Lognormal Severity Models

Why Use a Lognormal Severity Model?

A loss amount is nonnegative, often strongly right-skewed, and may vary over several orders of magnitude. A lognormal model represents such a loss by assuming that its logarithm follows a normal distribution. This makes normal probability tables useful for calculating probabilities and quantiles, while the model also has explicit formulas for its moments.

The distinction between the two scales is essential. The normal parameters describe the logarithm of the loss; they are not the mean and standard deviation of the loss measured in dollars. In particular, the exponentiated normal mean is the median of the loss, not generally its arithmetic mean.

Definition: A positive random variable $X$ has a lognormal distribution if $Y=\ln X$ is normally distributed. Write $\ln X\sim N(\mu,\sigma^2)$, where $\mu$ is the mean of $\ln X$, $\sigma^2$ is the variance of $\ln X$, and $\sigma>0$. Equivalently, $X=e^{\mu+\sigma Z}$, where $Z\sim N(0,1)$. The parameters $\mu$ and $\sigma$ are on the log scale.

For a severity measured in dollars, the logarithm is understood to be taken relative to one dollar. More generally, a logarithm should be applied to a dimensionless ratio such as $X/(1\text{ dollar})$. Changing the unit of measurement changes the log-scale location parameter, but does not change the underlying loss when the conversion is applied consistently.

The parameter $\sigma$ is the standard deviation of the logarithm, not a dollar standard deviation. Since $Z$ has median zero, the median loss is $e^\mu$. The loss itself is always positive under this model.

Probabilities and Quantiles

Let $\Phi(z)=P(Z\le z)$ denote the standard normal cumulative distribution function. For any $x>0$, taking logarithms preserves the inequality because the logarithm is increasing. Therefore, a loss probability can be calculated by converting the loss threshold to the log scale and standardizing.

F_X(x)=P(X\le x)= \begin{cases} 0, & x\le 0,\\ \Phi\!\left(\dfrac{\ln x-\mu}{\sigma}\right), & x>0. \end{cases}

The complementary probability is $P(X>x)=1-\Phi((\ln x-\mu)/\sigma)$, for $x>0$. For an interval with $0<a<b$, subtract the two cumulative probabilities:

P(a<X\le b) = \Phi\!\left(\frac{\ln b-\mu}{\sigma}\right) - \Phi\!\left(\frac{\ln a-\mu}{\sigma}\right).

Because the lognormal distribution is continuous, including or excluding either endpoint does not change this probability. The $p$th quantile, for $0<p<1$, is obtained by solving $F_X(x)=p$. If $z_p$ is the $p$th standard normal quantile, then

x_p=\exp(\mu+\sigma z_p).

This formula also shows why $e^\mu$ is the median: $z_{0.50}=0$, so $x_{0.50}=e^\mu$. Quantiles on the loss scale are positive, even when the corresponding normal quantile is negative.

Worked Example: Probability Between Two Loss Thresholds

A hypothetical insurer models a claim severity $X$, in dollars, using $\ln X\sim N(\ln 500,0.2^2)$. Find the probability that a claim exceeds $500 but is no greater than $500e^{0.2}$ dollars. Report the probability to four decimal places.

Method. Standardize the logarithms of both thresholds. The model has log-scale mean $\mu=\ln 500$ and log-scale standard deviation $\sigma=0.2$. The upper threshold is $500e^{0.2}$, approximately $610.70.

At the lower threshold, the standardized value is \[ \frac{\ln 500-\ln 500}{0.2}=0. \] At the upper threshold, it is \[ \frac{\ln(500e^{0.2})-\ln 500}{0.2} = \frac{\ln 500+0.2-\ln 500}{0.2} =1. \]

Thus \[ P(500<X\le 500e^{0.2}) =\Phi(1)-\Phi(0) =0.8413-0.5000 =0.3413. \] The probability is 0.3413. The method avoids treating the loss threshold itself as a normal value: it is the logarithm of the threshold that is standardized.

Check. The interval runs from the median to one log-scale standard deviation above the median. Its probability is therefore the standard normal probability between 0 and 1, consistent with the calculation.

Worked Example: Finding a High Loss Quantile

A hypothetical equipment portfolio has loss severity $X$ with $\ln X\sim N(\ln 1000,0.1^2)$, where $X$ is measured in dollars. Estimate the 90th percentile using $z_{0.90}=1.2816$, and report the amount to the nearest cent.

Method. Use the lognormal quantile formula. The log-scale parameters are $\mu=\ln 1000$ and $\sigma=0.1$; the 90th percentile is not found by adding a normal quantile directly to $1,000.

First calculate the log-scale quantile: \[ \ln x_{0.90} =\ln 1000+(0.1)(1.2816) =\ln 1000+0.12816. \] Exponentiating gives \[ x_{0.90}=1000e^{0.12816} =1000(1.136734865\ldots) =1136.734865\ldots\text{ dollars}. \]

Rounded only at the final step, the 90th percentile is $1,136.73 to the nearest cent. As a check, taking the logarithm of the unrounded result and standardizing gives $1.2816$, which corresponds to approximately the 90th percentile.

Moments of a Lognormal Loss

The moments follow from the moment-generating function of the normal variable $Y=\ln X$. This is not an assertion that the moment-generating function of $X$ exists: the calculation instead evaluates $E[e^{kY}]$, which gives the $k$th raw moment of $X$.

Theorem: If $\ln X\sim N(\mu,\sigma^2)$, then for any real $k$, \[ E[X^k]=\exp\!\left(k\mu+\frac{k^2\sigma^2}{2}\right). \] In particular, \[ E[X]=e^{\mu+\sigma^2/2}, \qquad E[X^2]=e^{2\mu+2\sigma^2}, \] and \[ \operatorname{Var}(X) =e^{2\mu+\sigma^2}\bigl(e^{\sigma^2}-1\bigr). \]

To justify the raw-moment formula, write $X^k=e^{kY}$. For $Y\sim N(\mu,\sigma^2)$, the normal exponential moment is $E[e^{tY}]=e^{t\mu+t^2\sigma^2/2}$. Substituting $t=k$ gives the stated result. Taking $k=1$ gives the mean, and taking $k=2$ gives the second raw moment. Subtracting the square of the mean yields

\begin{aligned} \operatorname{Var}(X) &=e^{2\mu+2\sigma^2}-\left(e^{\mu+\sigma^2/2}\right)^2\\ &=e^{2\mu+2\sigma^2}-e^{2\mu+\sigma^2}\\ &=e^{2\mu+\sigma^2}\left(e^{\sigma^2}-1\right). \end{aligned}

The variance is in squared loss units. If the loss is measured in dollars, its variance is in dollars squared; it is not itself a dollar amount. The mean is larger than the median whenever $\sigma>0$, reflecting the model’s right skew.

Worked Example: Mean and Variance Without Intermediate Rounding

For a hypothetical claim portfolio, suppose $\ln X\sim N(\ln 100,0.5^2)$, with $X$ measured in dollars. Calculate the mean and variance of the loss, reporting the mean to the nearest cent and the variance to the nearest cent squared.

Method. Apply the first and second raw-moment formulas, then use the factored variance formula. Here $\mu=\ln 100$, $\sigma=0.5$, $\sigma^2=0.25$, and $2\mu=\ln(10000)$.

The mean is \[ E[X] =e^{\ln 100+0.25/2} =100e^{0.125} =100(1.133148453\ldots) =113.3148453\ldots\text{ dollars}. \]

The second raw moment is \[ E[X^2] =e^{2\ln 100+2(0.25)} =10000e^{0.5} =16487.2127\ldots\text{ dollars}^2. \] For the variance, use the unrounded model parameters directly: \[ \begin{aligned} \operatorname{Var}(X) &=e^{2\ln 100+0.25}(e^{0.25}-1)\\ &=10000e^{0.25}(e^{0.25}-1)\\ &=10000(e^{0.5}-e^{0.25})\\ &=3646.9585\ldots\text{ dollars}^2. \end{aligned} \]

The mean is $113.31, and the variance is 3,646.96 dollars squared, with each rounded only at the final step. As a check, $E[X^2]$ is greater than $E[X]^2$, and their unrounded difference agrees with the variance formula.

Common Exam Traps

A common mistake is to report $e^\mu$ as the mean. For the model in the preceding example, that incorrect calculation would give $e^{\ln 100}=\$100$ as the mean. It is the median; the correct mean is $e^{\mu+\sigma^2/2}=\$113.31$. The extra factor $e^{\sigma^2/2}$ matters whenever $\sigma>0$.

Another error is to standardize a dollar threshold as though $X$ itself were normal. For example, using $(x-\mu)/\sigma$ is invalid: $\mu$ and $\sigma$ describe $\ln X$, not $X$. The correct standardized value is $(\ln x-\mu)/\sigma$, for a positive threshold $x$.

Keep the scales distinct. Exponentiate $\mu+\sigma z_p$ to obtain a loss quantile in the original units.
Use the right tail carefully. For an exceedance probability, calculate $1-\Phi((\ln x-\mu)/\sigma)$, not $\Phi((\ln x-\mu)/\sigma)$.
Do not round early. In particular, calculate variance from the unrounded parameters or moments; subtracting rounded values can produce a mismatched result.

The lognormal model is useful when normal calculations are convenient but negative loss values would be inappropriate. Correctly identifying the log-scale parameters lets an analyst use standard normal tables for loss probabilities and quantiles and derive moments without confusing dollars with logarithms. These techniques are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

Use the stated lognormal parameters, keep intermediate calculations unrounded, and report probabilities to four decimals and dollar amounts to the nearest cent where requested.

A hypothetical claim has $\ln X\sim N(\ln 250,0.3^2)$. Find $P(250<X\le 250e^{0.3})$. (Answer: 0.3413)
A hypothetical loss has $\ln X\sim N(\ln 400,0.2^2)$. Find the probability that it exceeds $400e^{0.4}$. (Answer: 0.0228)
A severity has median $600 and log-scale standard deviation $0.2$. Using $z_{0.8413}=1.0000$, find its 84.13th percentile. (Answer: $732.84)
A hypothetical loss has $\ln X\sim N(\ln 50,0.1^2)$, with $X$ measured in dollars. Find its mean and variance. (Answer: mean $50.25; variance 25.38 dollars squared)

What You'll Learn