Lognormal Parameters Versus Original-Scale Moments

Why Lognormal Parameters Need Careful Interpretation

A lognormal distribution is specified through the normal distribution of a logarithm. Its parameters therefore describe the logarithm of the variable, not the variable on its original scale. In particular, the parameter called the location is not generally the original-scale mean, and the parameter called the scale is not the original-scale standard deviation.

This distinction matters when interpreting fitted models, reporting population summaries, or translating between parameterizations. We will use \(X\) for the positive lognormal variable and \(Y=\ln X\) for its logarithm. The calculations below show how to move in either direction between their parameters and moments.

Definition: A random variable \(X\) has a lognormal distribution with log-scale parameters \(\mu\in\mathbb{R}\) and \(\sigma^2>0\) if \(Y=\ln X\sim N(\mu,\sigma^2)\). We write \(X\sim\operatorname{Lognormal}(\mu,\sigma^2)\). Thus \(\mu=E[\ln X]\) and \(\sigma^2=\operatorname{Var}(\ln X)\); they are not, in general, \(E[X]\) and \(\operatorname{Var}(X)\).

The condition \(\sigma^2>0\) describes a nondegenerate lognormal distribution. Since \(X=\exp(Y)\), \(X\) is positive. By the change-of-variables formula, its density is

f_X(x)=\frac{1}{x\sigma\sqrt{2\pi}} \exp\!\left[-\frac{(\ln x-\mu)^2}{2\sigma^2}\right], \qquad x>0,

and \(f_X(x)=0\) for \(x\leq 0\). The parameters \(\mu\) and \(\sigma\) are often called the log-scale location and log-scale standard deviation, respectively. In particular, \(\sigma\) is the standard deviation of \(\ln X\), not of \(X\).

Deriving Original-Scale Moments

The moment calculation follows from the normal exponential moment. For completeness, it can be obtained directly by completing the square in the normal density. For any real \(r\), the transformation \(X=\exp(Y)\) gives \(X^r=\exp(rY)\), so

\begin{aligned} E[X^r] &=\frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp\!\left(ry-\frac{(y-\mu)^2}{2\sigma^2}\right)\,dy\\ &=\exp\!\left(r\mu+\frac{r^2\sigma^2}{2}\right) \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp\!\left[-\frac{(y-\mu-r\sigma^2)^2}{2\sigma^2}\right]\,dy\\ &=\exp\!\left(r\mu+\frac{r^2\sigma^2}{2}\right). \end{aligned}

For the second equality, expand the square: \[ ry-\frac{(y-\mu)^2}{2\sigma^2} =-\frac{(y-\mu-r\sigma^2)^2}{2\sigma^2} +r\mu+\frac{r^2\sigma^2}{2}. \] The integral in the second line is the integral of a normal density with mean \(\mu+r\sigma^2\) and variance \(\sigma^2\), multiplied by its normalizing constant, so it equals \(1\). This also shows that the moment exists for every real \(r\), including negative \(r\).

Theorem: If \(X\sim\operatorname{Lognormal}(\mu,\sigma^2)\), then for every real \(r\), \[ E[X^r]=\exp\!\left(r\mu+\frac{r^2\sigma^2}{2}\right). \] In particular, \[ E[X]=\exp\!\left(\mu+\frac{\sigma^2}{2}\right), \qquad \operatorname{Var}(X)= \bigl(\exp(\sigma^2)-1\bigr)\exp(2\mu+\sigma^2). \]

To obtain the variance, apply the moment formula at \(r=1\) and \(r=2\):

\begin{aligned} E[X^2]&=\exp(2\mu+2\sigma^2),\\ \operatorname{Var}(X) &=E[X^2]-(E[X])^2\\ &=\exp(2\mu+2\sigma^2)-\exp(2\mu+\sigma^2)\\ &=\bigl(\exp(\sigma^2)-1\bigr)\exp(2\mu+\sigma^2). \end{aligned}

This variance is on the original scale. It depends on both log-scale parameters, and it is not equal to \(\sigma^2\). The original-scale standard deviation is the square root of this variance.

Worked Example: Computing Moments From Log-Scale Parameters

Suppose \(X\sim\operatorname{Lognormal}(2,0.36)\). Here \(E[\ln X]=2\) and \(\operatorname{Var}(\ln X)=0.36\), so the standard deviation of \(\ln X\) is \(0.6\).

The original-scale mean is \[ E[X]=\exp(2+0.36/2)=\exp(2.18)\approx 8.8463. \] It is not \(2\), and it is not \(\exp(2)\).

The original-scale variance is \[ \begin{aligned} \operatorname{Var}(X) &=\bigl(\exp(0.36)-1\bigr)\exp(4+0.36)\\ &\approx (1.433329-1)(78.257134)\\ &\approx 33.9111. \end{aligned} \] Thus the original-scale standard deviation is approximately \(\sqrt{33.9111}\approx 5.8233\), not \(0.6\). The median is \(\exp(2)\approx 7.3891\), so the mean is larger than the median in this example.

Converting Moments Back to Log-Scale Parameters

Sometimes the available summaries are the original-scale population mean and variance rather than the parameters of \(\ln X\). For a nondegenerate lognormal variable, write \(m=E[X]\) and \(v=\operatorname{Var}(X)\). Both are finite, and \(m>0\), \(v>0\). Taking logarithms of the mean formula and using the variance-to-mean-squared ratio gives

\begin{aligned} m&=\exp\!\left(\mu+\frac{\sigma^2}{2}\right),\\ \frac{v}{m^2} &=\frac{(\exp(\sigma^2)-1)\exp(2\mu+\sigma^2)} {\exp(2\mu+\sigma^2)} =\exp(\sigma^2)-1. \end{aligned}

The ratio \(v/m^2\) therefore determines the log-scale variance, after which the mean equation determines the log-scale location.

Theorem: If \(X\) is nondegenerate lognormal with original-scale mean \(m>0\) and variance \(v>0\), then its log-scale parameters are uniquely determined by \[ \sigma^2=\ln\!\left(1+\frac{v}{m^2}\right), \qquad \mu=\ln(m)-\frac{\sigma^2}{2}. \]

Proof. The ratio identity above gives \[ \exp(\sigma^2)=1+\frac{v}{m^2}. \] Taking logarithms yields \(\sigma^2=\ln(1+v/m^2)\). Since \(m>0\), taking the logarithm of \(m=\exp(\mu+\sigma^2/2)\) gives \(\ln(m)=\mu+\sigma^2/2\), hence \(\mu=\ln(m)-\sigma^2/2\). These steps determine both parameters uniquely. The conditions \(m>0\) and \(v>0\) ensure the formulas are defined and give \(\sigma^2>0\). \(\square\)

Worked Example: Recovering Parameters From a Mean and Variance

Suppose a positive variable is lognormal with original-scale mean \(m=50\) and variance \(v=400\). Its squared mean is \(m^2=2500\), so \[ \frac{v}{m^2}=\frac{400}{2500}=0.16. \]

The log-scale variance is \[ \sigma^2=\ln(1+0.16)=\ln(1.16)\approx 0.148420. \] The log-scale standard deviation is therefore \(\sigma\approx\sqrt{0.148420}\approx 0.3853\). Next, \[ \mu=\ln(50)-\frac{0.148420}{2} \approx 3.912023-0.074210 \approx 3.837813. \]

As a check, the recovered parameters give \[ E[X]=\exp(3.837813+0.148420/2)\approx \exp(3.912023)=50. \] The variance formula gives \[ \bigl(\exp(0.148420)-1\bigr)\exp(2(3.837813)+0.148420) \approx 0.16(2500)=400. \] The median, which is \(\exp(\mu)\), is approximately \(46.42\), below the mean of \(50\).

Location, Spread, and Original-Scale Summaries

The log-scale location has a direct interpretation through the median. Since \(Y=\ln X\) is normal with median \(\mu\), and the exponential function is strictly increasing, the median of \(X\) is \(\exp(\mu)\). More generally, if \(z_p\) is the \(p\)-quantile of \(N(0,1)\), for \(0<p<1\), then

Q_X(p)=\exp(\mu+\sigma z_p).

Thus, for \(0<p<1/2\), an interval from the \(p\)-quantile to the \((1-p)\)-quantile is obtained by exponentiating the corresponding normal quantiles. These expressions describe multiplicative distances from \(\exp(\mu)\): changing \(\ln X\) by \(\sigma\) multiplies \(X\) by \(\exp(\sigma)\). In contrast, the arithmetic mean and variance on the original scale follow the moment formulas, not the quantile formulas.

Worked Example: Checking the Meaning of Each Summary

Let \(X\sim\operatorname{Lognormal}(\ln 10,\ln 1.25)\). The log-scale mean is \(\ln 10\), and the log-scale variance is \(\ln 1.25\approx 0.223144\). The original-scale median is \[ \exp(\mu)=\exp(\ln 10)=10. \]

The original-scale mean is \[ E[X]=\exp\!\left(\ln 10+\frac{\ln 1.25}{2}\right) =10\sqrt{1.25}\approx 11.1803. \] For the variance, first calculate the second moment: \[ E[X^2]=\exp\!\left(2\ln 10+2\ln 1.25\right) =100(1.25)^2=156.25. \] Also, \((E[X])^2=100(1.25)=125\). Therefore \[ \operatorname{Var}(X)=156.25-125=31.25. \] Equivalently, the variance formula gives \[ (1.25-1)\exp(2\ln 10+\ln 1.25) =0.25(100)(1.25)=31.25. \] The log-scale variance \(0.223144\) is not the original-scale variance \(31.25\), and the log-scale location \(\ln 10\) is not the original-scale mean \(11.1803\).

Common Parameter-Interpretation Errors

A frequent error is to read \(\mu\) and \(\sigma^2\) as if they were the mean and variance of \(X\). They are the mean and variance of \(\ln X\). For example, if \(X\sim\operatorname{Lognormal}(2,0.36)\), then \(\operatorname{Var}(\ln X)=0.36\), while \(\operatorname{Var}(X)\approx 33.9111\). Replacing one by the other changes the scale of the quantity being described.

Another common error is to claim \(E[X]=\exp(\mu)\). The quantity \(\exp(\mu)\) is the median, whereas the mean is \(\exp(\mu+\sigma^2/2)\). They coincide only in the limiting zero-spread case, which is excluded from the nondegenerate lognormal model defined here.

When converting from original-scale summaries, use the variance \(v\), not the standard deviation, in \(v/m^2\). If a standard deviation \(s\) is supplied, first set \(v=s^2\). Also distinguish population moments from sample summaries: the conversion formulas are exact for the population mean and variance of a lognormal distribution; plugging sample estimates into them produces parameter estimates, not an identity for every finite sample.

Key takeaway: For \(X\sim\operatorname{Lognormal}(\mu,\sigma^2)\), \(\mu\) and \(\sigma^2\) describe \(\ln X\). The original-scale mean is \(\exp(\mu+\sigma^2/2)\), the original-scale variance is \((\exp(\sigma^2)-1)\exp(2\mu+\sigma^2)\), and the inverse conversion uses \(\sigma^2=\ln(1+v/m^2)\) and \(\mu=\ln(m)-\sigma^2/2\).

Check Your Understanding

Use the log-scale versus original-scale distinction in each answer.

If \(X\sim\operatorname{Lognormal}(1,0.25)\), what are \(E[\ln X]\) and \(\operatorname{Var}(\ln X)\)? Which formula gives \(E[X]\)?
Explain why \(\exp(\mu)\) is the median of a nondegenerate lognormal variable rather than its mean.
A lognormal variable has mean \(m=20\) and variance \(v=100\). Write the formulas for its log-scale variance and location, and calculate the log-scale variance.
A report gives an original-scale standard deviation \(s=8\) and mean \(m=40\). What quantity must be squared before using the moment-to-parameter conversion?
For \(X\sim\operatorname{Lognormal}(\mu,\sigma^2)\), explain why the variance of \(X\) cannot generally be reported as \(\sigma^2\).