Normal Mean Inference With Known Variance

Why the Known-Variance Model Is Useful

A continuous clinical endpoint might be a change in a symptom score, a laboratory measurement, or a functional scale. In a one-arm trial, the immediate inferential target can be the population mean endpoint value, such as the mean change from baseline among patients like those enrolled. Bayesian inference combines information about that mean from a prior distribution with the trial observations through their likelihood.

This tutorial treats the endpoint variance as known. That assumption makes the update especially transparent: the prior and the data each contribute precision, and the posterior remains normal. “Known” means specified as part of the model, not merely estimated from the same small trial and then treated as exact. The result is useful as a foundation and can be reasonable when a variance is supplied by a reliable external source or design assumption; its limitations matter when that information is uncertain.

Model and assumptions: Let \(X_1,\ldots,X_n\) be continuous endpoint measurements in one trial arm. Conditional on the unknown population mean \(\mu\), assume the observations are independent and identically distributed as \(X_i\mid\mu\sim N(\mu,\sigma^2)\), where the sampling variance \(\sigma^2\) is known and constant. Assign the mean a proper normal prior \(\mu\sim N(\mu_0,\tau_0^2)\), with specified \(\mu_0\) and \(\tau_0^2>0\). The prior describes uncertainty about the population mean before observing this trial’s endpoint data.

The normal model assumes a common mean and variance for the observations and, exactly, a normal distribution for individual endpoint values. The independence assumption means that one patient’s measurement does not provide additional information about another’s measurement once the model parameters are fixed. Trial design, repeated measurements, clustering, or substantial departures from normality may require a different model. For this model, the sample mean \(\bar{x}\) contains all the data information about \(\mu\) relevant to the update; the known variance determines how precise that information is.

Deriving the Posterior Update

The sample mean has sampling distribution \(\bar{X}\mid\mu\sim N(\mu,\sigma^2/n)\). Consequently, the likelihood for \(\mu\), after dropping factors that do not depend on \(\mu\), is proportional to a normal density centered at \(\bar{x}\), with precision \(n/\sigma^2\). The prior density has precision \(1/\tau_0^2\). Multiplying the two densities adds their quadratic terms in \(\mu\).

\begin{aligned} p(\mu\mid x_1,\ldots,x_n) &\propto \exp\left\{-\frac{1}{2}\left[ \frac{(\mu-\mu_0)^2}{\tau_0^2} +\frac{n(\mu-\bar{x})^2}{\sigma^2} \right]\right\}\\ &\propto \exp\left\{-\frac{1}{2}\left[ \left(\frac{1}{\tau_0^2}+\frac{n}{\sigma^2}\right)\mu^2 -2\left(\frac{\mu_0}{\tau_0^2}+\frac{n\bar{x}}{\sigma^2}\right)\mu \right]\right\}. \end{aligned}

To identify the resulting normal distribution, write its quadratic exponent in the form \((\mu-\mu_n)^2/v_n\), plus a term that does not depend on \(\mu\). Matching the coefficient of \(\mu^2\) gives the posterior precision \(1/v_n\); matching the coefficient of \(\mu\) gives \(\mu_n/v_n\). Thus:

Posterior result: Under the stated model and prior, the posterior distribution is normal:

\begin{aligned} \mu\mid x_1,\ldots,x_n&\sim N(\mu_n,v_n),\\ v_n&=\left(\frac{1}{\tau_0^2}+\frac{n}{\sigma^2}\right)^{-1},\\ \mu_n&=v_n\left(\frac{\mu_0}{\tau_0^2}+\frac{n\bar{x}}{\sigma^2}\right). \end{aligned}

Here \(v_n\) is the posterior variance, so the posterior standard deviation is \(\sqrt{v_n}\). The mean can also be expressed as a weighted average of the prior mean and the observed sample mean:

\mu_n= \frac{\sigma^2}{\sigma^2+n\tau_0^2}\mu_0+ \frac{n\tau_0^2}{\sigma^2+n\tau_0^2}\bar{x}.

The two nonnegative weights sum to one. More precise prior information (smaller \(\tau_0^2\)) gives the prior mean more weight; a larger sample size or smaller known sampling variance gives the data more weight. Posterior precision is the sum of prior and data precision. This addition of precisions explains why \(v_n\) is smaller than either \(\tau_0^2\) or \(\sigma^2/n\), provided both are positive.

Worked One-Arm Trial Examples

Worked Example: Mean Change With a Moderately Informative Prior

Hypothetical data. Suppose a one-arm trial measures change in a clinical score in points. Assume \(n=25\) independent patient measurements, a known within-arm standard deviation of \(\sigma=10\) points, and an observed sample mean change of \(\bar{x}=3\) points. Before the trial, take \(\mu\sim N(0,25)\), so the prior mean is 0 points and prior standard deviation is 5 points. The model assumes that individual changes are normally distributed with the stated known variance \(100\) points squared.

The data precision is \(n/\sigma^2=25/100=0.25\) per point squared, while the prior precision is \(1/\tau_0^2=1/25=0.04\) per point squared. Therefore:

v_n=\frac{1}{0.04+0.25}=\frac{100}{29}\approx 3.448\text{ points}^2, \qquad \mu_n=\frac{0(0.04)+3(0.25)}{0.29} =\frac{75}{29}\approx 2.586\text{ points}.

The posterior is \(N(2.586,3.448)\), where the second parameter is the variance. Its standard deviation is \(\sqrt{100/29}\approx 1.857\) points. For example, the posterior probability that the population mean change is greater than 0 points is \(P(\mu>0\mid x)=P\{Z>(0-2.586)/1.857\}\approx 0.918\), where \(Z\sim N(0,1)\). Under this model and prior, the trial data imply about a 91.8% posterior probability that the mean change is positive. This is a probability about \(\mu\) conditional on the data and model, not a probability that a null hypothesis is true.

Worked Example: A More Precise Prior Pulls the Mean Toward Its Center

Hypothetical data. Suppose a trial measures a continuous endpoint in points, with \(n=16\), known individual standard deviation \(\sigma=8\) points, and sample mean \(\bar{x}=6\) points. Let the prior be \(N(2,16)\): its mean is 2 points and its standard deviation is 4 points. Independence, normality, and the stipulated common known variance are assumed.

The prior precision is \(1/16=0.0625\), and the data precision is \(16/64=0.25\). Hence:

v_n=\frac{1}{0.0625+0.25}=3.2\text{ points}^2, \qquad \mu_n=\frac{2(0.0625)+6(0.25)}{0.3125} =\frac{1.625}{0.3125}=5.2\text{ points}.

The posterior is \(N(5.2,3.2)\), with standard deviation \(\sqrt{3.2}\approx1.789\) points. The posterior mean lies between the prior mean of 2 and the sample mean of 6, and is closer to the sample mean because the data precision is four times the prior precision. The update does not simply replace the sample mean with a different estimate; it combines two sources of uncertainty according to their precisions.

Worked Example: Posterior Probability Relative to a Clinical Threshold

Hypothetical data. Consider an endpoint measured in points for which a mean of 10 points is a prespecified reference threshold. Suppose \(n=9\), \(\sigma=6\) points is known, \(\bar{x}=7\) points, and the prior is \(N(10,9)\), with prior standard deviation 3 points. Assume the same independent normal sampling model as above.

The prior precision is \(1/9\), and the data precision is \(9/36=1/4\). Their sum is \(13/36\), giving \(v_n=36/13\approx2.769\) points squared. The posterior mean is:

\mu_n=\frac{10(1/9)+7(1/4)}{13/36} =\frac{103}{13}\approx7.923\text{ points}, \qquad \sqrt{v_n}=\sqrt{\frac{36}{13}}\approx1.664\text{ points}.

Because the posterior is normal, the probability that the mean is below 10 points is \(P(\mu<10\mid x)=\Phi\{(10-7.923)/1.664\}=\Phi(1.248)\approx0.894\), where \(\Phi\) is the standard normal cumulative distribution function. Thus, under the specified model and prior, the posterior probability that the population mean is below the reference threshold is about 89.4%. This calculation alone does not establish clinical benefit: the meaning of the endpoint, the direction of a desirable change, and the clinical relevance of the threshold must be justified separately.

Assumptions, Diagnostics, and Alternatives

A useful diagnostic is to ask whether the observed sample mean is surprising under the prior predictive distribution. Before observing the data, integrating over the normal prior gives \(\bar{X}\sim N(\mu_0,\tau_0^2+\sigma^2/n)\). This describes the range of sample means anticipated jointly from prior uncertainty about \(\mu\) and sampling variation. An observation far into its tails can indicate prior-data conflict, a data or scale problem, or a model that does not describe the trial well. It is a prompt for investigation, not an automatic instruction to discard the prior or the data.

In the first worked example, the prior predictive mean is 0 points and its variance is \(25+100/25=29\) points squared. The observed mean of 3 points is only \(3/\sqrt{29}\approx0.557\) prior predictive standard deviations above the prior center, so it is not unusual under this check. Also inspect the endpoint scale and units, verify that the stated variance is applicable to this trial population, and consider whether independence and approximate normality are credible. Strong skewness, outliers, clustering, or heterogeneous variances can make the model’s posterior precision misleading.

A common mistake is to treat the sample mean as the posterior mean, or to use \(\sigma^2/n\) as the posterior variance. The sample mean is the data-only estimate, and \(\sigma^2/n\) is its sampling variance under the model; neither is generally the posterior summary after incorporating a proper prior. The posterior variance is \((1/\tau_0^2+n/\sigma^2)^{-1}\), and the posterior mean is the precision-weighted combination derived above.

Approach	What it uses	Interpretation or limitation
Normal prior with known variance	Prior mean and variance, sample mean, sample size, and specified \(\sigma^2\)	Gives the normal posterior described here; conclusions may be sensitive to the prior or the assumed known variance.
Flat-prior limiting case	Negligible prior precision relative to data precision	The posterior mean approaches \(\bar{x}\) and its variance approaches \(\sigma^2/n\); this limiting comparison is not a substitute for justifying a prior in a finite-data analysis.
Frequentist known-variance inference	Sampling distribution under a fixed \(\mu\), without a prior distribution on \(\mu\)	Answers sampling-based questions about an estimator or interval; its interpretation is not a posterior probability statement about \(\mu\).

The known-variance assumption is the central limitation here. If the variance is not known and must be inferred from the trial, treating it as fixed can understate uncertainty; that case requires a model that accounts for uncertainty in the variance. Likewise, a prior can dominate when its precision is large relative to \(n/\sigma^2\). Sensitivity to plausible prior means and variances should then be examined rather than presenting a single posterior as though the prior were inconsequential.

The posterior standard deviation quantifies uncertainty about the population mean under the model. It is not the standard deviation of individual patient outcomes, which is \(\sigma\) under this model, and it is not the uncertainty for a future individual measurement. Keeping these targets distinct is important when communicating trial results.

Check Your Understanding

Use the known-variance normal model and show the precision calculation for each numerical question.

With a prior mean of 4 points and prior variance 9 points squared, what is the prior precision?
Suppose \(n=4\), \(\sigma=2\) points, \(\bar{x}=5\) points, and the prior is \(N(1,4)\). Find the posterior mean and variance.
For that same example, is the posterior mean closer to the prior mean or the sample mean? Explain using the precisions.
Why is it incorrect to interpret \(\sigma^2/n\) as the posterior variance when a proper normal prior is used?

Solutions. (1) The precision is the reciprocal of the variance: \(1/9\) per point squared. (2) Prior precision is \(1/4=0.25\); data precision is \(4/4=1\). Thus \(v_n=1/(1.25)=0.8\) points squared and \(\mu_n=\{1(0.25)+5(1)\}/1.25=4.2\) points. (3) The posterior mean is 4.2 points, closer to the sample mean 5 than to the prior mean 1 because the data precision, 1, exceeds prior precision, 0.25. (4) \(\sigma^2/n\) is the sampling variance of the sample mean conditional on a fixed \(\mu\); posterior variance also incorporates prior precision and equals \((1/\tau_0^2+n/\sigma^2)^{-1}\).

Normal Mean Inference With Known Variance

What You'll Learn