Lognormal Quantiles And Multiplicative Spread

Why Lognormal Quantiles Are Useful

A severity model is often used not only to describe an average claim, but also to answer questions such as “What claim amount is exceeded by 10% of losses?” For a lognormal model, those questions are answered by working on the logarithmic scale and then transforming back to dollars. The same approach gives a useful interpretation of the distribution’s median and multiplicative spread.

A lognormal distribution is appropriate for a positive-valued quantity whose logarithm is modeled as normal. Its two parameters describe the logarithm of severity, not the dollar amount directly. Keeping that distinction clear is essential when calculating percentiles or explaining how variable claims are.

Definition: A positive loss $X$ is lognormal with log-scale parameters $\mu$ and $\sigma>0$ if $\ln X\sim N(\mu,\sigma^2)$. Write $\Phi$ for the standard normal cumulative distribution function and $z_p=\Phi^{-1}(p)$ for its $p$th quantile. The $p$th quantile of $X$ is $Q_p=\inf\{x:F_X(x)\ge p\}$.

Quantiles From Log-Scale Parameters

Because the logarithm is strictly increasing, the event $X\le x$ is equivalent to $\ln X\le\ln x$ for $x>0$. Standardizing $\ln X$ therefore gives the lognormal cumulative distribution function. Solving that relationship for $x$ gives the quantile formula.

F_X(x)=\Phi\!\left(\frac{\ln x-\mu}{\sigma}\right),\qquad x>0, \qquad Q_p=\exp(\mu+\sigma z_p),\quad 0<p<1.

The formula is for a continuous, nondegenerate lognormal severity. It does not apply directly to a distribution with a point mass at zero, or to a model in which the logarithm of severity is not normal. For the lognormal model, however, the calculation is systematic: find $z_p$, form $\mu+\sigma z_p$, and exponentiate the entire result.

Choose the percentile.
For example, a 90th percentile uses $p=0.90$, so $z_p$ is positive.

Calculate on the log scale.
Evaluate $\mu+\sigma z_p$, using the parameters of $\ln X$.

Transform back to severity.
Exponentiate the result to obtain $Q_p$ in the original units, such as dollars.

Worked Example: Finding a High Claim Percentile

A hypothetical insurer models the amount $X$, in dollars, of a particular type of claim by $\ln X\sim N(\ln 800,(\ln 2)^2)$. Find the 84.13th and 90th percentiles. There is no discounting, so take $i=0$; mortality is not modeled. Use $z_{0.8413}=1$ and $z_{0.90}=1.2816$. Round dollar amounts to the nearest cent.

Method. Apply the lognormal quantile formula to each percentile. The specified distribution has $\mu=\ln800$ and $\sigma=\ln2$.

For the 84.13th percentile, $z_{0.8413}=1$, so

Q_{0.8413} =\exp(\ln800+\ln2) =\exp(\ln1600) =1600.

For the 90th percentile, first calculate the log-scale value:

\ln800+(1.2816)(\ln2) =6.6846+0.8883 =7.5729.

Exponentiating gives $Q_{0.90}=\exp(7.5729)\approx \$1{,}944.80$, using unrounded values in the calculation and rounding only the final dollar amount to cents.

Check. The 90th percentile, about $\$1{,}944.80$, is above the 84.13th percentile of $\$1{,}600$, as it must be. The method also avoids a common error: the expression $7.5729$ is a log claim amount, not a dollar claim amount.

Median and Geometric Standard Deviation

Two summaries make the lognormal model easier to interpret on the original severity scale. Its median is the central claim amount: half the modeled claims are at or below it and half are at or above it. The geometric standard deviation (GSD) describes spread as a multiplicative factor rather than as an additive dollar amount.

Result: If $\ln X\sim N(\mu,\sigma^2)$, then the median of $X$ is $\exp(\mu)$, and its geometric standard deviation is $\exp(\sigma)$. Writing $M=\exp(\mu)$ for the median and $G=\exp(\sigma)$ for the GSD, the quantile formula becomes $Q_p=M G^{z_p}$.

To see why, the median corresponds to $z_{0.50}=0$, so $Q_{0.50}=\exp(\mu)=M$. At one standard deviation above the mean on the log scale, the severity is $\exp(\mu+\sigma)=MG$; one standard deviation below it gives $\exp(\mu-\sigma)=M/G$. Thus $G$ is a factor: moving one standard deviation on the log scale multiplies or divides the median by $G$. It is not the standard deviation of $X$ measured in dollars.

More generally, taking the ratio of two quantiles cancels the median:

\frac{Q_p}{Q_r} =\frac{\exp(\mu+\sigma z_p)}{\exp(\mu+\sigma z_r)} =\exp\!\left(\sigma(z_p-z_r)\right) =G^{\,z_p-z_r}.

This identity explains multiplicative spread. If two lognormal models have the same median but different GSDs, the model with the larger GSD has more widely separated quantiles as ratios. The dollar difference between two percentiles also depends on the median; the ratio isolates the spread factor.

Worked Example: Recovering the Median and GSD From Quantiles

A hypothetical portfolio’s positive claim severity is lognormal. Its 15.87th percentile is $\$600$, and its 84.13th percentile is $\$2{,}400$. Find the median and GSD, then find the 90th percentile. Assume no discounting ($i=0$) and no mortality model. Use $z_{0.1587}=-1$, $z_{0.8413}=1$, and $z_{0.90}=1.2816$. Round dollar amounts to cents and the GSD to four decimals.

Method. At the two specified percentiles, the lognormal quantile equations are $Q_{0.1587}=\exp(\mu-\sigma)$ and $Q_{0.8413}=\exp(\mu+\sigma)$. Multiplying these equations gives the square of the median:

M=\exp(\mu)=\sqrt{600(2400)} =\sqrt{1{,}440{,}000} =1200.

Dividing the upper quantile by the lower one gives $\exp(2\sigma)=2400/600=4$. Hence $\sigma=\tfrac12\ln4=\ln2$, and

G=\exp(\sigma)=2.0000.

For the 90th percentile, use the median-GSD form:

Q_{0.90} =1200(2)^{1.2816} \approx 1200(2.4310) =\$2{,}917.20.

Check. The inferred median lies between $\$600$ and $\$2{,}400$, and the GSD of 2 means those supplied quantiles are $M/2$ and $2M$, respectively. This confirms both recovered parameters.

Exceedance Probabilities and Multiplicative Comparisons

Quantiles can also be used in reverse to find the probability of exceeding a threshold $c>0$. Standardize $\ln c$, then use the upper tail of the standard normal distribution. In terms of the median and GSD, this standardization is especially interpretable: it measures how many log-scale standard deviations the threshold lies above or below the median.

P(X>c) =1-\Phi\!\left(\frac{\ln c-\mu}{\sigma}\right) =1-\Phi\!\left(\frac{\ln(c/M)}{\ln G}\right).

The ratio $c/M$ matters here, not just the dollar difference $c-M$. For example, doubling the median gives the same standardized threshold for any model with the same GSD, even if the models’ medians are different. This is one reason the GSD is useful when claim amounts vary across substantially different scales.

Worked Example: Probability of Exceeding a Multiple of the Median

For a hypothetical severity model, $\ln X\sim N(\ln2500,(\ln2)^2)$, where $X$ is measured in dollars. Find the probability that a claim exceeds $\$10{,}000$. Assume $i=0$, with no discounting, and no mortality model. Use $\Phi(2)=0.9772$. Give the probability to four decimals.

Method. The median is $M=\exp(\ln2500)=\$2{,}500$, and the GSD is $G=\exp(\ln2)=2$. The threshold is four times the median: $10{,}000/2{,}500=4$. Its standardized log value is

z=\frac{\ln(10{,}000/2500)}{\ln2} =\frac{\ln4}{\ln2} =2.

Therefore,

P(X>10{,}000)=1-\Phi(2)=1-0.9772=0.0228.

Check. The result is a small upper-tail probability because the threshold is two log-scale standard deviations above the median. The calculation uses the ratio to the median and the GSD, rather than treating the dollar threshold as a normal value.

Common Mistakes and Exam Checks

The most important check is whether the calculation is on the correct scale. In the model $\ln X\sim N(\ln800,(\ln2)^2)$, a common incorrect calculation for the 90th percentile is to treat $\mu+\sigma z_{0.90}=7.5729$ as the answer in dollars. That value is the logarithm of the percentile. The correction is to exponentiate it: $Q_{0.90}=\exp(7.5729)\approx\$1{,}944.80$.

A related mistake is to call $\sigma$ the GSD. In this model $\sigma=\ln2$, whereas the GSD is $\exp(\sigma)=2$. The GSD is dimensionless and expresses a factor of spread; it is not a dollar-scale standard deviation. When comparing percentiles, check that a higher percentile produces a higher severity, and confirm that the final answer is expressed on the original scale.

These calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. In exam-style work, state the parameterization explicitly, identify the percentile or threshold, and keep the logarithm and exponential steps visible. Those checks help prevent errors in both interpretation and arithmetic.

Check Your Understanding

For each hypothetical severity model, use the stated lognormal assumptions. Unless specified otherwise, assume $i=0$, no discounting, and no mortality model. Round dollar amounts to cents and probabilities to four decimals.

A claim severity satisfies $\ln X\sim N(\ln900,(\ln1.25)^2)$. Find its median and its 84.13th percentile, using $z_{0.8413}=1$. (Answer: median $\$900$; 84.13th percentile $\$1{,}125$.)
A severity has $\ln X\sim N(\ln400,0.25)$, where $0.25$ is the variance of $\ln X$. Find the 15.87th percentile, using $z_{0.1587}=-1$. (Answer: $\$242.61$.)
A lognormal severity has 15.87th percentile $\$300$ and 84.13th percentile $\$1{,}200$. Find the median and GSD. (Answer: median $\$600$; GSD $2.0000$.)
A lognormal severity has median $\$2{,}000$ and GSD $1.5$. Find $P(X>\$3{,}000)$, using $\Phi(1)=0.8413$. (Answer: $0.1587$.)
A lognormal severity has median $\$1{,}000$ and GSD $2$. Find the ratio $Q_{0.90}/Q_{0.10}$, using $z_{0.90}=1.2816$ and $z_{0.10}=-1.2816$. (Answer: approximately $5.9098$.)

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