Lognormal Parameters: Typical Values, Medians, And Means

Why Three Values Can Describe One Parameter

In a population PK model, a positive parameter such as clearance may vary across individuals. With a lognormal random-effects model, the individual parameter is formed by adding a random effect on the log scale and then exponentiating. This simple transformation gives rise to three quantities that are easy to confuse: the typical value associated with a zero random effect, the population median, and the population arithmetic mean.

The typical value and median coincide under the usual model, but the arithmetic mean is generally larger when there is nonzero variability. This distinction matters when reporting parameter values, interpreting model output, or selecting a fixed effect to achieve a desired population mean. All numerical examples below use hypothetical values.

Definition: For a positive parameter $P_i$ in individual $i$, the scalar lognormal random-effects model is $$ P_i=\exp(\theta+\eta_i), \qquad \eta_i\sim N(0,\omega^2), $$ where $\theta$ is the fixed effect on the log scale and $\eta_i$ is the individual random effect. The typical value is the parameter value at $\eta_i=0$, namely $\exp(\theta)$. The population median and arithmetic mean are, respectively, the median and expectation of $P_i$ over the modeled random-effects distribution.

Here $\omega^2$ is the variance of the random effect on the log scale, and $\omega$ is its standard deviation. Both describe variability in the log parameter, not in the original parameter units. The parameter $P_i$ is positive because it is an exponential. The model assumes that the random effect has a normal distribution with mean zero and variance $\omega^2$; when $\omega^2=0$, the distribution is degenerate at zero. We consider one random effect at a time, so no covariance structure is needed.

The value $\exp(\theta)$ is called typical because it corresponds to the reference individual with random effect zero. It does not mean that every individual has this value, nor does it generally equal the population arithmetic mean. The distinction is about which summary of the modeled population is being requested.

Median and Arithmetic Mean

The median follows from the symmetry of the normal distribution. If $\omega^2>0$, the random effect has median zero. Since exponentiation is strictly increasing, applying it to $\theta+\eta_i$ preserves the ordering of values, so the median of $P_i$ is $\exp(\theta)$. When $\omega^2=0$, all modeled values equal $\exp(\theta)$, which is still the median. Thus the typical value and population median agree in both cases.

The arithmetic mean requires averaging after exponentiation. In general, exponentiating an average is not the same as averaging exponentiated values. The normal distribution’s exponential moment gives the required result.

Write the parameter as a product.
From $P_i=\exp(\theta+\eta_i)$, separate the fixed effect and random effect: $P_i=\exp(\theta)\exp(\eta_i)$.

Take the expectation.
Because $\exp(\theta)$ is constant, $E(P_i)=\exp(\theta)E[\exp(\eta_i)]$.

Use the normal exponential moment.
For $\eta_i\sim N(0,\omega^2)$, $E[\exp(\eta_i)]=\exp(\omega^2/2)$. Therefore $E(P_i)=\exp(\theta+\omega^2/2)$.

Result: Under the stated model, the typical value and population median are $\exp(\theta)$, while the population arithmetic mean is $\exp(\theta+\omega^2/2)$. If $\omega^2>0$, the mean exceeds the median because $\exp(\omega^2/2)>1$. If $\omega^2=0$, all three quantities are equal.

The result can also be justified from the moment-generating function of a normal random variable. If $Z\sim N(0,\omega^2)$, then $E[\exp(tZ)]=\exp(t^2\omega^2/2)$. Setting $t=1$ gives $E[\exp(\eta_i)]=\exp(\omega^2/2)$, which completes the mean calculation. This derivation depends on the normal distribution for the log-scale random effect; it is not a general identity for every possible random-effects distribution.

More generally, if $z_q$ is the $q$-quantile of the standard normal distribution, the $q$-quantile of the parameter is $\exp(\theta+\omega z_q)$. The median follows by using $z_{0.5}=0$. The mode, which is a different distributional summary, is $\exp(\theta-\omega^2)$ when $\omega^2>0$. Therefore, $\exp(\theta)$ should not be called the mode of the lognormal distribution.

Calculating the Three Summaries

Worked Example: Clearance With a Given Fixed Effect

Suppose a hypothetical clearance parameter follows the lognormal model with $\theta=\log(5.0)$ and $\omega^2=0.25$. The clearance unit is L/h. Calculate its typical value, population median, and arithmetic mean.

The typical value is $$ \exp(\theta)=\exp(\log(5.0))=5.0\ \text{L/h}. $$ The median is the same, $5.0$ L/h, because the random effect is centered at zero and the exponential transformation preserves order.

For the arithmetic mean, use the log-scale variance, $0.25$: $$ E(P_i)=\exp\left(\log(5.0)+\frac{0.25}{2}\right) =5.0\exp(0.125) =5.666\ \text{L/h}. $$ Here $\exp(0.125)\approx1.13315$. The mean is larger than the median because $\omega^2>0$. These are summaries of the modeled population distribution, not a claim that a particular finite sample must have a mean of exactly $5.666$ L/h.

Worked Example: Volume With a Given Log-Scale Variance

For a hypothetical volume parameter, suppose the typical value is $40$ L and the log-scale random-effect variance is $\omega^2=0.16$. Find $\theta$, the population median, and the arithmetic mean. Confirm that the variance, rather than the standard deviation, is used in the mean formula.

Since $\exp(\theta)=40$ L, the fixed effect is $\theta=\log(40)$. The population median is $40$ L. The standard deviation on the log scale is $\omega=\sqrt{0.16}=0.4$, but the mean formula uses $\omega^2=0.16$: $$ E(P_i)=\exp\left(\log(40)+\frac{0.16}{2}\right) =40\exp(0.08) =43.33\ \text{L}. $$ The multiplication is $40\times1.08329\approx43.3315$ L. The arithmetic mean is thus about $43.33$ L, compared with a median and typical value of $40$ L. Substituting $0.4$ in place of $0.16$ would incorrectly use the log-scale standard deviation as though it were the variance.

Choosing a Fixed Effect for a Target Mean

Sometimes a model specification or a scientific target gives an arithmetic population mean, while the model is parameterized by $\theta$. Rearranging the mean formula gives the fixed effect needed for that target. If the desired arithmetic mean is $m$, then $\exp(\theta+\omega^2/2)=m$, so $\theta=\log(m)-\omega^2/2$. The resulting typical value and median are $m\exp(-\omega^2/2)$, not $m$, unless the variance is zero.

Worked Example: Matching a Target Arithmetic Mean

A hypothetical clearance model should have a population arithmetic mean of $8.0$ L/h, with log-scale variance $\omega^2=0.16$. Find the required fixed effect and the corresponding typical value and median.

Set the target mean to $m=8.0$ L/h and solve for the fixed effect: $$ \theta=\log(8.0)-\frac{0.16}{2} =2.07944-0.08 =1.99944. $$ The typical value, and hence the median, is $$ \exp(\theta)=\exp(1.99944) =8.0\exp(-0.08) =7.385\ \text{L/h}. $$ Checking the arithmetic mean, $$ \exp(\theta+0.16/2)=\exp(1.99944+0.08) =\exp(\log(8.0)) =8.0\ \text{L/h}. $$ By contrast, if the fixed effect were set to $\theta=\log(8.0)$, the typical value and median would be $8.0$ L/h, but the arithmetic mean would be $8.0\exp(0.08)=8.666$ L/h. The two parameterizations answer different questions: one sets the median to $8.0$ L/h, and the other sets the arithmetic mean to $8.0$ L/h.

Interpretation, Diagnostics, and Common Pitfalls

Before interpreting a reported parameter value, identify what it represents. A fixed effect $\theta$ is on the log scale in this model. Exponentiating it gives the typical value and median on the original scale. The population arithmetic mean requires the additional $\omega^2/2$ adjustment on the log scale. Keeping the units in view helps: $\theta$ is a log-scale quantity, whereas $\exp(\theta)$ and $\exp(\theta+\omega^2/2)$ have the parameter’s original units.

A practical check is to calculate the mean-to-median ratio: $$ \frac{E(P_i)}{\operatorname{median}(P_i)} =\frac{\exp(\theta+\omega^2/2)}{\exp(\theta)} =\exp(\omega^2/2). $$ This ratio is at least one, equals one only when $\omega^2=0$, and increases with the log-scale variance. If a calculation with positive variance gives an arithmetic mean below the median, check the sign in the adjustment. If the ratio seems unexpectedly large, confirm that the input was the variance and that it is on the log scale.

Common mistake: “The fixed effect is $\log(8)$, so the population arithmetic mean is $8$.” Under this model, that reasoning identifies the typical value and median, not the mean. With $\omega^2=0.16$, the arithmetic mean is $8\exp(0.08)=8.666$ L/h. To make the arithmetic mean $8$ L/h instead, use $\theta=\log(8)-0.08$, giving a typical value and median of $7.385$ L/h.

The formula also has limits. It presumes a normal distribution for the log random effect and a finite variance $\omega^2$; a different distribution on the log scale can yield a different arithmetic mean. It describes the model-implied population distribution, not uncertainty in the estimated fixed effect or variance. An estimated $\theta$ and $\omega^2$ have their own estimation uncertainty, which is separate from the distinction among the three population summaries. Finally, this is a scalar calculation: relationships among multiple random effects are not needed to derive it.

The alternative summaries are useful for different purposes. The typical value gives the parameter for a zero random effect, and the median divides the modeled population into two equal halves (with all values tied when variance is zero). The arithmetic mean is the expected parameter value across the modeled random-effects distribution. Reporting which summary is intended prevents an apparent discrepancy from being mistaken for a model inconsistency.

Check Your Understanding

Answer each question, then check the short solutions below.

A hypothetical parameter has $\theta=\log(12)$ and $\omega^2=0.09$. What are its typical value, median, and arithmetic mean?
For $\omega^2=0$, do the typical value and median still agree? What is the arithmetic mean?
A model has a typical value of $20$ mg and log-scale variance $0.25$. Calculate the arithmetic mean.
A target arithmetic mean is $10$ L/h and $\omega^2=0.20$. Find the fixed effect and typical value.
Explain the error in using $\omega=0.3$ directly in the adjustment $\omega^2/2$ when the stated log-scale variance is $0.09$.

Solutions. (1) The typical value and median are $12$. The mean is $12\exp(0.09/2)=12\exp(0.045)=12.55$, in the parameter’s units. (2) Yes. With zero variance, $\eta_i=0$ for every modeled individual, so all three values are $\exp(\theta)$. (3) $20\exp(0.25/2)=20\exp(0.125)=22.66$ mg. (4) $\theta=\log(10)-0.20/2=2.20259$. The typical value is $10\exp(-0.10)=9.048$ L/h, and the mean checks to $9.048\exp(0.10)=10.00$ L/h. (5) The stated variance is $0.09$; $0.3$ is its square root, the standard deviation. The correct adjustment is $0.09/2=0.045$, not $0.3/2$.