in this way, the likelihood and prior must be specified, the observed data must be represented appropriately, and the posterior must be proper for the quantities being summarized. In computation, posterior draws must adequately represent the posterior. These conditions do not require $g$ to be linear or one-to-one. They do require that $g(\theta)$ be defined for posterior parameter values with probability one.

The transformation principle follows directly from probability. For any function $h$ for which the expectation exists,

E\{h(\Psi)\mid D\} =\int h(g(\theta))p(\theta\mid D)\,d\theta.

Thus, if $\theta^{(1)},\ldots,\theta^{(M)}$ are representative posterior draws, calculate $\Psi^{(m)}=g(\theta^{(m)})$ for each draw. The resulting values approximate the posterior distribution of $\Psi$. For example, their average approximates $E(\Psi\mid D)$, and their empirical quantiles approximate posterior quantiles. The approximation improves as the draws provide better coverage of the posterior; a large draw count alone does not guarantee adequate sampling.

For a one-to-one differentiable transformation $\Psi=g(\theta)$ with inverse $\theta=g^{-1}(\Psi)$, the density changes according to the Jacobian:

p(\psi\mid D) = p\bigl(g^{-1}(\psi)\mid D\bigr) \left|\frac{d\,g^{-1}(\psi)}{d\psi}\right|.

This formula is useful when a density can be derived directly. In trial analyses, the estimand is often a function of several parameters or is not one-to-one. The posterior-draw method still applies in those cases without needing an explicit density formula.

Worked Examples

Worked Example: A Binary-Outcome Treatment Contrast

Consider a hypothetical randomized trial with a binary endpoint: response by the specified assessment time. In 20 participants assigned to treatment, 13 respond; in 20 assigned to control, 8 respond. Suppose the estimand is the difference between the response probabilities in these two randomized groups, with the treatment probability minus the control probability.

Let $p_1$ and $p_0$ denote those probabilities. Assume independent binomial sampling within groups, with a constant response probability within each group, and independent uniform priors $p_1,p_0\sim\operatorname{Beta}(1,1)$. These assumptions give posterior distributions $p_1\mid D\sim\operatorname{Beta}(14,8)$ and $p_0\mid D\sim\operatorname{Beta}(9,13)$: the beta parameters are the prior parameters updated by the response and nonresponse counts.

The estimand is the function $\Psi=p_1-p_0$. Its posterior is the distribution of that difference across paired draws from the joint posterior. Because the arm-specific posteriors are independent under this model, independently drawing $p_1$ and $p_0$ from their respective posteriors gives the appropriate pairing. The posterior means are $14/22$ and $9/22$, so the posterior mean of the estimand is

E(\Psi\mid D)=E(p_1\mid D)-E(p_0\mid D) =\frac{14}{22}-\frac{9}{22} =\frac{5}{22}\approx 0.2273.

For illustration, suppose four posterior draw pairs are $(0.61,0.37)$, $(0.72,0.45)$, $(0.55,0.32)$, and $(0.68,0.49)$, with treatment probability listed first. Applying the function gives differences $0.24$, $0.27$, $0.23$, and $0.19$. In a real analysis, many adequately sampled draws would be transformed; these four merely show the calculation. The posterior mean above is an exact calculation under the stated model, not the average of these illustrative pairs.

The binomial model assumes independent participant outcomes conditional on the group probabilities, complete endpoint classification for these 40 participants, and the specified treatment of the target groups. If clustering, missing outcomes, or a different target population matters, the model and estimand mapping would need to address it. This example uses a binary contrast to illustrate transformation; the key step is evaluating the defined function for every posterior draw.

Worked Example: A Continuous-Outcome Mean Difference

Suppose a hypothetical trial measures a continuous outcome in units of points, with $n=10$ participants per group. The sample means are $\bar{x}_1=5.2$ points for treatment and $\bar{x}_0=3.8$ points for control. Consider the population mean difference, treatment minus control. Assume independent normal outcomes within each group, known standard deviation $\sigma=2$ points, and independent priors $\mu_1,\mu_0\sim N(0,100)$, where 100 is the prior variance in squared points.

For either group $j$, the likelihood is based on $\bar{x}_j\mid\mu_j\sim N(\mu_j,4/10)$. Combining this normal likelihood with the stated normal prior gives posterior variance $1/(1/100+10/4)=1/2.51$, and posterior mean $(10/4)\bar{x}_j/2.51$. Therefore the treatment and control posterior means are $2.5(5.2)/2.51\approx5.1793$ and $2.5(3.8)/2.51\approx3.7849$ points.

The estimand is $\Psi=\mu_1-\mu_0$. Its posterior mean is the difference of the two posterior means, approximately $1.3944$ points. Since the group parameters are independent under both the likelihood and prior here, the posterior variance of the difference is the sum of the posterior variances:

\operatorname{Var}(\Psi\mid D) =\frac{1}{2.51}+\frac{1}{2.51} =\frac{2}{2.51}\approx 0.7968\text{ points}^2.

The posterior standard deviation is $\sqrt{0.7968}\approx0.8926$ points. Alternatively, draw $\mu_1$ and $\mu_0$ jointly from their posterior and subtract for each draw. That procedure produces the full posterior for the mean difference, not just its mean and variance. The normal likelihood, known common variance, independent groups, and prior are substantive assumptions; if the variance were unknown or outcomes were clustered, the parameter model and posterior calculation would change.

Worked Example: Transforming a Continuous Outcome to a Threshold Probability

In a hypothetical analysis, suppose the outcome $Y$, measured in points, is modeled as $Y\mid\mu\sim N(\mu,3^2)$, with known standard deviation 3 points. The estimand is the probability that the outcome exceeds 10 points in the target population. Conditional on $\mu$, this probability is $q=g(\mu)=P(Y>10\mid\mu)$.

Standardizing the normal outcome gives

q=1-\Phi\left(\frac{10-\mu}{3}\right) =\Phi\left(\frac{\mu-10}{3}\right),

where $\Phi$ is the standard normal cumulative distribution function. Suppose three illustrative posterior draws for $\mu$ are 11, 12, and 13 points. Transform each draw: the corresponding probabilities are $\Phi(1/3)\approx0.6306$, $\Phi(2/3)\approx0.7475$, and $\Phi(1)\approx0.8413$. The average of these three transformed draws is approximately $(0.6306+0.7475+0.8413)/3=0.7398$. This average illustrates the Monte Carlo calculation; three draws are not enough to characterize a real posterior reliably.

The calculation depends on the normal outcome model and its known standard deviation. If the outcome distribution were skewed, or the standard deviation were uncertain, the probability would need to be evaluated from the corresponding model parameters for each posterior draw. The estimand is the threshold probability; the normal mean is only an input to its calculation.

Diagnostics, Alternatives, and Limitations

Transformation is usually computationally simple, but the result can still be wrong or unreliable. Check first that the function matches the estimand exactly: treatment ordering, outcome units, target population, and intercurrent-event strategy must all be represented correctly. Then check whether each transformed draw is mathematically valid and within the range the estimand permits. A probability outside $[0,1]$, for example, signals a coding or model-to-estimand error.

When the parameter posterior is obtained by simulation, assess convergence and mixing using appropriate chain diagnostics, including effective sample sizes and Monte Carlo error for the estimand summaries of interest. Diagnostics should be applied to the transformed quantity as well as to important model parameters: a well-behaved parameter summary does not guarantee that a nonlinear tail functional is estimated precisely. If repeated runs or chains give materially different estimand summaries, increase or improve sampling before interpreting them.

A plug-in calculation evaluates $g$ at a point estimate, such as the posterior mean parameter. It is quick, but generally does not give the posterior mean of the estimand: in general, $E\{g(\theta)\mid D\}\ne g\{E(\theta\mid D)\}$. A delta-method approximation can account for local uncertainty around an estimate, but may be inaccurate for strongly nonlinear functions, bounded quantities, or skewed posteriors. Direct analytic integration is attractive when available; transforming posterior draws is flexible and usually retains the full posterior shape, at the cost of simulation error.

No transformation can compensate for a poorly chosen estimand, a misspecified likelihood, inappropriate prior information, or inadequate representation of the target population. The posterior is conditional on the model and prior. Uncertainty about those choices requires additional modeling or sensitivity analysis rather than merely generating more draws from the same posterior.

Common pitfall: Reporting $g$ evaluated at a posterior mean parameter as though it were the posterior mean or full posterior distribution of the estimand. The correction is to calculate $g(\theta^{(m)})$ for every posterior draw and summarize those transformed draws. Equality holds in special cases, such as a linear $g$, but not in general.

Check Your Understanding

Answer each question, then compare with the solutions below.

If $\Psi=\exp(\theta)$, what calculation gives posterior draws for $\Psi$ from draws of $\theta$?
Suppose two independent posterior draws for arm probabilities are $p_1=0.70$ and $p_0=0.45$. What is the draw for the treatment-minus-control probability difference?
For $Y\mid\mu\sim N(\mu,2^2)$, write the model-based probability that $Y$ exceeds 8 as a function of $\mu$.
Why can a plug-in value $g\{E(\theta\mid D)\}$ differ from $E\{g(\theta)\mid D\}$?
Name two checks that are needed before interpreting a posterior distribution obtained by transforming draws.

Solutions. (1) For every draw $\theta^{(m)}$, calculate $\Psi^{(m)}=\exp(\theta^{(m)})$; the resulting values approximate the posterior distribution of $\Psi$. (2) The transformed draw is $0.70-0.45=0.25$. (3) It is $P(Y>8\mid\mu)=1-\Phi((8-\mu)/2)=\Phi((\mu-8)/2)$. (4) A nonlinear function does not generally commute with averaging, so applying it to an average need not equal averaging its values across the posterior. (5) Examples include verifying that the function matches the defined estimand and checking posterior sampling convergence or Monte Carlo error for the transformed quantity.

A Bayesian Estimand as a Posterior Distribution

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