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Choosing and Eliciting Priors · Tutorial 64 of 696

Translating Expert Judgments Into Distribution Parameters

Translate an expert’s stated center and uncertainty into distribution parameters, then check the implied summaries against the original judgment.

Intermediate 11 min read

What You'll Learn

  • Distinguish elicited means, medians, modes, standard deviations, and quantiles before choosing a parameterization.
  • Recover normal-distribution parameters from a central value and a symmetric probability interval.
  • Convert a beta prior’s elicited mean and standard deviation into its two shape parameters.
  • Recover lognormal parameters from a median and a stated percentile.
  • Verify fitted distribution summaries and identify incompatible or ambiguous elicitation targets.
  • Compare direct moment matching with quantile matching and explain when numerical fitting is needed.

From a Judgment to a Distribution

An expert may describe a plausible central value, give a range, or state how likely a value is to fall below a threshold. A prior distribution turns such judgments into a mathematical representation, but the translation is not automatic: “center” could mean a mean, median, or mode, while “uncertainty” could mean a standard deviation, a range, or a pair of quantiles. The distribution’s parameters must be chosen to match the particular summaries the expert intended.

This tutorial focuses on parameterizing normal, beta, and lognormal distributions from elicited centers and uncertainty statements. It is not enough to calculate parameters once. The resulting distribution must be checked by calculating its implied mean, standard deviation, median, or quantiles and comparing those values with the elicited judgments.

Definition: Distribution parameter matching is the process of choosing parameters so that specified summaries of a probability distribution agree with specified expert judgments. A match is defined by the summaries chosen in advance; matching a mean and standard deviation does not, by itself, ensure that particular quantiles also match.

Before doing any calculation, record the quantity being elicited, its scale and units, the meaning of the stated center, and the precise meaning of the uncertainty statement. For example, a “95% range” should be clarified as a central 95% interval, not merely a range the expert considers plausible. The probability statement also refers to the elicited uncertainty distribution, not to a guarantee about future trial results.

Normal Distributions: Center and Symmetric Interval

A normal distribution is suitable for a parameter that is naturally represented on the entire real line and for which symmetric uncertainty around a central value is a reasonable approximation. If \(X\sim N(\mu,\sigma^2)\), then its mean and median are both \(\mu\), its standard deviation is \(\sigma\), and its quantiles are determined by the standard normal distribution. In particular, if \(z_q\) is the \(q\)-quantile of a standard normal distribution, then the \(q\)-quantile of \(X\) is \(\mu+z_q\sigma\).

Suppose the expert identifies a central value \(c\) and a central \(100(1-\gamma)\%\) interval with endpoints \(L\) and \(U\). For a symmetric normal distribution, the interval is centered at \(c\), and each endpoint is the same number of standard deviations from it. Thus the elicitation is internally consistent with this model only if \(c=(L+U)/2\). When it is consistent, \(\mu=c\), and \(\sigma\) can be recovered from either endpoint.

$$ \mu=c=\frac{L+U}{2}, \qquad \sigma=\frac{U-L}{2z_{1-\gamma/2}}. $$

This conversion assumes the interval is central and symmetric on the original measurement scale. A lower and upper plausible bound that is not intended as a probability interval cannot be substituted into this formula without an additional assumption. If the stated center is not the midpoint, a symmetric normal distribution cannot match all three stated summaries exactly.

Worked Example: Normal Prior From a Central 95% Interval

Suppose an expert gives a hypothetical expected change in a clinical symptom score of 3.0 points, and says that a central 95% interval for the parameter is 1.2 to 4.8 points. Assume a normal distribution and interpret 3.0 as the mean and the interval as equal-tailed.

The interval midpoint is \((1.2+4.8)/2=3.0\) points, agreeing with the stated center. For a central 95% normal interval, \(z_{0.975}=1.960\). Therefore,

$$ \mu=3.0, \qquad \sigma=\frac{4.8-1.2}{2(1.960)} =\frac{3.6}{3.920} =0.9184\text{ points}. $$

The fitted distribution is \(X\sim N(3.0,0.9184^2)\), with variance in squared points. Checking the implied central interval gives \(3.0\pm1.960(0.9184)=3.0\pm1.8\), or 1.2 to 4.8 points, as elicited. This confirms the interval match; it does not establish that a normal model is appropriate for every possible value of this parameter.

Beta Distributions: Matching a Probability and Its Spread

A beta distribution is defined on the interval from 0 to 1 and is often used for an unknown probability or proportion. Write \(X\sim\operatorname{Beta}(\alpha,\beta)\), where \(\alpha>0\) and \(\beta>0\). Its mean and variance are

$$ E(X)=\frac{\alpha}{\alpha+\beta}, \qquad \operatorname{Var}(X)= \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}. $$

If the expert’s central judgment is the mean \(m\), and the stated uncertainty is a standard deviation \(s\), set \(v=s^2\). The two parameters can be recovered by defining the concentration \(\kappa=\alpha+\beta\). The mean equation gives \(\alpha=m\kappa\) and \(\beta=(1-m)\kappa\). Substituting these expressions into the variance equation gives \(v=m(1-m)/(\kappa+1)\), which can be solved for \(\kappa\).

$$ \kappa=\frac{m(1-m)}{v}-1, \qquad \alpha=m\kappa, \qquad \beta=(1-m)\kappa. $$

For a non-degenerate beta distribution, \(0<m<1\) and \(0<v<m(1-m)\). If an elicited variance does not satisfy this inequality, no beta distribution can have both the stated mean and variance. These formulas match the mean and variance; beta distributions can be asymmetric, so the mean need not equal the median or mode.

Worked Example: Beta Prior From a Mean and Standard Deviation

For a hypothetical baseline event probability, an expert’s mean judgment is 0.30 and the elicited standard deviation is 0.10. Treat both as summaries of the prior distribution, so \(m=0.30\) and \(v=0.10^2=0.0100\).

First check feasibility: \(m(1-m)=0.30(0.70)=0.210\), and \(0.0100<0.210\). The mean and variance are therefore compatible with a non-degenerate beta distribution. The concentration is

$$ \kappa=\frac{0.30(0.70)}{0.0100}-1 =21-1=20. $$

Thus \(\alpha=0.30(20)=6\) and \(\beta=0.70(20)=14\), giving \(X\sim\operatorname{Beta}(6,14)\). Check the summaries using the beta formulas:

$$ E(X)=\frac{6}{20}=0.300, \qquad \operatorname{Var}(X)=\frac{6(14)}{20^2(21)} =\frac{84}{8400}=0.0100, \qquad SD(X)=\sqrt{0.0100}=0.100. $$

The fitted beta distribution reproduces the elicited mean probability and standard deviation. These checks do not imply that an additional elicited quantile will also match; a separate quantile check would be needed if that quantile is important to the judgment.

If the expert instead supplies two probability quantiles, such as a 10th and 90th percentile, there is generally no simple algebraic formula for \(\alpha\) and \(\beta\). One can numerically choose positive parameters so that the beta quantile function matches both elicited values, then calculate the implied mean and other quantiles. The fit should be checked because a numerical solution can match the selected pair while still implying a center the expert did not intend.

Lognormal Distributions: Elicitation on a Positive Scale

A lognormal distribution is appropriate for a strictly positive quantity when uncertainty is naturally described multiplicatively or when the logarithm of the quantity is reasonably modeled as normal. By definition, \(X\) is lognormal with parameters \(\mu_L\) and \(\sigma_L^2\) when \(\log(X)\sim N(\mu_L,\sigma_L^2)\). The subscript distinguishes the normal parameters on the log scale from summaries on the original scale. The median of \(X\) is \(\exp(\mu_L)\), while its \(q\)-quantile is \(\exp(\mu_L+z_q\sigma_L)\). Its arithmetic mean is \(\exp(\mu_L+\sigma_L^2/2)\), which exceeds the median whenever \(\sigma_L>0\).

A median and one percentile above it provide a direct conversion. If the median is \(M>0\) and the \(q\)-quantile is \(Q_q>M\), then the median sets the log-scale location, and the percentile difference sets the log-scale standard deviation.

$$ \mu_L=\log(M), \qquad \sigma_L=\frac{\log(Q_q)-\log(M)}{z_q}. $$

This requires the stated percentile to be above the median, with \(q>0.5\). If the elicited value is below the median, use a lower percentile \(q<0.5\), for which both the numerator and \(z_q\) are negative. The ratio is then positive. The calculation assumes the elicited percentile is a quantile of the lognormal uncertainty distribution and that the positive, multiplicative form is sensible for the quantity.

Worked Example: Lognormal Prior From a Median and 90th Percentile

For a hypothetical positive time-to-resolution parameter, an expert gives a median of 12 days and a 90th percentile of 20 days. Assume a lognormal distribution and use \(z_{0.90}=1.2816\).

The log-scale location and standard deviation are

$$ \mu_L=\log(12)=2.4849, \qquad \sigma_L= \frac{\log(20)-\log(12)}{1.2816} =\frac{\log(20/12)}{1.2816} =\frac{0.5108}{1.2816} =0.3986. $$

Thus \(\log(X)\sim N(2.4849,0.3986^2)\), with \(X\) measured in days on the original scale. The fitted median is \(\exp(2.4849)=12.0\) days, and its 90th percentile is \(\exp(2.4849+1.2816(0.3986))=20.0\) days.

As a further check, the arithmetic mean is \(\exp(\mu_L+\sigma_L^2/2)=12\exp(0.3986^2/2)\approx12.99\) days. This is not the elicited median; it is a different summary implied by the fitted distribution. The distinction matters when an expert’s “typical value” was intended to mean an arithmetic mean rather than a median.

Checking the Match and Choosing a Method

The calculations above use different information and make different assumptions. The appropriate method is determined by the elicited summaries and the parameter’s scale, not simply by which formula is easiest to apply.

DistributionTypical elicited informationParameter-matching methodKey check
NormalMean or midpoint and a symmetric central intervalSet the location to the center and solve for the standard deviation from a normal quantileCheck that the interval midpoint agrees with the center
BetaMean and standard deviation for a probabilityUse the mean and variance equations to solve for \(\alpha\) and \(\beta\)Check \(0<m<1\) and \(0<v<m(1-m)\)
LognormalPositive-scale median and a percentileTransform the elicited summaries with logarithms and solve on the normal scaleCheck the fitted median and percentile on the original scale

The most important diagnostic is to recompute the exact summaries that were elicited, using the fitted parameters. Also inspect any other summary that the expert considers important: for instance, whether a matched mean implies a median that is materially different from the stated “typical” value. If these summaries conflict, do not silently treat one as another. Clarify the judgment, revise the target summaries, or use a method that can fit the summaries the expert actually wishes to express.

A second diagnostic is parameter feasibility and support. A beta distribution cannot represent a probability outside 0 to 1, and its mean and variance must satisfy the stated constraint. A lognormal distribution excludes zero and negative values. A normal distribution allows every real value, so it may be unsuitable when values outside a constrained range would be inconsistent with the parameter’s meaning. These are checks on whether the distribution can represent the judgment, not a substitute for checking the matched summaries.

For normal or lognormal distributions, quantile matching can be convenient when an expert is more comfortable with plausible percentiles than with a standard deviation. Moment matching is direct for a beta distribution when a mean and variance are available. When the elicitation gives several quantiles, or summaries that do not have a closed-form conversion, numerical fitting is an alternative. In every case, the fitted distribution should be reported with the elicited quantities it was designed to match.

Common Mistakes and Limitations

  • Treating every “center” as a mean. For a normal distribution, mean and median coincide; for a beta distribution they need not, and for a lognormal distribution the mean generally exceeds the median. Confirm the intended summary before matching parameters.
  • Using a range without its probability meaning. A pair of plausible bounds is not automatically a 95% interval. The normal interval conversion requires a specified central probability and symmetry.
  • Confusing standard deviation with variance. The beta formulas use \(v=s^2\), not \(s\). A standard deviation of 0.10 corresponds to variance 0.0100.
  • Checking only the parameters rather than their implications. A numerical parameter pair is not evidence of a successful match. Substitute it back into the relevant summary formulas or quantile function.
  • Assuming a fitted distribution captures every judgment. Matching two summaries usually leaves other features implied by the chosen distribution. If those implied summaries contradict the expert’s meaning, the distribution or elicitation targets need reconsideration.
Key takeaway: State exactly which elicited summaries are targets, convert them using the distribution’s own parameterization, and then calculate the implied summaries to verify the match. A valid conversion is both mathematically feasible and interpretable on the parameter’s original scale.

Check Your Understanding

Answer each question, then compare with the brief solutions.

  1. A normal distribution is intended to have a central 90% interval from 4 to 10 units. What are its mean and standard deviation?
  2. For a beta distribution with elicited mean \(m=0.40\) and standard deviation \(s=0.10\), calculate \(\alpha\) and \(\beta\).
  3. A lognormal distribution has a median of 5 and a 75th percentile of 8. Write formulas for \(\mu_L\) and \(\sigma_L\). Which normal quantile is needed?
  4. An expert gives a beta-distribution mean of 0.20 and variance 0.18. Can a non-degenerate beta distribution match both? Explain.

Solutions. (1) The midpoint is \((4+10)/2=7\), so \(\mu=7\). With \(z_{0.95}=1.6449\), \(\sigma=(10-4)/(2(1.6449))=1.824\) units. (2) The variance is \(v=0.0100\), so \(\kappa=0.40(0.60)/0.0100-1=23\); therefore \(\alpha=0.40(23)=9.2\) and \(\beta=0.60(23)=13.8\). (3) Use \(\mu_L=\log(5)\) and \(\sigma_L=[\log(8)-\log(5)]/z_{0.75}\), with \(z_{0.75}\approx0.6745\). (4) No. The maximum permissible variance at mean 0.20 is \(m(1-m)=0.20(0.80)=0.16\), and the stated variance 0.18 exceeds that bound.

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