Lognormal Means, Medians, and Geometric Means

Why Three Different Averages Matter

For a positive variable with a right-skewed distribution, a few large values can pull the arithmetic mean well above the value seen near the middle of the distribution. A lognormal model makes this distinction precise: its median and population geometric mean are determined by the location of the logarithms, while its arithmetic mean also depends on log-scale variance.

This difference matters when describing hypothetical claim sizes, repair costs, or other positive measurements. The median describes a central point on the original scale; the arithmetic mean describes expected amount per observation. Treating one as the other can lead to a misleading summary, particularly when variability is substantial.

Definition: A positive random variable $X$ has a lognormal distribution with parameters $\mu$ and $\sigma^2$, written $X\sim\operatorname{Lognormal}(\mu,\sigma^2)$, when $Z=\ln X\sim N(\mu,\sigma^2)$. Here $\ln$ is the natural logarithm, $\mu$ is the log-scale location, and $\sigma^2$ is the log-scale variance. Unless stated otherwise, take $\sigma>0$. The degenerate case $\sigma=0$ gives $X=e^\mu$ with probability 1.

The parameters describe $Z=\ln X$, not the original variable $X$. In particular, $\mu$ is the mean of the logarithms, $E[\ln X]$, rather than the mean of $X$. Exponentiating $\mu$ gives an important value on the original scale, but it does not generally give the arithmetic mean.

Median and Population Geometric Mean

Definition: For a positive random variable $X$ with $E[|\ln X|]<\infty$, its population geometric mean is $\exp(E[\ln X])$. The median is a value $m$ for which $P(X\le m)\ge 1/2$ and $P(X\ge m)\ge 1/2$. For a continuous distribution with a unique median, this is the value at which the cumulative probability is $1/2$.

For the lognormal model, $E[\ln X]=E[Z]=\mu$, so the population geometric mean is $e^\mu$. The median is also $e^\mu$: because $Z$ is normally distributed and symmetric about $\mu$, $P(Z\le\mu)=1/2$. Since the exponential function is increasing, $X\le e^\mu$ exactly when $Z\le\mu$. Thus $P(X\le e^\mu)=1/2$.

\operatorname{Median}(X)=\exp(E[\ln X])=e^\mu.

The equality of median and geometric mean is a feature of the lognormal model, not a general rule for positive distributions. For a finite sample of positive observations $x_1,\ldots,x_n$, the sample geometric mean is $(\prod_{j=1}^n x_j)^{1/n}=\exp(n^{-1}\sum_{j=1}^n\ln x_j)$. It is the sample counterpart to exponentiating an average on the log scale.

Arithmetic Mean and the Effect of Variance

The arithmetic mean $E[X]$ is not just the exponential of $E[\ln X]$. It can be derived from the normal density of $Z=\ln X$. The derivation below shows why the log-scale variance contributes an additional term.

Write the expectation on the log scale.
Since $X=e^Z$, its arithmetic mean is $E[X]=E[e^Z]$. Use the density of $Z\sim N(\mu,\sigma^2)$ to express this expectation as an integral.

Complete the square.
In the exponent, combine $z-(z-\mu)^2/(2\sigma^2)$ into a shifted normal quadratic and a constant term.

Integrate the shifted normal density.
The shifted normal kernel integrates to 1 after its normalizing constant is included, leaving the constant factor $e^{\mu+\sigma^2/2}$.

\begin{aligned} E[X] &=E[e^Z]\\ &=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left(z-\frac{(z-\mu)^2}{2\sigma^2}\right)\,dz\\ z-\frac{(z-\mu)^2}{2\sigma^2} &=\mu+\frac{\sigma^2}{2} -\frac{(z-\mu-\sigma^2)^2}{2\sigma^2}\\ E[X] &=e^{\mu+\sigma^2/2} \frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left(-\frac{(z-\mu-\sigma^2)^2}{2\sigma^2}\right)\,dz\\ &=e^{\mu+\sigma^2/2}. \end{aligned}

The final integral is 1 because its integrand, including the factor $1/(\sigma\sqrt{2\pi})$, is the density of a normal variable with mean $\mu+\sigma^2$ and variance $\sigma^2$. Comparing this arithmetic mean with the median and geometric mean gives

\frac{E[X]}{\operatorname{Median}(X)} =\frac{e^{\mu+\sigma^2/2}}{e^\mu} =e^{\sigma^2/2}.

The ratio depends only on $\sigma^2$, not on $\mu$. For any fixed log-scale variance, shifting the distribution changes the median and mean in the same proportion. Increasing the variance raises the arithmetic mean relative to the median and geometric mean. The mean remains at least as large as either value, with equality when $\sigma^2=0$.

Quantity	Lognormal value	Dependence on parameters
Median	$e^\mu$	Depends on log-scale location $\mu$
Population geometric mean	$e^\mu$	Depends on log-scale location $\mu$
Arithmetic mean	$e^{\mu+\sigma^2/2}$	Depends on both $\mu$ and $\sigma^2$
Mean-to-median ratio	$e^{\sigma^2/2}$	Depends only on log-scale variance

Worked Examples

Worked Example: Comparing Typical and Expected Repair Costs

Suppose a hypothetical repair cost $X$, measured in dollars, has model $X\sim\operatorname{Lognormal}(\ln 400,0.25)$. Find its median, population geometric mean, and arithmetic mean, then compare the two means. Give dollar amounts to the nearest cent.

Method. Identify $\mu=\ln 400$ and $\sigma^2=0.25$. Use $e^\mu$ for the median and geometric mean, and $e^{\mu+\sigma^2/2}$ for the arithmetic mean.

Median and geometric mean.

e^\mu=e^{\ln 400}=400.

Thus the median and population geometric mean are both $\$400.00$.

Arithmetic mean. The variance contribution in the exponent is $0.25/2=0.125$, so

E[X]=e^{\ln 400+0.125} =400e^{0.125} =400(1.13314845) =\$453.25938\ldots

Rounded to the nearest cent, $E[X]=\$453.26$, which is $\$53.26$ above the median. The ratio is $e^{0.125}=1.1331$ to four decimals, so the arithmetic mean is about 13.31% greater than the median. Check: $\$453.26/\$400.00\approx1.1332$, consistent with the unrounded ratio $e^{0.125}$. The common mistake is to report $\$400$ as the arithmetic mean simply because $e^\mu=\$400$; that value is the median and geometric mean here.

Worked Example: Comparing Variability at a Fixed Median

Two hypothetical lognormal models describe positive claim sizes in dollars. Model A has median $\$1{,}200$ and log-scale variance $0.09$; Model B has the same median and log-scale variance $0.49$. Find each arithmetic mean, the dollar difference between them, and the ratio of their arithmetic means. Round dollar amounts to the nearest cent and the ratio to four decimals.

Method. Since both medians equal $e^\mu=\$1{,}200$, use $E[X]=\operatorname{Median}(X)e^{\sigma^2/2}$ for each model.

For Model A, $\sigma^2/2=0.09/2=0.045$:

E[X_A]=1{,}200e^{0.045} =1{,}200(1.04602786) =\$1{,}255.23.

For Model B, $\sigma^2/2=0.49/2=0.245$:

E[X_B]=1{,}200e^{0.245} =1{,}200(1.27762131) =\$1{,}533.15.

The difference, calculated from the unrounded means and then rounded to cents, is

E[X_B]-E[X_A] =1{,}200(e^{0.245}-e^{0.045}) =\$277.91.

The ratio of arithmetic means cancels the common median:

\frac{E[X_B]}{E[X_A]} =e^{(0.49-0.09)/2} =e^{0.20} =1.2214.

Model B’s arithmetic mean is therefore 1.2214 times Model A’s, even though their medians are identical. Check: $\$1{,}533.15-\$1{,}255.23=\$277.92$ when rounded values are subtracted; the one-cent difference from $\$277.91$ comes from rounding each mean separately. The stated difference is correctly rounded from the unrounded calculations. The trap is to compare only medians and conclude the models have equal expected claim sizes.

Worked Example: Recovering Log-Scale Variance From the Mean

A hypothetical positive measurement is modeled as lognormal with median $\$500$. Its arithmetic mean is specified exactly as $\$500e^{0.5}$ (approximately $\$824.36$). Find the log-scale variance and standard deviation, and verify the model’s geometric mean. Give the standard deviation to four decimals.

Method. Use the ratio of arithmetic mean to median to isolate $\sigma^2$:

\frac{E[X]}{\operatorname{Median}(X)} =e^{\sigma^2/2}.

The given ratio is $(500e^{0.5})/500=e^{0.5}$. Taking natural logarithms gives

\frac{\sigma^2}{2}=\ln(e^{0.5})=0.5, \qquad \sigma^2=1, \qquad \sigma=\sqrt{1}=1.0000.

The median is $e^\mu=\$500$, so $\mu=\ln 500$. The population geometric mean is therefore $e^\mu=\$500$, as required. The arithmetic mean is $500e^{1/2}\approx\$824.36$, matching the stated value. Check: the mean-to-median ratio is $e^{1/2}$, and $e^{\sigma^2/2}=e^{1/2}$ when $\sigma^2=1$. The common mistake is to use $\ln(E[X]/\operatorname{Median}(X))$ as the variance; it equals $\sigma^2/2$, so it must be multiplied by 2.

Interpretation and Common Traps

The median and geometric mean describe the model’s central scale on the original measurement scale. The arithmetic mean includes the effect of the right tail and is the relevant average when the question concerns expected dollars per observation. Which summary is useful depends on the quantity being summarized, not on which value is easier to calculate.

Confusing log-scale and original-scale means. The model gives $E[\ln X]=\mu$, but $E[X]=e^{\mu+\sigma^2/2}$, not $e^\mu$.
Forgetting that the second parameter is variance. If a model is stated with $\sigma^2=0.25$, use $0.25/2$ in the arithmetic-mean exponent. Do not square $0.25$ again.
Assuming the median and arithmetic mean are equal. They coincide only in the zero-variance degenerate case. With positive log-scale variance, the arithmetic mean is greater.
Claiming that log-scale variance changes the median. At fixed $\mu$, changing $\sigma^2$ leaves the median and geometric mean at $e^\mu$, while it changes the arithmetic mean.
Ignoring the measurement scale. The logarithm must be taken of a positive numerical measurement in a consistently chosen unit. If dollars are converted to cents, the log-scale location changes; the original-scale median and mean convert by the same factor, while $\sigma^2$ and the mean-to-median ratio do not change.

Key takeaway: If $X\sim\operatorname{Lognormal}(\mu,\sigma^2)$, then its median and population geometric mean are both $e^\mu$, while its arithmetic mean is $e^{\mu+\sigma^2/2}$. The relative gap is exactly $e^{\sigma^2/2}$, so log-scale variance controls how far the arithmetic mean rises above the central value.

Check Your Understanding

For each problem, identify whether the requested quantity depends on the log-scale location, the log-scale variance, or both.

A hypothetical positive expense has $X\sim\operatorname{Lognormal}(\ln 250,0.16)$. Find its median, population geometric mean, and arithmetic mean. Round the arithmetic mean to cents. (Answer: median and geometric mean $=\$250.00$; arithmetic mean $=\$270.82$.)
A lognormal measurement has median $900$ units and log-scale variance $0.64$. Find its arithmetic mean, rounded to two decimals. (Answer: $900e^{0.32}=1{,}239.41$ units.)
Two lognormal models have the same median, while their log-scale variances are $0.04$ and $0.36$. By what factor is the second model’s arithmetic mean larger than the first model’s? Give the factor to four decimals. (Answer: $e^{(0.36-0.04)/2}=e^{0.16}=1.1735$.)
A hypothetical lognormal cost has arithmetic mean $\$700$ and median $\$500$. Find its log-scale variance and standard deviation, rounded to four decimals. (Answer: $\sigma^2=2\ln(700/500)=0.6729$, $\sigma=0.8203$.)
Three positive observed amounts are $\$25$, $\$100$, and $\$400$. Compute their sample geometric mean. (Answer: $(25\cdot100\cdot400)^{1/3}=\$100$.)

Quantity	Lognormal value	Dependence on parameters
Median	\(e^\mu\)	Depends on log-scale location \(\mu\)
Population geometric mean	\(e^\mu\)	Depends on log-scale location \(\mu\)
Arithmetic mean	\(e^{\mu+\sigma^2/2}\)	Depends on both \(\mu\) and \(\sigma^2\)
Mean-to-median ratio	\(e^{\sigma^2/2}\)	Depends only on log-scale variance

What You'll Learn