What the Lognormal Model Describes
Many positive quantities vary multiplicatively: a value may be several times as large as another, rather than a fixed amount larger. A lognormal model is useful when taking logarithms turns such a quantity into a normal random variable. The model is used, for example, as a hypothetical description of claim sizes, repair costs, or biological measurements. Its parameters belong to the logarithmic scale, so they must be interpreted with care.
Throughout, logarithms written as \(\ln\) are natural logarithms. A logarithm of a measured quantity is understood to use its numerical value in a specified unit, or equivalently the ratio of the quantity to a fixed reference unit. Changing that unit changes the logarithmic location parameter, even though it does not change the underlying physical quantity.
An equivalent construction is to take \(Z\sim N(0,1)\) and set \(X=\exp(\mu+\sigma Z)\). This form makes the model’s central transformation explicit: first a normal value is formed on the log scale, then exponentiation returns a value on the original scale. Since an exponential is always positive, \(X\) cannot be zero or negative.
Interpreting the Parameters
The value \(\mu\) locates the normal distribution of \(\ln X\), while \(\sigma\) controls its spread. Both parameters are specified on the log scale. In particular, \(\sigma\), not \(\sigma^2\), is the standard deviation of \(\ln X\). A larger \(\sigma\) means greater dispersion in logarithmic values and typically a more pronounced right tail after exponentiation.
The parameters also depend on how \(X\) is measured. If \(X\) is replaced by \(cX\), where \(c>0\) is a fixed conversion factor, then $$ \ln(cX)=\ln c+\ln X. $$ Thus multiplication by \(c\) adds \(\ln c\) to the log-scale location, while leaving the log-scale variance unchanged. For instance, a cost recorded in dollars rather than thousands of dollars has a different location parameter because the numerical values differ by a factor of 1,000. It is essential to record the measurement unit with the model.
The logarithm base matters as well. If a model is stated using base-10 logarithms, converting it to natural logarithms multiplies the logarithm by \(\ln 10\). A shift and a rescaling both affect the parameters; the variance changes by the square of the rescaling factor.
Worked Example: Reading Log-Scale Parameters for Claim Costs
Suppose a hypothetical claim cost \(X\), recorded as a numerical dollar amount, satisfies \(\ln(X/1000)\sim N(0.20,0.09)\). Identify the lognormal parameters for \(X/1000\), then express the model for the numerical dollar amount \(X\). No interest or mortality assumption is involved; this is a distributional model for one positive cost.
Method. First use the ratio \(R=X/1000\), measured in thousands of dollars. The given statement is directly \(\ln R\sim N(0.20,0.09)\). Therefore \(R\) is lognormal with log-scale mean \(0.20\) and log-scale variance \(0.09\). Its log-scale standard deviation is \(\sqrt{0.09}=0.30\).
Next, express the model in terms of the numerical dollar amount. Since \(X=1000R\), $$ \ln X=\ln(1000)+\ln R. $$ Adding a constant shifts a normal mean but does not change its variance. Hence $$ \ln X\sim N\bigl(\ln(1000)+0.20,\;0.09\bigr). $$ Using \(\ln(1000)=6.9078\) to four decimal places, the dollar-scale numerical value has log-scale parameters \(\mu=7.1078\) and \(\sigma^2=0.09\), with \(\sigma=0.30\).
Check. The two specifications describe the same cost because converting thousands of dollars to dollars multiplies the numerical value by 1,000. The variance remains \(0.09\), as required for adding the constant \(\ln(1000)\) to the logarithm. The common mistake is to retain \(0.20\) as the log-scale mean for dollar amounts; that value applies to \(X/1000\), not to \(X\).
From a Normal Logarithm to a Density
The density follows by transforming the normal density. This derivation is also a useful check on the support and the factor that appears in the lognormal formula.
Let \(Y=\ln X\), with \(Y\sim N(\mu,\sigma^2)\). Its density is \(f_Y(y)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left[-\frac{(y-\mu)^2}{2\sigma^2}\right]\), for real \(y\).
Since \(Y=\ln X\), the inverse is \(X=e^Y\), or \(Y=\ln X\). This inverse is defined for \(x>0\), which is why the transformed variable has positive support.
For \(x>0\), the derivative of the inverse transformation is \(\frac{d}{dx}\ln x=1/x\). Therefore \(f_X(x)=f_Y(\ln x)\left|1/x\right|\).
Substituting the normal density at \(y=\ln x\) gives
For \(x\leq 0\), \(f_X(x)=0\). The factor \(1/x\) is the absolute derivative of the inverse transformation; omitting it does not give the density of \(X\). This formula assumes \(\sigma>0\) and the natural-log parameterization above. If a different logarithm base or a different unit is used, convert the model first or derive the density using that transformation consistently.
Worked Example: Deriving and Checking a Density in Thousands of Dollars
Let \(R\) be a hypothetical repair cost measured in thousands of dollars, with \(\ln R\sim N(\ln 2,0.25)\). Find its density and evaluate the density at \(r=2\). Then state the density for the numerical dollar amount \(X=1000R\). No interest or mortality assumption is involved; the model concerns a positive repair cost.
Method. The log-scale parameters are \(\mu=\ln 2\) and \(\sigma^2=0.25\), so \(\sigma=0.5\). Substituting these values in the change-of-variables density gives, for \(r>0\), $$ f_R(r)=\frac{1}{0.5r\sqrt{2\pi}} \exp\left[-\frac{(\ln r-\ln 2)^2}{2(0.25)}\right]. $$
At \(r=2\), the logarithmic difference is \(\ln 2-\ln 2=0\), so the exponential factor is \(e^0=1\). Thus $$ f_R(2)=\frac{1}{(0.5)(2)\sqrt{2\pi}} =\frac{1}{\sqrt{2\pi}} \approx 0.3989. $$ This is a density per unit of \(R\), where one unit represents one thousand dollars; it is not a probability at exactly \(r=2\).
To express the density for \(X=1000R\), use \(r=x/1000\) and the derivative \(dr/dx=1/1000\). Therefore $$ f_X(x)=\frac{1}{1000}f_R(x/1000),\qquad x>0. $$ In particular, at \(x=2000\), \(f_X(2000)=0.3989/1000\approx0.0003989\) per dollar. This is consistent with the density per thousand-dollar unit: the numerical density is divided by 1,000 when the measurement unit is changed from thousands of dollars to dollars.
Check. The model has positive support because \(R=\exp(\ln R)>0\), and the density at the corresponding point \(x=2000\) is scaled by the absolute derivative \(1/1000\). The common mistake is to copy \(f_R(2)\) unchanged as a density per dollar; density values must transform with the unit.
Why the Distribution Is Right-Skewed
A normal variable is symmetric around its mean, but exponentiation does not preserve additive distances. If \(Y=\ln X\), equal changes of \(c\) above and below a log value correspond to multiplying or dividing \(X\) by \(e^c\). For example, a change of \(c\) on the log scale changes the original-scale value by a factor \(e^c\), not by a fixed additive amount.
The shape can be seen directly from the density. As \(x\) approaches zero from above, \(\ln x\) decreases without bound, so the density approaches zero. As \(x\) becomes very large, the normal exponential term decreases and the density again approaches zero. Between these extremes, the distribution has its mass on positive values and a tail extending toward large values. The exponential transformation stretches the upper side in additive units, producing right-skewness.
This is not a claim that every positive, right-skewed data set must be lognormal. The model specifically requires the logarithm to be normally distributed. A positive support and a long upper tail may make the model worth considering, but they do not by themselves establish that the lognormal assumption is suitable.
Common Parameterization Traps
| Given statement | Correct interpretation | Frequent error |
|---|---|---|
| \(\ln X\sim N(\mu,\sigma^2)\) | \(\mu\) and \(\sigma^2\) describe \(\ln X\) | Calling \(\mu\) the mean of \(X\) |
| \(\log_{10}X\sim N(a,b^2)\) | Convert the logarithm base before using a natural-log formula | Using \(a,b^2\) unchanged as natural-log parameters |
| \(\ln(cX)=\ln c+\ln X\) | Multiplying by a unit-conversion factor shifts the log location | Changing units but keeping the location parameter fixed |
| \(X=e^Y\) | \(X>0\) whenever \(Y\) is finite | Assigning positive probability to zero or negative values under the continuous lognormal model |
When a problem gives a normal model for a logarithm, write down the exact variable being logged before identifying the parameters. Check the logarithm base, the measurement unit, whether the second normal parameter is a variance or standard deviation, and whether the original quantity is strictly positive. These checks prevent most parameterization errors.
Check Your Understanding
For each problem, identify the log-scale parameters carefully and track any transformation before interpreting the model.
- A positive quantity \(A\) satisfies \(\ln(A/10)\sim N(0.4,0.16)\). State the lognormal parameters for \(A/10\), its log-scale standard deviation, and the variance parameter for \(A\). (Answer: For \(A/10\), \(\mu=0.4\), \(\sigma^2=0.16\), and \(\sigma=0.4\); for \(A\), \(\mu=\ln 10+0.4\), \(\sigma^2=0.16\).)
- A hypothetical cost \(C\) satisfies \(\log_{10}(C/200)\sim N(0,0.09)\). Convert this to a natural-log model for \(C/200\), giving the variance and standard deviation to four decimal places. (Answer: \(\ln(C/200)\sim N(0,0.4772)\), with standard deviation \(0.6908\).)
- A model gives \(\ln Q\sim N(1.2,0.49)\), where \(Q\) is a numerical value in a specified unit. State the log-scale standard deviation and the support of \(Q\). Does the model permit \(Q=0\)? (Answer: \(\sigma=0.7\); support \(q>0\); it does not permit \(Q=0\).)
- A positive duration \(T\), measured in days, has \(\ln T\sim N(2,0.25)\). A new variable is recorded in hours as \(H=24T\). Give the natural-log distribution of \(H\), using \(\ln 24\approx3.1781\). (Answer: \(\ln H\sim N(5.1781,0.25)\).)
- A variable \(W\) has density \(f_W(w)=\frac{1}{0.6w\sqrt{2\pi}}\exp[-(\ln w-0.3)^2/(2(0.36))]\) for \(w>0\). Identify its lognormal parameters and explain why the density includes a factor \(1/w\). (Answer: \(W\sim\operatorname{Lognormal}(0.3,0.36)\), with \(\sigma=0.6\); the factor \(1/w\) is the absolute derivative of \(d(\ln w)/dw\) in the change of variables.)