Lognormal Sampling From Normal Variates

Why the Lognormal Model Uses a Normal Draw

A loss amount must be positive, and its distribution may be strongly right-skewed: many losses are moderate, while a smaller number are much larger. A lognormal model represents this by assuming that the natural logarithm of the loss follows a normal distribution. To simulate a loss, draw a normal variate, use it to obtain a value on the log scale, and exponentiate.

The normal distribution’s parameters describe the logarithm of the loss, not the loss itself. This distinction matters in simulation: exponentiating the normal mean gives the median loss, not the arithmetic mean loss. The generated observations are positive, but their arithmetic mean can be substantially larger than their median.

Definition: A positive random variable $X$ has a lognormal distribution with log-scale parameters $\mu$ and $\sigma^2$ if $\ln X\sim N(\mu,\sigma^2)$, where $\sigma>0$. Equivalently, $X=\exp(\mu+\sigma Z)$, where $Z\sim N(0,1)$. Here $\mu$ is the mean of $\ln X$, and $\sigma$ is the standard deviation of $\ln X$; neither is generally the mean or standard deviation of $X$.

The logarithm requires a positive quantity. In a loss model, $X$ is therefore a positive loss amount, not a signed net loss that could be zero or negative. The parameter values also depend on the unit used for the amount: if $X$ is recorded in dollars, $\ln X$ means the logarithm of its numerical dollar amount. State the unit consistently when reporting simulated losses.

Generating a Lognormal Loss

Suppose a standard normal variate $Z$ is available. If the desired loss model is $\ln X\sim N(\mu,\sigma^2)$, form $\mu+\sigma Z$ and exponentiate. If instead a normal variate $W\sim N(a,b^2)$ is supplied, first standardize it as $Z=(W-a)/b$, then apply the same construction. This assumes $b>0$.

X=\exp(\mu+\sigma Z),\qquad Z\sim N(0,1).

For repeated simulations, use a separate standard normal draw for each simulated loss. If the draws are independent, the resulting losses are independent as well, because each loss is a function of its corresponding draw. The method generates the specified lognormal model; it does not establish that the model is appropriate for any particular portfolio.

Specify the log-scale parameters.
Record $\mu$ and $\sigma$, or convert the given loss-scale information to these parameters.

Obtain a standard normal draw.
Use the supplied $Z\sim N(0,1)$. The method for generating normal variates is a separate question.

Form the log loss.
Calculate $Y=\mu+\sigma Z$. This is the simulated value of $\ln X$, not yet the loss amount.

Exponentiate and report the amount.
Calculate $X=e^Y$, then report it in the stated monetary unit.

Interpreting Lognormal Parameters

The principal loss-scale summaries follow from the normal distribution of $\ln X$. For $\ln X\sim N(\mu,\sigma^2)$, the median is $e^\mu$, because the median of the normal variable $\ln X$ is $\mu$ and exponentiation preserves order. The mean is $e^{\mu+\sigma^2/2}$, which exceeds the median whenever $\sigma>0$. The mode is $e^{\mu-\sigma^2}$.

Theorem: If $\ln X\sim N(\mu,\sigma^2)$, then

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

The variance is $\operatorname{Var}(X)=(e^{\sigma^2}-1)e^{2\mu+\sigma^2}$. These are summaries of $X$, not of $\ln X$.

A short derivation clarifies why the mean contains the extra factor $e^{\sigma^2/2}$. The moment-generating function of a normal variable $Y\sim N(\mu,\sigma^2)$ is $E[e^{tY}]=e^{t\mu+t^2\sigma^2/2}$. Since $X=e^Y$, setting $t=1$ gives the mean of $X$. Setting $t=2$ gives its second moment, from which the variance follows.

E[X]=E[e^Y]=e^{\mu+\sigma^2/2}, \qquad E[X^2]=E[e^{2Y}]=e^{2\mu+2\sigma^2},

\operatorname{Var}(X)=E[X^2]-E[X]^2 =e^{2\mu+2\sigma^2}-e^{2\mu+\sigma^2} =(e^{\sigma^2}-1)e^{2\mu+\sigma^2}.

The log-scale parameters can also be recovered from a positive arithmetic mean $m$ and standard deviation $s$. This conversion assumes that the loss is lognormal and that $m$ and $s$ describe its arithmetic mean and standard deviation.

\sigma^2=\ln\left(1+\frac{s^2}{m^2}\right), \qquad \mu=\ln(m)-\frac{\sigma^2}{2}.

These equations come from dividing the variance formula by the square of the mean, which gives $\operatorname{Var}(X)/E[X]^2=e^{\sigma^2}-1$. Solving for $\sigma^2$, then using $m=e^{\mu+\sigma^2/2}$, gives the stated conversion. If supplied values are not positive, or if the variance and mean cannot satisfy these equations, they do not define a nondegenerate lognormal model.

Worked Simulations and Parameter Conversions

Worked Example: Simulating from an Arithmetic Mean and Standard Deviation

A hypothetical property-loss model is lognormal. Its arithmetic mean is $\$2{,}000$, its arithmetic standard deviation is $\$1{,}000$, and a standard normal draw is $Z=\sqrt{\ln 2}$. Find the log-scale parameters and the simulated loss. Report money to the cent.

Method. Convert the arithmetic mean and standard deviation to $\mu$ and $\sigma$, then apply the normal-draw construction. The coefficient of variation is $s/m=1000/2000=0.5$, so

\sigma^2=\ln(1+0.5^2)=\ln(1.25)\approx0.2231435513, \qquad \sigma=\sqrt{\ln(1.25)}\approx0.4723807.

Next,

\mu=\ln(2000)-\frac{\ln(1.25)}{2} \approx7.489330684.

The simulated log loss can be written relative to the arithmetic mean:

\mu+\sigma Z =\ln(2000)-\frac{\ln(1.25)}{2} +\sqrt{\ln(1.25)}\sqrt{\ln(2)} \approx\ln(2000)+0.28171098.

Exponentiating gives $X=2000e^{0.28171098}\approx2650.79$. Therefore, the simulated loss is $\$2{,}650.79$, rounded to the nearest cent. This value is one simulated observation; it need not equal the model’s arithmetic mean.

Check. The input mean and standard deviation imply a positive log-scale variance, and exponentiation produces a positive amount. The draw $Z>0$ is above the normal center, so a loss above the median is reasonable.

Common mistake avoided: using $2000+\sigma Z$ as the loss. That adds a log-scale standard deviation to a dollar amount. The correct operation is to form $\mu+\sigma Z$ on the log scale and then exponentiate.

Worked Example: Separating the Median, Mean, and Simulated Loss

A hypothetical specialty-claims model has $\ln X\sim N(\ln 800,0.6^2)$, with $X$ measured in dollars. For the supplied draw $Z=-1.25$, calculate the model median, model arithmetic mean, and simulated loss.

Median. The log-scale location is $\mu=\ln 800$, so

\operatorname{Median}(X)=e^\mu=e^{\ln 800}=\$800.00.

Arithmetic mean. Here $\sigma^2=0.36$, so

E[X]=e^{\ln 800+0.36/2} =800e^{0.18} \approx\$957.77.

Simulated loss. The simulated log loss is $\ln 800+0.6(-1.25)=\ln 800-0.75$. Thus

X=e^{\ln 800-0.75}=800e^{-0.75} \approx\$377.89.

The draw is below zero, so the generated loss is below the median, as expected. The model mean is above the median because the right tail raises the arithmetic average.

Common mistake avoided: calling $\$800$ the arithmetic mean because it equals $e^\mu$. It is the median; the arithmetic mean is $\$957.77$.

Worked Example: Simulating from a Median-Based Specification

A hypothetical equipment-loss model specifies a median loss of $\$1{,}000$ and log-scale standard deviation $\sigma=0.5$. A standard normal draw is $Z=1$. Find the log-scale location, arithmetic mean, and simulated loss.

Find the location. Since the median is $e^\mu$, $\mu=\ln(1000)$.

Find the arithmetic mean. The log-scale variance is $0.5^2=0.25$. Therefore,

E[X]=e^{\ln(1000)+0.25/2} =1000e^{0.125} \approx\$1{,}133.15.

Generate the loss. The log loss for this draw is $\ln(1000)+0.5(1)=\ln(1000)+0.5$. Hence

X=e^{\ln(1000)+0.5}=1000e^{0.5} \approx\$1{,}648.72.

The simulated loss exceeds the median because $Z=1$ is above the normal center. The mean, median, and simulated value are different quantities: they are approximately $\$1{,}133.15$, $\$1{,}000.00$, and $\$1{,}648.72$, respectively.

Common mistake avoided: substituting the arithmetic mean for $e^\mu$ when setting the log-scale location. The specified median directly gives $\mu=\ln(1000)$; the arithmetic mean is a separate summary.

Common Exam Traps and Model Checks

The sampling calculation is short, but several interpretation errors can change the answer substantially. Check the scale and role of every parameter before exponentiating.

Confusing the normal mean with the loss mean. $\mu=E[\ln X]$, whereas $E[X]=e^{\mu+\sigma^2/2}$. In particular, $e^\mu$ is the median, not the arithmetic mean.
Forgetting to multiply the draw by $\sigma$. The log loss is $\mu+\sigma Z$, not $\mu+Z$, unless $\sigma=1$.
Exponentiating too early or too late. Calculate the full log loss first, then exponentiate once. The draw is not itself a loss amount.
Reporting a nonpositive loss. A lognormal draw is positive. A negative output usually signals that the log-scale value was mistakenly reported as the loss.
Mixing units. If the model is parameterized for numerical amounts in thousands of dollars, convert the resulting simulated amount back to dollars before reporting it in dollars.

A useful check is to compare a draw with zero. When $Z=0$, the simulated loss is $e^\mu$, the median. A positive draw produces a value above that median, and a negative draw produces one below it. This check concerns the median, not whether a draw lies above or below the arithmetic mean.

Lognormal sampling is useful when a simulation needs positive, right-skewed severities and the model is specified on the log scale. It is especially important to keep the parameterization explicit when a problem supplies loss-scale summaries but requests a simulated observation. The lognormal assumption determines the conversion; it should not be inferred merely because losses are positive.

Check Your Understanding

Use the lognormal sampling and parameter relationships above. Unless stated otherwise, report monetary amounts to the nearest cent and probabilities to four decimal places.

A hypothetical repair-loss model has $\ln X\sim N(\ln 500,0.3^2)$. Find its median and arithmetic mean. (Answer: median $\$500.00$; mean $500e^{0.045}\approx\$523.01$.)
A standard normal draw $Z=-1.5$ is used for a loss with $\ln X\sim N(\ln 2500,0.2^2)$. Calculate the simulated loss. (Answer: $2500e^{-0.3}\approx\$1{,}852.05$.)
A hypothetical lognormal loss has arithmetic mean $\$4{,}000$ and standard deviation $\$2{,}000$. Find $\sigma^2$ and the median. (Answer: $\sigma^2=\ln(1.25)\approx0.2231$; median $4000e^{-\ln(1.25)/2}\approx\$3{,}577.71$.)
For a loss with $\ln X\sim N(\ln 900,0.4^2)$, find the probability that $X$ exceeds $\$1{,}350$. (Answer: $P(Z>\ln(1.5)/0.4)\approx0.1554$.)
A hypothetical portfolio model has arithmetic mean $\$10{,}000$ and median $\$8{,}000$. Find its log-scale variance and standard deviation. (Answer: $\sigma^2=2\ln(10{,}000/8{,}000)=2\ln(1.25)\approx0.4463$, so $\sigma\approx0.6680$.)

What You'll Learn