Lognormal Severity From Log-Scale Parameters

Lognormal Severity Begins on the Log Scale

Claim severities are positive and can be strongly right-skewed: most losses may be moderate, while a small number are much larger. A lognormal model represents this pattern by assuming that the logarithm of the loss follows a normal distribution. Its parameters therefore describe log loss, not the mean and variance of the loss itself.

Be explicit about the units used inside the logarithm. For example, if $X$ is a loss measured in dollars, the conventional shorthand $\ln X$ means the natural logarithm of its numerical value in dollars. More formally, one can write $\ln(X/\$1)$. Changing the unit changes the log-scale location parameter, so a parameter pair is meaningful only together with its measurement scale.

Definition: A positive loss $X$ is lognormal with log-scale parameters $\mu$ and $\sigma^2$, written $X\sim\operatorname{Lognormal}(\mu,\sigma^2)$, if $\ln X\sim N(\mu,\sigma^2)$, with $X$ measured in the stated unit. Here $\mu=E[\ln X]$ and $\sigma^2=\operatorname{Var}(\ln X)$. The parameter $\sigma$ is the standard deviation of log loss; $\sigma^2$, not $\sigma$, is its variance.

The lognormal distribution has support $x>0$. If $\sigma>0$, 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.

The factor $1/x$ is part of the density because the model is defined through the transformation from $\ln X$ back to $X$. A lognormal model is not a normal model for $X$: a normal loss can be negative, while a lognormal loss is strictly positive.

Converting to the Loss Scale

Three loss-scale summaries are especially useful: the mean, variance, and median. With the stated parameterization, they are

E[X]=e^{\mu+\sigma^2/2},\qquad \operatorname{Var}(X)=e^{2\mu+\sigma^2}\left(e^{\sigma^2}-1\right),\qquad \operatorname{Median}(X)=e^\mu.

These formulas use $\sigma^2$, the variance of log loss. The median is obtained by transforming the median of the normal distribution: because the exponential function is strictly increasing, the value $e^\mu$ has half the lognormal probability below it and half above it. The mean exceeds the median whenever $\sigma>0$, reflecting the right tail.

Useful conversion: If $m=E[X]$ and $v_X=\operatorname{Var}(X)$ are known and $X$ is lognormal, then $$ \sigma^2=\ln\left(1+\frac{v_X}{m^2}\right),\qquad \mu=\ln m-\frac{\sigma^2}{2}. $$ The median can then be recovered as $e^\mu=m e^{-\sigma^2/2}$. These conversions require $m>0$, $v_X\geq 0$, and the assumption that the loss is lognormal.

The inversion follows by first dividing the variance formula by the square of the mean. The resulting ratio is the squared coefficient of variation, and equals $e^{\sigma^2}-1$. Solving for $\sigma^2$, then using the mean formula to solve for $\mu$, gives the expressions above.

Why the Moment Formulas Hold

A short derivation clarifies why the loss-scale mean is not $e^\mu$. Let $Z=\ln X$, so $Z\sim N(\mu,\sigma^2)$ and $X=e^Z$. For any real $t$, completing the square in the normal density gives

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

The integral in the second line is one: its integrand is a normal density with mean $\mu+t\sigma^2$ and variance $\sigma^2$. Setting $t=1$ gives $E[X]=e^{\mu+\sigma^2/2}$, and setting $t=2$ gives $E[X^2]=e^{2\mu+2\sigma^2}$. Therefore

\begin{aligned} \operatorname{Var}(X) &=E[X^2]-\{E[X]\}^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 term $\sigma^2/2$ in the exponent of the mean is consequently essential. It accounts for the effect of the spread of log loss when exponentiating.

Worked Example: Converting Log Parameters to Claim Summaries

A hypothetical insurer models a claim amount $X$, in dollars, by $\ln X\sim N(\ln 1000,\ln 1.25)$. Find the median, mean, and variance of the claim amount. State the units of each result.

Method. Identify $\mu=\ln 1000$ and $\sigma^2=\ln 1.25$. In particular, the normal variance is $\ln 1.25$; the normal standard deviation is $\sqrt{\ln 1.25}$.

Median. The median is

e^\mu=e^{\ln 1000}=\$1{,}000.

Mean. Since $e^{\sigma^2/2}=\sqrt{1.25}$,

E[X]=e^{\ln 1000+(\ln 1.25)/2} =1000\sqrt{1.25} =1000(1.1180339887\ldots) =\$1{,}118.03.

Variance. Here $e^{2\mu}=1{,}000{,}000$, $e^{\sigma^2}=1.25$, and $e^{\sigma^2}-1=0.25$. Thus

\operatorname{Var}(X) =1{,}000{,}000(1.25)(0.25) =312{,}500\text{ dollars}^2.

The median is $\$1{,}000$, the mean is $\$1{,}118.03$, and the variance is $312{,}500\text{ dollars}^2$. The mean exceeds the median, as expected for a non-degenerate lognormal distribution. The calculations are checked by the variance-to-mean-squared ratio: $312{,}500/(1{,}118.0339887\ldots)^2=0.25/1.25=0.20$, which equals $e^{\sigma^2}-1$. The common mistake this method avoids is treating $e^\mu$ as the mean; it is the median.

Worked Example: Recovering Parameters From the Mean and Variance

A hypothetical portfolio has lognormal claim severity measured in dollars, with mean $\$2{,}500$ and variance $3{,}515{,}625\text{ dollars}^2$. Recover the log-scale parameters and the median.

Method. Use the variance-to-mean-squared ratio to find the log variance. First,

\frac{\operatorname{Var}(X)}{\{E[X]\}^2} =\frac{3{,}515{,}625}{2{,}500^2} =\frac{3{,}515{,}625}{6{,}250{,}000} =0.5625.

Therefore,

\sigma^2=\ln(1+0.5625)=\ln(1.5625)=\ln(25/16).

Use the mean conversion to solve for $\mu$:

\begin{aligned} \mu &=\ln(2500)-\frac12\ln(25/16)\\ &=\ln(2500)-\ln(5/4)\\ &=\ln(2000). \end{aligned}

Thus $\ln X\sim N(\ln 2000,\ln(25/16))$, and the median is $e^\mu=\$2{,}000$. As a check, $e^{\sigma^2}-1=25/16-1=9/16=0.5625$, so the fitted variance-to-mean-squared ratio agrees with the supplied data. This method avoids trying to take a logarithm of the variance alone; the ratio $\operatorname{Var}(X)/\{E[X]\}^2$ is the quantity that isolates $\sigma^2$.

Measurement Units and Parameter Changes

Suppose the numerical loss amount is multiplied by a positive constant $c$, as when converting from dollars to cents or from thousands of dollars to dollars. If $W=cX$, then

\ln W=\ln X+\ln c.

Consequently, if $X\sim\operatorname{Lognormal}(\mu,\sigma^2)$, then $W\sim\operatorname{Lognormal}(\mu+\ln c,\sigma^2)$. The log-scale mean shifts by $\ln c$, while the log-scale variance is unchanged. On the loss scale, the mean and median are multiplied by $c$, and the variance is multiplied by $c^2$. A variance always has squared monetary units.

Worked Example: Converting a Model From Thousands to Dollars

A hypothetical claim amount $X$ is recorded in thousands of dollars, with $\ln X\sim N(\ln 2,\ln 1.21)$. Find the median, mean, and variance in dollars. Do not treat the parameters as though $X$ were already recorded in dollars.

Method. First work with the stated unit, thousands of dollars. The log variance is $\ln 1.21$, so $e^{\sigma^2}=1.21$ and $e^{\sigma^2/2}=1.1$.

The median in thousands of dollars is $e^{\ln 2}=2$, so the median is $\$2{,}000$. The mean in thousands of dollars is $2(1.1)=2.2$, so the mean is $\$2{,}200$.

For the variance, first calculate it in squared thousands of dollars:

\operatorname{Var}(X) =e^{2\ln 2+\ln 1.21}(1.21-1) =(4)(1.21)(0.21) =1.0164

in $(\text{thousands of dollars})^2$. Since one squared thousand dollars equals $1{,}000{,}000$ dollars squared, the variance is $1.0164(1{,}000{,}000)=1{,}016{,}400\text{ dollars}^2$. As a check, the mean and variance in thousands give $1.0164/(2.2)^2=0.21/1.21$, the same ratio $e^{\sigma^2}-1$. The common mistake avoided here is multiplying a variance by $1{,}000$ rather than by $1{,}000^2$.

Common Exam Traps

Confusing the log standard deviation and variance. If the model states $\ln X\sim N(\mu,\sigma^2)$, use $\sigma^2$ in the moment formulas. If it gives $\sigma$, square it first.
Calling $e^\mu$ the mean. It is the median. The mean is $e^\mu e^{\sigma^2/2}$, which is larger when $\sigma>0$.
Mixing up units. If the model is for losses in thousands, its direct summaries are also in thousands (and squared thousands for the variance). Convert each summary with the appropriate scale factor.
Using the wrong variance units. If losses are converted by $c$, variance changes by $c^2$, not $c$.

Key takeaway: Read the model as a normal distribution for log loss, identify whether the supplied spread is $\sigma$ or $\sigma^2$, and keep the measurement unit attached to the calculation. The loss-scale median is $e^\mu$, while the mean and variance also depend on $\sigma^2$.

Check Your Understanding

For each hypothetical loss model, identify the log-scale parameters before calculating. Round monetary amounts to the nearest cent where needed.

A loss in dollars satisfies $\ln X\sim N(\ln 500,\ln 1.44)$. Find its median, mean, and variance in dollars and dollars squared. (Answer: median $\$500$; mean $\$600$; variance $158{,}400\text{ dollars}^2$.)
A lognormal loss has mean $\$1{,}800$ and variance $680{,}400\text{ dollars}^2$. Find $\sigma^2$, $\mu$, and the median. (Answer: $\sigma^2=\ln 1.21$, $\mu=\ln(1800/1.1)$, median $\$1{,}636.36$.)
A loss in dollars satisfies $\ln X\sim N(\ln 3000,\ln 1.21)$. Find its mean and variance in dollars and dollars squared. (Answer: mean $\$3{,}300$; variance $2{,}286{,}900\text{ dollars}^2$.)
A loss $X$, recorded in thousands of dollars, satisfies $\ln X\sim N(\ln 1.5,\ln 1.44)$. Convert its median, mean, and variance to dollars and dollars squared. (Answer: median $\$1{,}500$; mean $\$1{,}800$; variance $1{,}425{,}600\text{ dollars}^2$.)
A lognormal loss has median $\$4{,}000$ and mean $\$5{,}000$. Find $\mu$, $\sigma^2$, and the variance of the loss. (Answer: $\mu=\ln 4000$, $\sigma^2=2\ln 1.25=\ln 1.5625$, variance $14{,}062{,}500\text{ dollars}^2$.)

What You'll Learn