Bounded Distributions for Fractions and Probabilities

Why bounded parameters need bounded distributions

Many PBPK inputs are fractions or probabilities: a tissue composition fraction, a fraction unbound, or the probability assigned to a particular event. Such a quantity cannot be negative or exceed one. A distribution that assigns substantial probability outside this range can therefore generate biologically impossible parameter values, even if its mean is plausible. The earlier material on lognormal distributions addresses positive parameters; positivity alone does not enforce an upper bound of one.

Two useful choices for uncertainty about a quantity strictly between zero and one are the beta distribution and a logit-normal distribution. Both respect that support, but their shapes and parameter interpretations differ. Choosing and parameterizing one requires more than matching a mean: the implied variability and behavior near zero and one must also be plausible.

Definition: A random variable $X$ has a beta distribution with shape parameters $a>0$ and $b>0$, written $X\sim\operatorname{Beta}(a,b)$, if its density for $0<x<1$ is $$ f(x)=\frac{x^{a-1}(1-x)^{b-1}}{B(a,b)}, $$ where $B(a,b)$ is the normalizing beta function. Its mean and variance are $$ E(X)=\frac{a}{a+b},\qquad \operatorname{Var}(X)=\frac{ab}{(a+b)^2(a+b+1)}. $$

Here $a$ and $b$ are shape parameters, not observations or sample sizes. The density is continuous on the open interval $(0,1)$, so a beta model does not assign positive probability to exactly zero or exactly one. Its behavior near either boundary depends on the shape parameters.

Matching a beta distribution to a mean and variance

Suppose the desired mean is $m$, where $0<m<1$, and the desired variance is $v$. Define the concentration $k=a+b$. Since the beta mean is $a/(a+b)$, setting it equal to $m$ gives $a=mk$ and $b=(1-m)k$. Substituting those expressions into the beta variance gives

v=\frac{m(1-m)}{k+1}.

Solving for $k$, then recovering the two shape parameters, produces the moment-matching formulas below. These formulas provide an exact match to the specified mean and variance when the inputs are feasible.

Result: For $0<m<1$ and $0<v<m(1-m)$, the beta distribution matching mean $m$ and variance $v$ has $$ k=\frac{m(1-m)}{v}-1,\qquad a=mk,\qquad b=(1-m)k. $$ The condition $v<m(1-m)$ ensures $k>0$ and therefore positive shape parameters.

The upper bound $m(1-m)$ is a mathematical feasibility bound for a variable restricted to $[0,1]$; a beta distribution has strictly smaller variance for an interior mean. As $v$ decreases at a fixed mean, $k$ increases and the distribution becomes more concentrated around that mean. At a fixed $k$, changing the balance between $a$ and $b$ changes the mean and usually the skew.

Worked Example: Matching a Tissue Fraction’s Mean and SD

Assume, hypothetically, that uncertainty about a tissue composition fraction is summarized by mean $m=0.25$ and standard deviation $0.10$. Find the beta distribution with these moments. First square the SD to obtain the variance: $v=0.10^2=0.0100$. The feasibility check is $0.0100<0.25(0.75)=0.1875$, so moment matching is possible.

The concentration is $$ k=\frac{0.25(0.75)}{0.0100}-1 =\frac{0.1875}{0.0100}-1 =18.75-1=17.75. $$ Therefore $$ a=0.25(17.75)=4.4375,\qquad b=0.75(17.75)=13.3125. $$

Thus a suitable model is $X\sim\operatorname{Beta}(4.4375,13.3125)$. Checking the moments, $a/(a+b)=4.4375/17.75=0.25$, and $ab/[(a+b)^2(a+b+1)]=0.25(0.75)/(18.75)=0.0100$, giving SD $0.10$. These are hypothetical uncertainty inputs, not measurements from a study.

The beta shape is not determined by the mean alone. When $a>1$ and $b>1$, the density has an interior mode at $(a-1)/(a+b-2)$. If both shape parameters are below one, density rises toward both boundaries; if one is below one, density can be especially high near its corresponding boundary. Such boundary behavior can have a substantive interpretation, but it should not be accepted accidentally as a side effect of matching two summaries.

Worked Example: A Probability With Small Relative Uncertainty

Suppose, hypothetically, a PBPK model input is a probability with a plausible mean of $0.80$ and SD $0.08$. The variance is $v=0.08^2=0.0064$. It is feasible because $0.0064<0.80(0.20)=0.16$. The concentration and shapes are $$ k=\frac{0.16}{0.0064}-1=25-1=24, \qquad a=0.80(24)=19.2,\qquad b=0.20(24)=4.8. $$

The resulting model is $X\sim\operatorname{Beta}(19.2,4.8)$. It has mean $19.2/24=0.80$, and variance $0.16/(24+1)=0.0064$, hence SD $0.08$. Because $a$ is larger than $b$, the distribution places its central mass toward higher probabilities. The density has an interior mode at $(19.2-1)/(24-2)=18.2/22\approx0.827$, distinct from its mean of $0.80$.

Modeling uncertainty on the logit scale

A second option is to transform the bounded quantity to the real line. The logit transformation is $\operatorname{logit}(x)=\ln[x/(1-x)]$; its inverse is the logistic function $\operatorname{logistic}(z)=1/[1+\exp(-z)]$. A logit-normal model is defined by $Z\sim N(\mu_z,\sigma_z^2)$ and $X=\operatorname{logistic}(Z)$. Every finite value of $Z$ maps to a value strictly between zero and one.

The normal model describes uncertainty in log-odds, not in the original fraction. In general, $\operatorname{logistic}(\mu_z)$ is the median of $X$, not its mean. The mean and variance of $X$ usually have no simple closed-form expressions, so setting $\mu_z=\operatorname{logit}(m)$ does not in general match a desired mean $m$. For substantial uncertainty, obtain moments by numerical integration or simulation if moment matching is required.

Worked Example: Constructing a Logit-Normal Model From Quantiles

Suppose, hypothetically, a fraction is judged to have median $0.20$ and central 90% interval from $0.08$ to $0.42$. For a logit-normal model, the median corresponds to $\mu_z=\operatorname{logit}(0.20)$, because the median of $Z$ is $\mu_z$ and the logistic transformation is increasing. Thus $$ \mu_z=\ln(0.20/0.80)=\ln(0.25)\approx-1.3863. $$

The 5th and 95th percentiles of a normal distribution are approximately $\mu_z-1.645\sigma_z$ and $\mu_z+1.645\sigma_z$. Transforming the stated fraction limits to logits gives $\operatorname{logit}(0.08)=\ln(0.08/0.92)\approx-2.4423$ and $\operatorname{logit}(0.42)=\ln(0.42/0.58)\approx-0.3228$. Therefore $$ \sigma_z=\frac{-0.3228-(-2.4423)}{2(1.645)} =\frac{2.1195}{3.290} \approx0.6445. $$

A corresponding model is $Z\sim N(-1.3863,0.6445^2)$, with $X=\operatorname{logistic}(Z)$. By construction, its median is $0.20$ and its approximate central 90% interval is the specified $0.08$ to $0.42$. Its mean is not thereby guaranteed to be $0.20$; the elicited summaries here are a median and quantiles, not a mean and SD.

Choosing and checking the representation

Both distributions respect the bounds, but they impose different shapes. A beta distribution is convenient when a mean and variance are the primary summaries and its implied density is plausible. A logit-normal model is convenient when uncertainty is naturally described in log-odds or by quantiles on the fraction scale. Neither is universally superior. In particular, a logit-normal model is not simply a beta distribution with different notation, and matching the same mean and variance does not generally make the two shapes identical.

Feature	Beta distribution	Logit-normal distribution
Construction	Directly defined on values between zero and one	Normal model for the logit, transformed back to the fraction scale
Moment interpretation	Mean and variance have direct formulas	Moments on the fraction scale generally require numerical calculation
Shape control	Two shape parameters control mean, spread, skew, and boundary behavior	Normal location and scale control logit-scale location and spread
Quantiles	Can be calculated from the beta distribution	Follow directly from normal quantiles transformed by the logistic function
Exact endpoints	No point mass at zero or one	No point mass at zero or one for finite normal values

A useful diagnostic is to compare the fitted distribution with all the information used to motivate it, not just the summaries matched by construction. Check the mean, SD, relevant quantiles, and probability near important boundaries. Plotting the density or examining simulated draws can expose a shape that is overly concentrated, excessively skewed, or unexpectedly U-shaped. Also check that draws remain meaningful in the PBPK model: a valid fraction considered alone may still conflict with a sum constraint when several fractions must add to one.

A continuous beta or logit-normal model is not appropriate if the parameter is known to equal zero or one with positive probability. A point mass at an endpoint, or a model that explicitly separates endpoint cases from interior values, may be needed. Likewise, if moment matching yields $v\ge m(1-m)$, do not force the beta formulas to produce invalid shapes: reconsider the summaries, the assumed support, or the model. The earlier material on physiological bounds is relevant when checking whether proposed values are possible in the biological setting.

Common mistake

Wrong reasoning: “The desired mean is 0.20, so a logit-normal model should use $\mu_z=\operatorname{logit}(0.20)$, which makes its mean 0.20.”
Correction: That choice makes the median $0.20$, not generally the mean. State which summary is being matched, and calculate or numerically evaluate the fraction-scale mean if the mean is required.
Another warning: Matching a beta mean does not ensure a plausible distribution. Inspect its variance and boundary behavior rather than treating the mean as a complete description.

Key takeaway: Use beta moment matching when its feasibility condition holds and its implied shape is credible. Use a logit-normal model when log-odds or transformed quantiles provide a natural description. For either choice, check the full distribution against the biological bounds and the intended interpretation of uncertainty.

Check Your Understanding

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

A fraction has hypothetical mean $0.40$ and SD $0.10$. What beta shape parameters match these moments?
Why can a beta distribution not match a mean $m=0.50$ and variance $v=0.30$?
In a logit-normal model, what does $\operatorname{logistic}(\mu_z)$ represent?
A proposed continuous beta model is used for a parameter that is exactly zero in 15% of cases. What is the modeling issue?

Solutions. 1. The variance is $0.10^2=0.0100$, and $m(1-m)=0.40(0.60)=0.24$. Thus $k=0.24/0.0100-1=23$, $a=0.40(23)=9.2$, and $b=0.60(23)=13.8$. 2. A beta variance must satisfy $v<m(1-m)=0.25$. Since $0.30$ exceeds this bound, no positive beta shape parameters match those moments. 3. It is the median of $X$, because $\mu_z$ is the median of the normal variable $Z$ and the logistic transformation is increasing; it is not generally the mean. 4. A continuous beta distribution assigns zero probability to the exact endpoint. The model must represent the 15% endpoint probability explicitly, for example with a point mass at zero and a separate model for interior values.