Lognormal Severity: Moments and Quantiles

Introduction

Claim severities are often positive and right-skewed: most claims may be moderate, while a small number are much larger. A lognormal model represents this shape by assuming that the natural logarithm of the severity is normally distributed. This lets an actuary use normal probabilities on the log scale and then transform results back to the original claim scale.

This tutorial focuses on three calculations: the mean, variance, and specified quantiles of a lognormal severity. The key is to read the model parameters correctly. In particular, the second parameter of a normal distribution is a variance, not a standard deviation.

Definition: A positive severity $X$ has a lognormal distribution with parameters $\mu$ and $\sigma^2$ if $\ln X\sim N(\mu,\sigma^2)$, where $\sigma>0$. Equivalently, $X=\exp(\mu+\sigma Z)$ for $Z\sim N(0,1)$. The parameter $\mu$ is the mean of $\ln X$, and $\sigma$ is the standard deviation of $\ln X$.

The parameters $\mu$ and $\sigma$ describe logarithms of claim amounts, not claim amounts themselves. If $X$ is measured in dollars, then $\mu$ is the mean of the logarithm of its numerical dollar value. Changing the unit in which $X$ is expressed changes $\mu$, while leaving $\sigma$ unchanged.

Moments of a Lognormal Severity

The lognormal raw moments follow from the normal distribution of $\ln X$. For any real number $r$, the $r$th raw moment is

E[X^r]=\exp\left(r\mu+\frac{r^2\sigma^2}{2}\right).

In particular, setting $r=1$ gives the mean. Setting $r=2$ gives the second raw moment; subtracting the square of the mean then gives the variance:

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

Here is a short derivation of the raw-moment result. Since $X=\exp(\mu+\sigma Z)$,

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

The second equality comes from completing the square: $r\sigma z-z^2/2=-(z-r\sigma)^2/2+r^2\sigma^2/2$. The remaining integral is $1$, because its integrand is a normal density shifted by $r\sigma$. For $r=1$, the mean is $\exp(\mu+\sigma^2/2)$. For $r=2$, $E[X^2]=\exp(2\mu+2\sigma^2)$. Therefore,

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

These formulas apply to the stated lognormal model with finite real $\mu$ and $0<\sigma<\infty$. In particular, both the mean and variance are finite under this model.

Worked Example: Calculating the Mean and Variance

A hypothetical portfolio models an individual claim severity $X$, in dollars, by $\ln X\sim N(\ln 1000,0.16)$. Find $E[X]$ and $\operatorname{Var}(X)$. This is a severity calculation only; no interest rate, discounting, or mortality assumption applies.

Step 1: Read the parameters. Here $\mu=\ln 1000$ and $\sigma^2=0.16$. Thus $\sigma=0.4$, although the moment formulas can be evaluated directly using $\sigma^2=0.16$.

Step 2: Calculate the mean.

E[X]=\exp\left(\ln 1000+\frac{0.16}{2}\right) =1000\exp(0.08) \approx \$1{,}083.29.

Step 3: Calculate the variance.

\begin{aligned} \operatorname{Var}(X) &=\exp\left(2\ln 1000+0.16\right)\left(\exp(0.16)-1\right)\\ &=1{,}000{,}000\left(\exp(0.32)-\exp(0.16)\right)\\ &=1{,}000{,}000(1.3771278-1.1735109)\\ &\approx 203{,}616.89\text{ dollars}^2. \end{aligned}

The mean is greater than the median, which is $\exp(\mu)=\$1{,}000$, consistent with the right-skewed lognormal shape. The variance is positive, as it must be. Money amounts are rounded to the nearest cent; the variance is rounded to the nearest cent in squared-dollar units.

Quantiles and Percentiles

For a continuous severity, the $p$th quantile $x_p$, for $0<p<1$, is the value satisfying $P(X\leq x_p)=p$. Write $z_p$ for the $p$th quantile of a standard normal distribution, so that $P(Z\leq z_p)=p$. Since taking a logarithm preserves order for positive values,

P(X\leq x) =\Phi\left(\frac{\ln x-\mu}{\sigma}\right), \qquad x>0,

where $\Phi$ is the standard normal cumulative distribution function. Setting this probability equal to $p$ gives the quantile formula:

Quantile formula: If $\ln X\sim N(\mu,\sigma^2)$, then the $p$th quantile of $X$ is $x_p=\exp(\mu+\sigma z_p)$, where $z_p$ is the $p$th standard normal quantile. The formula requires $\sigma>0$ and $0<p<1$.

For example, $z_{0.50}=0$, so the median is $\exp(\mu)$. For $p>0.50$, $z_p>0$, and the corresponding quantile exceeds the median. Quantiles are calculated on the log scale first and then exponentiated; a log-scale value is not itself a claim amount.

Worked Example: Finding a 90th Percentile

A hypothetical equipment-loss severity has $\ln X\sim N(\ln 1500,0.25)$, with $X$ measured in dollars. Find its 90th percentile using $z_{0.90}=1.28155$. There is no interest or mortality assumption because the model describes one claim severity without discounting.

Step 1: Identify the log-scale standard deviation. The stated normal variance is $0.25$, so $\sigma=\sqrt{0.25}=0.5$.

Step 2: Apply the quantile formula.

\begin{aligned} x_{0.90} &=\exp\left(\ln 1500+0.5(1.28155)\right)\\ &=1500\exp(0.640775)\\ &\approx \$2{,}846.93. \end{aligned}

The answer is a claim amount in dollars, not $0.640775$, which is the increment on the log scale. Rounded to the nearest cent, the 90th percentile is $\$2{,}846.93$.

Recovering Parameters from Percentiles

Percentile information can also be used in reverse. Because the median is $\exp(\mu)$, a known median $m$ gives $\mu=\ln m$. If another percentile $x_p$ is known, then

\sigma=\frac{\ln(x_p/m)}{z_p}.

For an upper percentile $p>0.50$, both $\ln(x_p/m)$ and $z_p$ are positive. This is a useful sign check. Once $\mu$ and $\sigma$ are recovered, the moment and quantile formulas can be used as usual.

Worked Example: Recovering the Log-Scale Spread

A hypothetical property severity model has a median of $\$1{,}200$ and a 75th percentile of $\$1{,}800$. Using $z_{0.75}=0.67449$, find $\mu$, $\sigma$, and the 90th percentile, using $z_{0.90}=1.28155$. The model is for undiscounted claim severity; interest and mortality are not applicable.

Step 1: Recover the location parameter. The median is $\exp(\mu)$, so $\mu=\ln(1200)$.

Step 2: Recover the log-scale standard deviation.

\begin{aligned} \sigma &=\frac{\ln(1800)-\ln(1200)}{0.67449}\\ &=\frac{\ln(1.5)}{0.67449}\\ &=\frac{0.4054651}{0.67449}\\ &\approx 0.6011. \end{aligned}

Step 3: Find the 90th percentile. Using the unrounded ratio for the calculation,

\begin{aligned} x_{0.90} &=1200\exp\left(\frac{\ln(1.5)}{0.67449}(1.28155)\right)\\ &\approx \$2{,}592.74. \end{aligned}

The recovered $\sigma$ is positive, and the 90th percentile exceeds the given 75th percentile, as expected. The reported estimate of $\sigma$ is rounded to four decimal places; the percentile is rounded to the nearest cent.

Common Exam Traps and Checks

A frequent error is to use the variance of $\ln X$ as though it were its standard deviation. In the 90th-percentile example above, an incorrect substitution would produce

1500\exp(0.25\times1.28155) \approx \$2{,}066.49,

because it uses $0.25$ in place of $\sigma=0.5$. The correct exponent is $0.5\times1.28155=0.640775$, giving $\$2{,}846.93$. Always take the square root of the stated normal variance before using a formula containing $\sigma$.

Other useful checks include:

The median is $\exp(\mu)$, whereas the mean is $\exp(\mu+\sigma^2/2)$; for $\sigma>0$, the mean is greater than the median.
The $p$th percentile for $p>0.50$ must exceed the median. A result below the median usually signals a sign or percentile error.
Exponentiation returns a positive severity. If an answer is nonpositive, a transformation or parameter has been mishandled.
Use the same monetary units throughout a calculation. A parameter $\mu=\ln(1500)$ corresponds to a median of 1,500 in the units used to define $X$.
Keep enough precision in recovered parameters until the final step, especially when those parameters are used again to calculate a quantile.

These checks do not replace the formulas, but they can reveal a parameter-interpretation error before it affects several subsequent calculations.

Check Your Understanding

Use the stated normal quantiles where needed. Round money to the nearest cent and report $\sigma$ to four decimal places.

A hypothetical severity has $\ln X\sim N(\ln 800,0.09)$, with $X$ in dollars. Find its mean and variance. (Answer: $E[X]\approx \$836.82$; $\operatorname{Var}(X)\approx 65{,}947.57\text{ dollars}^2$.)
A hypothetical severity has $\ln X\sim N(\ln 2500,0.04)$. Using $z_{0.95}=1.64485$, find the 95th percentile. (Answer: $2500\exp(0.2(1.64485))\approx \$3{,}473.84$.)
A lognormal severity has median $\$900$ and 75th percentile $\$1{,}350$. Using $z_{0.75}=0.67449$, find $\mu$ and $\sigma$. (Answer: $\mu=\ln(900)$; $\sigma=\ln(1.5)/0.67449\approx 0.6011$.)
A severity has $\ln X\sim N(\ln 5000,0.09)$. Find its mean and 80th percentile, using $z_{0.80}=0.84162$. (Answer: $E[X]=5000e^{0.045}\approx \$5{,}230.14$; $x_{0.80}=5000e^{0.3(0.84162)}\approx \$6{,}436.15$.)
A lognormal severity has median $\$1{,}000$ and mean $\$1{,}250$. Recover $\sigma$, then find its 90th percentile using $z_{0.90}=1.28155$. (Answer: $\sigma=\sqrt{2\ln(1.25)}\approx 0.6680$; $x_{0.90}=1000\exp(\sqrt{2\ln(1.25)}(1.28155))\approx \$2{,}354.05$.)

What You'll Learn