What a Credible Interval Says
A point estimate such as a posterior mean summarizes a distribution, but it does not show how uncertain the endpoint mean remains after observing trial data. A posterior credible interval gives a range of parameter values with a specified probability under the posterior distribution. Its interpretation is conditional on the observed data, the likelihood, and the prior.
For a continuous endpoint, let \(X_1,\ldots,X_n\) be the observed values for \(n\) trial participants, and let \(\mu\) be their population mean under the model. Values might be changes in systolic blood pressure, measured in mmHg. The intervals below concern \(\mu\), not the range containing a specified proportion of individual participants.
The phrase “equal-tailed” means the posterior probability outside the interval is split equally between its two tails. Other interval constructions are possible, but the equal-tailed form is especially direct when the posterior distribution is symmetric. A credible interval is not automatically a frequentist confidence interval: the two can have the same endpoints in particular models, while their probability statements remain different.
Known Variance: Normal Posterior Quantiles
Suppose the endpoint observations are conditionally independent and normally distributed, \(X_i\mid\mu\sim N(\mu,\sigma^2)\), where the variance \(\sigma^2\) is known. With a normal prior \(\mu\sim N(\mu_0,\tau_0^2)\), the posterior is normal. Its variance \(V_n\) and mean \(m_n\) are
The posterior variance combines prior precision \(1/\tau_0^2\) and data precision \(n/\sigma^2\). For a central \(100(1-\alpha)\%\) interval, let \(z_{1-\alpha/2}\) be the \(1-\alpha/2\) quantile of the standard normal distribution. The interval is \(m_n\pm z_{1-\alpha/2}\sqrt{V_n}\). This calculation assumes that the normal sampling model is adequate, that the variance really is treated as known, and that the stated prior represents the intended information about \(\mu\).
Worked Example: A Known-Variance Endpoint Mean
Suppose a hypothetical trial measures change in systolic blood pressure in mmHg. Assume independent normal observations with known standard deviation \(\sigma=12\) mmHg. For \(n=36\) participants, the observed mean change is \(\bar{x}=-4.2\) mmHg. Use the prior \(\mu\sim N(0,10^2)\), with both prior and endpoint measured in mmHg, and calculate a 95% equal-tailed credible interval.
The posterior variance is \(V_n=(1/100+36/144)^{-1}=(0.01+0.25)^{-1}=3.8462\) mmHg\(^2\). The posterior standard deviation is \(\sqrt{3.8462}=1.961\) mmHg. The posterior mean is \(m_n=3.8462[(0/100)+(36(-4.2)/144)]=-4.038\) mmHg.
Using the unrounded values \(m_n=-4.03846\), \(\sqrt{V_n}=1.96116\), and \(z_{0.975}=1.95996\), the interval is approximately \([-7.882,-0.195]\) mmHg.
Under this model and prior, the posterior probability that the population mean change lies between \(-7.881\) and \(-0.195\) mmHg is 0.95. The interval is below zero, so the posterior assigns high probability to a negative mean change. This conclusion describes the modeled mean, not a guarantee about each participant's change.
Unknown Variance: A Student t Posterior
In many endpoint analyses the variance is not known. A common reference-prior analysis uses the normal sampling model \(X_i\mid\mu,\sigma^2\sim N(\mu,\sigma^2)\), independently, with prior density proportional to \(1/\sigma^2\) for \(\mu\in\mathbb{R}\) and \(\sigma^2>0\). This prior is improper, so it is essential to check that the resulting posterior is proper before interpreting posterior probabilities.
Write \(SSE=\sum_{i=1}^n(X_i-\bar{x})^2=(n-1)s^2\). Under this model and prior, when \(n\geq2\) and \(s^2>0\), the marginal posterior for \(\mu\) is a Student t distribution with \(n-1\) degrees of freedom, location \(\bar{x}\), and scale \(s/\sqrt{n}\). In this tutorial, that statement means \((\mu-\bar{x})/(s/\sqrt n)\) has a standard \(t_{n-1}\) posterior distribution.
The scale \(s/\sqrt n\) is the posterior scale parameter here; it is not the standard deviation of the Student t distribution. When \(n>3\), the latter is \(s/\sqrt n\sqrt{(n-1)/(n-3)}\). The interval quantiles must use the t distribution, not a normal distribution, particularly when the sample is small.
For a central \(100(1-\alpha)\%\) interval, let \(t_{1-\alpha/2,n-1}\) be the corresponding standard Student t quantile. Then
The interval is centered at the sample mean because the marginal posterior is symmetric about \(\bar{x}\). Its width reflects both the observed variability and the sample size. As \(n\) increases, the scale \(s/\sqrt n\) tends to decrease, though the actual interval also depends on the observed \(s\) and the appropriate t quantile.
Why the Unknown-Variance Result Holds
A short derivation shows both where the Student t distribution enters and why zero sample variance is a serious boundary case. The normal likelihood can be factored using \(\sum_i(X_i-\mu)^2=SSE+n(\mu-\bar{x})^2\). Multiplying by the prior gives a joint posterior kernel proportional to
Integrating over \(\mu\) contributes a factor proportional to \(\sqrt{\sigma^2/n}\). Thus the marginal kernel for \(\sigma^2\) is proportional to
If \(SSE>0\), the exponential term controls the behavior near zero, and the kernel is integrable at both zero and infinity for \(n\geq2\). If \(SSE=0\), the exponential term is 1, leaving \((\sigma^2)^{-(n+1)/2}\), which is not integrable near zero. Thus an all-identical sample gives an improper posterior, even if it contains at least two observations.
To obtain the marginal distribution of \(\mu\), integrate the joint kernel over \(\sigma^2\). For \(SSE>0\), the resulting kernel in \(\mu\) is proportional to \([SSE+n(\mu-\bar{x})^2]^{-n/2}\). Since \(SSE=(n-1)s^2\), this is proportional to \(\left[1+\{(\mu-\bar{x})/(s/\sqrt n)\}^2/(n-1)\right]^{-n/2}\), the kernel of a \(t_{n-1}\) distribution with the stated location and scale. This derivation also identifies the conditions needed before using its quantiles.
Worked Example: An Unknown-Variance Credible Interval
In a hypothetical trial, assume independent normal endpoint observations. A sample of \(n=16\) participants has mean change \(\bar{x}=-3.1\) mmHg and sample standard deviation \(s=4.0\) mmHg. The sample variance is positive, so the reference-prior posterior is proper. Find a 95% equal-tailed credible interval for the mean change.
The posterior has \(15\) degrees of freedom and scale \(s/\sqrt n=4.0/\sqrt{16}=1.0\) mmHg. The t quantile is \(t_{0.975,15}\approx2.131\). Therefore, \(\bar{x}\pm t_{0.975,15}s/\sqrt n=-3.1\pm2.131(1.0)\), giving \([-5.231,-0.969]\) mmHg.
Conditional on the model and these observations, the posterior probability that the mean change lies in this interval is 0.95. The interval is entirely negative. The normal-sampling assumption and independence assumption are not verified simply by performing this arithmetic; they must be judged using trial design, endpoint measurement, and appropriate data checks.
Worked Example: A Posterior Probability at a Benchmark
For another hypothetical endpoint dataset, suppose \(n=10\), \(\bar{x}=-2.5\) mmHg, and \(s=3.0\) mmHg, with independent normal observations and the same reference prior. The positive sample variance ensures a proper posterior. Calculate the posterior probability that the mean change is below zero.
The posterior scale is \(3.0/\sqrt{10}=0.9487\) mmHg and there are \(9\) degrees of freedom. Standardizing the benchmark gives \((0-\bar{x})/(s/\sqrt n)=2.5/0.9487=2.635\). Thus \(P(\mu<0\mid\text{data})=P(T_9<2.635)\approx0.986\), where \(T_9\) has a standard t distribution with 9 degrees of freedom.
Under this model and prior, the posterior probability that the population mean change is below zero is approximately 98.6%. This probability statement is not the probability that the null hypothesis is true, and it does not by itself establish a treatment effect: the example concerns a single modeled mean and makes no comparison with a control group.
Posterior Probability and Frequentist Coverage
The unknown-variance interval above has the same algebraic form as the usual two-sided t confidence interval for a normal mean. Under independent normal sampling, its repeated-sampling coverage is exactly \(1-\alpha\): the statistic \((\bar{x}-\mu)/(s/\sqrt n)\) has a \(t_{n-1}\) distribution when the true \(\mu\) is fixed. Therefore, over repeated samples generated with that fixed mean, the interval procedure contains \(\mu\) in a proportion \(1-\alpha\) of samples.
That coverage statement is not the interpretation of the Bayesian interval. In the Bayesian analysis, after observing these data, the parameter \(\mu\) has posterior probability \(1-\alpha\) of falling in the particular interval computed from them. In the frequentist interpretation, \(\mu\) is fixed and the interval varies across repeated samples. Identical endpoints do not make these statements interchangeable. With an informative proper prior, the Bayesian interval will generally have different endpoints and need not have exact frequentist coverage.
A central equal-tailed interval is also not the only possible Bayesian interval. For a symmetric normal or Student t posterior, the equal-tailed interval is centered at the posterior location and is also a highest posterior density interval. For an asymmetric posterior, equal-tailed and highest posterior density intervals may differ. The intended construction should therefore be stated, rather than simply calling an interval “95%” without explaining its convention.
Diagnostics, Pitfalls, and Limitations
Before reporting an interval, check whether its model and calculation are coherent. For the unknown-variance reference-prior result, verify that there are at least two observations and that \(s^2>0\). Inspect the endpoint distribution for marked skewness, outliers, or features incompatible with a normal model; with small samples, departures can materially affect the t-based posterior. Confirm that participants can reasonably be treated as independent observations. For the known-variance analysis, also ask whether treating \(\sigma^2\) as known is defensible rather than merely convenient.
- Using a normal quantile with small \(n\). Under the unknown-variance reference-prior model, the posterior is Student t with \(n-1\) degrees of freedom. Replacing its quantile with 1.96 generally understates tail uncertainty for small samples.
- Ignoring posterior propriety. If all observations are identical, \(s^2=0\); the stated improper prior does not yield a proper posterior. A t interval cannot be rescued by inserting zero into its formula.
- Calling the interval a probability about repeated samples. The 95% Bayesian statement concerns the parameter conditional on the observed data and model. Repeated-sampling coverage is a separate property of an interval procedure.
- Confusing the mean with individual outcomes. The interval quantifies uncertainty in \(\mu\), not a range expected to contain 95% of participants' endpoint values.
A Bayesian interval is sensitive to the likelihood and prior. The flat prior on \(\mu\) and \(1/\sigma^2\) prior on variance are reference choices, not universal defaults that erase modeling assumptions. This tutorial addresses a single normal endpoint mean; it does not establish how to compare two treatment groups or how to model a variance as the scientific target. When normality, independence, prior choice, or the handling of missing endpoint values is questionable, the interval should be accompanied by suitable sensitivity analyses and limitations rather than treated as self-validating.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- In a known-variance normal model with a normal prior, what posterior quantity determines the center of an equal-tailed interval?
- A trial sample has \(n=8\) observations, all exactly equal. Can the unknown-variance reference-prior t interval be used? Why or why not?
- For a 90% equal-tailed interval under the unknown-variance model, which degrees of freedom and which quantile are used?
- What does a 95% posterior credible interval mean, and how does that differ from 95% frequentist coverage?
- A reported interval for a mean endpoint is in mmHg. Does it give a range containing 95% of individual participants' changes? Explain.
Solutions. (1) The posterior mean \(m_n\) is its center for a symmetric normal posterior; the interval endpoints are \(m_n\pm z_{0.95}\sqrt{V_n}\) for 90%, for example. (2) No. Although \(n\geq2\), the sample variance is zero and the posterior under the stated improper prior is improper. (3) Use \(n-1=7\) degrees of freedom and the quantile \(t_{0.95,7}\), splitting 5% into each tail. (4) It assigns posterior probability 0.95 to the parameter lying in the computed interval, conditional on the data and model. Frequentist coverage says that a procedure built this way contains a fixed parameter in 95% of repeated samples under its assumptions. (5) No. It concerns uncertainty about the population mean change, not the distribution or spread of individual changes.