Method-of-Moments Estimation for Lognormal Losses

Introduction

A lognormal model is often used for positive loss amounts when the logarithms of losses are modeled as normally distributed. Its parameters describe the distribution of log losses, while actuarial quantities such as the expected loss and variance are on the original dollar scale. Method of moments provides a direct way to move between these scales using the first two raw moments.

The central calculation is short, but it is easy to mix up a raw second moment with a sample variance, or to report a log-scale parameter as though it were an arithmetic-scale quantity. The method below keeps those steps separate and checks that the fitted moments match the sample moments used to estimate them.

Definition: A positive random variable $X$ has a lognormal distribution with location parameter $\mu$ and scale parameter $\sigma>0$ if $\ln X\sim N(\mu,\sigma^2)$. Thus $\mu=E[\ln X]$ and $\sigma^2=\operatorname{Var}(\ln X)$. The parameter $\sigma^2$ is the variance on the logarithmic scale; it is not the variance of $X$.

Population Moments and Method-of-Moments Estimates

Assume $X_1,\ldots,X_n$ are independent, identically distributed, complete observations from one lognormal distribution, with all losses strictly positive. The first two population raw moments are

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

For a sample, define the first and second raw moments using divisor $n$:

m_1=\frac{1}{n}\sum_{j=1}^{n}X_j, \qquad m_2=\frac{1}{n}\sum_{j=1}^{n}X_j^2.

The method of moments sets $m_1$ equal to the model's first raw moment and $m_2$ equal to its second raw moment. These are the raw sample moments, not the sample mean and unbiased sample variance as a pair. Solving the resulting equations gives the estimates below.

Method-of-moments result: When $m_1>0$ and $m_2/m_1^2>1$, the lognormal method-of-moments estimates are $$ \widehat{\sigma}_{\mathrm{MM}}^{\,2} =\ln\left(\frac{m_2}{m_1^2}\right), \qquad \widehat{\mu}_{\mathrm{MM}} =\ln(m_1)-\frac{1}{2}\widehat{\sigma}_{\mathrm{MM}}^{\,2}. $$ The fitted arithmetic mean and variance are $$ \widehat{E[X]}_{\mathrm{MM}}=m_1, \qquad \widehat{\operatorname{Var}(X)}_{\mathrm{MM}} =m_1^2\left(e^{\widehat{\sigma}_{\mathrm{MM}}^{\,2}}-1\right) =m_2-m_1^2. $$

For positive observations, $m_2\geq m_1^2$, so the estimated log-scale variance cannot be negative. If all observations are identical, equality holds and the estimate is zero, which is a boundary case rather than a nondegenerate lognormal fit with $\sigma>0$. The fitted arithmetic variance is expressed in the square of the loss measurement unit.

Why the Moment Equations Give These Estimates

Dividing the second population raw moment by the square of the first eliminates the location parameter. This isolates the scale parameter; once that scale is found, the first moment determines the location.

\frac{E[X^2]}{(E[X])^2} = \frac{e^{2\mu+2\sigma^2}}{e^{2\mu+\sigma^2}} =e^{\sigma^2}.

Matching the ratio of sample raw moments to this population ratio gives $m_2/m_1^2=e^{\widehat{\sigma}_{\mathrm{MM}}^{\,2}}$. Taking natural logarithms yields the scale estimate. Substituting it into the first-moment equation and taking logarithms gives $\widehat{\mu}_{\mathrm{MM}}=\ln(m_1)-\widehat{\sigma}_{\mathrm{MM}}^{\,2}/2$. Finally, for a lognormal variable, $\operatorname{Var}(X)=E[X^2]-(E[X])^2$; matching both raw moments therefore gives the fitted arithmetic variance $m_2-m_1^2$.

Worked Applications

Worked Example: Fitting From Four Complete Losses

A hypothetical portfolio has four complete positive losses, recorded in thousands of dollars: $4,4,8,8$. Assume the losses are an independent sample from a lognormal distribution. Estimate $\mu$, $\sigma^2$, the arithmetic mean, and the arithmetic variance. The method is raw-moment matching; the main trap is using the unbiased sample variance instead of the second raw moment.

Step 1: Calculate the first raw moment. The sample total is $4+4+8+8=24$, so $m_1=24/4=6$ thousand dollars.

Step 2: Calculate the second raw moment. The sum of squared losses is $4^2+4^2+8^2+8^2=16+16+64+64=160$. Thus $m_2=160/4=40$, in squared thousands of dollars.

Step 3: Estimate the log-scale variance.

\widehat{\sigma}_{\mathrm{MM}}^{\,2} =\ln\left(\frac{40}{6^2}\right) =\ln\left(\frac{10}{9}\right) =0.1053605\ldots \approx 0.1054.

Step 4: Estimate the log-location.

\widehat{\mu}_{\mathrm{MM}} =\ln(6)-\frac{1}{2}\ln\left(\frac{10}{9}\right) =1.7917595-0.0526803 =1.7390792\ldots \approx 1.7391.

Step 5: Report the fitted arithmetic moments. The fitted mean is $m_1=6$ thousand dollars. The fitted variance is $m_2-m_1^2=40-36=4$ squared thousands of dollars, or $(\$4{,}000)^2$.

Check. The ratio of fitted second raw moment to squared fitted mean is $40/36=10/9$, which equals $e^{\widehat{\sigma}_{\mathrm{MM}}^{\,2}}$. The method avoids the common error of treating the unbiased sample variance, $16/3$, as the variance based on divisor $n$; that substitution would not reproduce the raw-moment equations.

Worked Example: Using Reported Sums and Sums of Squares

A hypothetical study contains six complete positive claim amounts, measured in thousands of dollars. The sum of the six amounts is $36$, and the sum of their squares is $360$. Assuming an independent lognormal sample, find the two method-of-moments parameter estimates and the fitted arithmetic mean and variance. Use the reported sums directly to form raw moments.

Step 1: Recover the sample raw moments.

m_1=\frac{36}{6}=6, \qquad m_2=\frac{360}{6}=60.

Step 2: Estimate the log-scale variance.

\widehat{\sigma}_{\mathrm{MM}}^{\,2} =\ln\left(\frac{60}{6^2}\right) =\ln\left(\frac{5}{3}\right) =0.5108256\ldots \approx 0.5108.

Step 3: Estimate the log-location.

\widehat{\mu}_{\mathrm{MM}} =\ln(6)-\frac{1}{2}\ln\left(\frac{5}{3}\right) =1.7917595-0.2554128 =1.5363467\ldots \approx 1.5363.

Step 4: Transform to the loss scale. The fitted arithmetic mean is $6$ thousand dollars. The fitted arithmetic variance is $60-36=24$ squared thousands of dollars.

Check. Since $e^{\widehat{\sigma}_{\mathrm{MM}}^{\,2}}=5/3$, the fitted second raw moment is $6^2(5/3)=60$, as required. A common mistake is to divide the provided sum of squares by $n-1$; the method-of-moments raw moment is $360/6=60$, not $360/5$.

Worked Example: Converting a Central Moment to Raw Moments

For a hypothetical sample of positive complete losses, the sample mean is $12$ dollars and the empirical central second moment, calculated with divisor $n$, is $18$ dollars squared. Assuming the lognormal model, estimate its parameters and fitted arithmetic moments. The key step is to convert the supplied central moment into the raw second moment before matching.

Step 1: Convert to the second raw moment. The identity between the central and raw second moments is $m_2=(m_2-m_1^2)+m_1^2$. Therefore,

m_1=12, \qquad m_2=18+12^2=18+144=162.

Step 2: Estimate the scale and location.

\widehat{\sigma}_{\mathrm{MM}}^{\,2} =\ln\left(\frac{162}{144}\right) =\ln\left(\frac{9}{8}\right) =0.1177830\ldots \approx 0.1178, $$ $$ \widehat{\mu}_{\mathrm{MM}} =\ln(12)-\frac{1}{2}\ln\left(\frac{9}{8}\right) =2.4849066-0.0588915 =2.4260152\ldots \approx 2.4260.

Step 3: State the fitted arithmetic moments. The fitted mean is $m_1=\$12$, and the fitted variance is $m_2-m_1^2=162-144=18$ dollars squared.

Check. The fitted second raw moment is $12^2e^{\ln(9/8)}=144(9/8)=162$. Using the supplied central moment of $18$ as though it were $m_2$ would give the wrong ratio and incorrect parameter estimates. The parameter calculations are rounded to four decimal places only at the final step.

Interpretation, Conditions, and Exam Traps

The lognormal parameters and the fitted arithmetic moments answer different questions. The location estimate is on the logarithmic scale, and the scale estimate is the variance of log losses. Neither is the fitted mean or variance in dollars. To report an expected loss or a variance of loss amounts, use the transformations derived above.

Match raw moments consistently. Use $m_1=\sum X_j/n$ and $m_2=\sum X_j^2/n$. If a central moment with divisor $n$ is supplied, add $m_1^2$ to obtain $m_2$.
Do not substitute the unbiased sample variance. That statistic uses divisor $n-1$; it is not the second central moment used in the raw-moment equations.
Keep the log-scale variance distinct from its standard deviation. The estimate in the formula is $\widehat{\sigma}^{\,2}$; if a scale standard deviation is requested, it is $\widehat{\sigma}=\sqrt{\widehat{\sigma}^{\,2}}$.
Respect units. Changing the loss unit changes the numerical log-location and arithmetic moments, but not the estimated log-scale variance. A variance always has squared loss units.
Check positivity and feasibility. The formulas assume positive complete observations from a common lognormal model. Censored, truncated, or zero-valued observations cannot simply be inserted as complete positive losses. Also, if $m_2=m_1^2$, the formula gives a zero scale variance, the boundary case of identical observations rather than a nondegenerate lognormal fit.

A reliable exam calculation therefore follows a fixed order: form the raw moments, estimate the log-scale variance, estimate the log-location, transform to arithmetic moments, and check that the fitted raw moments reproduce the sample raw moments. This method is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use raw moments with divisor $n$, keep track of the measurement units, and round parameter estimates to four decimal places.

A hypothetical sample of losses in dollars is $2,2,6$. Find $m_1$, $m_2$, both lognormal method-of-moments parameter estimates, and the fitted arithmetic variance. (Answer: $m_1=10/3$, $m_2=44/3$, $\widehat{\sigma}_{\mathrm{MM}}^{\,2}=\ln(33/25)=0.2776$, $\widehat{\mu}_{\mathrm{MM}}=1.0652$, fitted variance $=32/9$ dollars squared.)
A positive loss sample has mean $8$ dollars and empirical central second moment $12$ dollars squared, calculated with divisor $n$. Find the second raw moment, parameter estimates, and fitted arithmetic variance. (Answer: $m_2=76$, $\widehat{\sigma}_{\mathrm{MM}}^{\,2}=\ln(19/16)=0.1719$, $\widehat{\mu}_{\mathrm{MM}}=1.9935$, fitted variance $=\$12$ squared.)
Complete losses measured in thousands of dollars have $m_1=5$ and $m_2=35$. Find the fitted mean and variance and the two lognormal parameter estimates. (Answer: $\widehat{\sigma}_{\mathrm{MM}}^{\,2}=\ln(7/5)=0.3365$, $\widehat{\mu}_{\mathrm{MM}}=1.4412$, fitted mean $=5$ thousand dollars, fitted variance $=10$ squared thousands of dollars.)
For five complete positive observations, the sum is $20$ and the sum of squares is $100$. Find the two parameter estimates and fitted arithmetic moments. (Answer: $m_1=4$, $m_2=20$, $\widehat{\sigma}_{\mathrm{MM}}^{\,2}=\ln(5/4)=0.2231$, $\widehat{\mu}_{\mathrm{MM}}=1.2747$, fitted mean $=4$, fitted variance $=4$ squared loss units.)
A sample consists of three identical positive losses, each equal to $\$3$. What do the moment equations imply, and is the resulting fit a nondegenerate lognormal model? (Answer: $m_1=3$, $m_2=9$, $\widehat{\sigma}_{\mathrm{MM}}^{\,2}=0$, $\widehat{\mu}_{\mathrm{MM}}=\ln 3$, fitted variance $=0$; this is a boundary, degenerate fit, not a nondegenerate lognormal distribution.)

What You'll Learn