Worked Comparison of Moment and Likelihood Uncertainty

Why Compare Estimators?

A fitted parameter is only part of an actuarial estimate. Two methods can produce different parameter values from the same loss observations, and their uncertainty estimates can differ as well. A useful comparison therefore checks the fitted values, the assumptions behind each standard error, and the resulting confidence intervals.

This tutorial uses a Pareto severity model with known scale and unknown shape. It is a focused setting in which both the method-of-moments estimator and the maximum-likelihood estimator have simple forms, but the moment estimator’s uncertainty depends on a stronger condition than the likelihood estimator’s. The calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Model: Suppose \(X_1,\ldots,X_n\) are independent and identically distributed Pareto Type I losses, with known scale \(\theta>0\) and unknown shape \(\alpha>0\). The density is \(f_X(x)=\alpha\theta^\alpha x^{-(\alpha+1)}\) for \(x\geq\theta\), and zero otherwise. Equivalently, \(S_X(x)=(\theta/x)^\alpha\) for \(x\geq\theta\). The losses are assumed to be fully observed, with no truncation or censoring.

For this model, \(E[X]=\alpha\theta/(\alpha-1)\) only when \(\alpha>1\), and \(\operatorname{Var}(X)=\alpha\theta^2/((\alpha-1)^2(\alpha-2))\) only when \(\alpha>2\). These restrictions matter: the mean-based method-of-moments estimator can be defined when the mean exists, while its usual asymptotic variance calculation also needs a finite variance.

Two Estimators for the Shape

The method of moments sets the sample mean equal to the model mean and solves for \(\alpha\). The likelihood method instead maximizes the joint density of the observed losses. For a fixed known \(\theta\), the log likelihood is \(\ell(\alpha)=n\ln\alpha+n\alpha\ln\theta-(\alpha+1)\sum_{j=1}^n\ln X_j\).

Estimator formulas: If \(\bar X>\theta\), the mean-based method-of-moments estimator is \(\widehat{\alpha}_{\mathrm{MM}}=\bar X/(\bar X-\theta)\). The maximum-likelihood estimator is \(\widehat{\alpha}_{\mathrm{ML}}=n/\sum_{j=1}^n\ln(X_j/\theta)\). The moment estimator targets a finite model mean, so its population interpretation requires \(\alpha>1\). The likelihood estimator is defined for \(\alpha>0\).

To see why the likelihood formula holds, differentiate the log likelihood with respect to \(\alpha\): \(\ell'(\alpha)=n/\alpha-\sum_{j=1}^n\ln(X_j/\theta)\). Setting this score equal to zero gives the stated estimator. The second derivative is \(\ell''(\alpha)=-n/\alpha^2<0\), so the stationary point is a maximum.

For the uncertainty comparison, two large-sample variance formulas are useful. The Fisher information for \(\alpha\) in \(n\) independent observations is \(I_n(\alpha)=n/\alpha^2\), giving the approximate likelihood-estimator variance \(\alpha^2/n\). For the moment estimator, the delta method applies the derivative of \(g(m)=m/(m-\theta)\) to the sample mean. Since \(g'(m)=-\theta/(m-\theta)^2\), the resulting large-sample variance is \(\alpha(\alpha-1)^2/((\alpha-2)n)\), requiring \(\alpha>2\).

\widehat{\operatorname{Var}}(\widehat{\alpha}_{\mathrm{MM}}) = \frac{\widehat{\alpha}_{\mathrm{MM}}(\widehat{\alpha}_{\mathrm{MM}}-1)^2} {(\widehat{\alpha}_{\mathrm{MM}}-2)n}, \qquad \widehat{\operatorname{Var}}(\widehat{\alpha}_{\mathrm{ML}}) = \frac{\widehat{\alpha}_{\mathrm{ML}}^2}{n}.

Why the Moment-Estimator Variance Has a Restriction

The variance calculation for \(\widehat{\alpha}_{\mathrm{MM}}\) is not obtained merely by substituting an estimate into a formula. Its justification uses the sampling variance of \(\bar X\), which is \(\operatorname{Var}(X)/n\) when that variance is finite. The delta method then propagates this uncertainty through the transformation from \(\bar X\) to \(\widehat{\alpha}_{\mathrm{MM}}\).

Start with the sample mean.
For \(\alpha>2\), \(\operatorname{Var}(\bar X)=\alpha\theta^2/[n(\alpha-1)^2(\alpha-2)]\).

Differentiate the estimator function.
With \(g(m)=m/(m-\theta)\), \(g'(m)=-\theta/(m-\theta)^2\). At the model mean \(m=\alpha\theta/(\alpha-1)\), this derivative is \(-(\alpha-1)^2/\theta\).

Apply the delta method.
Multiply \([g'(m)]^2\) by \(\operatorname{Var}(\bar X)\). The result is \(\alpha(\alpha-1)^2/[n(\alpha-2)]\).

The likelihood-based variance has a different source: the inverse Fisher information. It does not require the severity variance to be finite. Consequently, a likelihood standard error for \(\alpha\) can be available even when the fitted Pareto model has infinite variance. Neither formula removes the need to assess whether a large-sample approximation is suitable for the sample at hand.

Worked Comparisons

Worked Example: Comparing Both Methods on One Loss Sample

A hypothetical portfolio has losses modeled as Pareto Type I with known scale \(\theta=1\) thousand dollars. There are \(n=40\) observed losses: ten observations each of \(1.1,\ 1.2,\ 1.3,\) and \(1.4\) thousand dollars. Treat these reported values as the observations for the calculation. Find both estimates of \(\alpha\), their estimated variances and standard errors, and approximate 95% confidence intervals. Also compare the fitted model means.

Method and sample summaries. The sample mean is \(\bar X=10(1.1+1.2+1.3+1.4)/40=50/40=1.25\) thousand dollars. The sum of log scale ratios is \(\sum\ln(X_j/\theta)=10\ln(1.1\cdot1.2\cdot1.3\cdot1.4)=10\ln(2.4024)\approx8.764682\).

Method-of-moments estimate. Substitution gives \(\widehat{\alpha}_{\mathrm{MM}}=1.25/(1.25-1)=1.25/0.25=5\). The fitted mean is \(5/(5-1)=1.25\) thousand dollars, exactly matching the sample mean as the moment equation requires.

Maximum-likelihood estimate. The estimate is \(\widehat{\alpha}_{\mathrm{ML}}=40/8.764682\approx4.563771\). Its fitted mean is \(4.563771/(4.563771-1)\approx1.2806\) thousand dollars.

Estimated uncertainty. The moment estimate exceeds 2, so its usual large-sample variance formula applies. It gives \(\widehat{\operatorname{Var}}(\widehat{\alpha}_{\mathrm{MM}})=5(5-1)^2/[(5-2)40]=80/120=0.666667\), with standard error \(\sqrt{0.666667}\approx0.816497\). For the MLE, \(\widehat{\operatorname{Var}}(\widehat{\alpha}_{\mathrm{ML}})=4.563771^2/40\approx0.52070\), with standard error approximately \(0.7216\).

Approximate 95% intervals. Using estimate \(\pm1.96\) standard errors, the moment interval is \(5\pm1.96(0.816497)=[3.3997,6.6003]\). The likelihood interval is \(4.563771\pm1.96(0.7216)\approx[3.1495,5.9781]\). Endpoints are rounded to four decimal places.

Check. Both intervals lie above 2, consistent with the fitted shapes being in the finite-variance region. The two estimates differ, and the moment estimate reproduces the sample mean while the likelihood estimate does not. The common mistake avoided here is treating the two estimation methods as if their point estimates must agree.

Worked Example: When the Moment Standard Error Does Not Apply

For a hypothetical severity sample, let \(\theta=1\), \(n=4\), and observed losses be \(1.2,\ 1.8,\ 2.5,\) and \(4.0\). Compute both shape estimates and the likelihood standard error. Explain why the usual moment-estimator variance is not reported.

Method-of-moments estimate. The sample mean is \((1.2+1.8+2.5+4.0)/4=9.5/4=2.375\). Thus \(\widehat{\alpha}_{\mathrm{MM}}=2.375/(2.375-1)=2.375/1.375=19/11\approx1.7273\). This estimate is above 1, so the fitted model has a finite mean, but it is below 2 and the fitted model has infinite variance.

Maximum-likelihood estimate. The product of the four losses is \(1.2(1.8)(2.5)(4.0)=21.6\), so the log-ratio sum is \(\ln(21.6)\approx3.072693\). Therefore \(\widehat{\alpha}_{\mathrm{ML}}=4/3.072693\approx1.3020\), and its estimated variance is \(1.3020^2/4\approx0.4238\). The standard error is approximately \(0.6510\).

A normal-approximation 95% interval for the MLE is \(1.3020\pm1.96(0.6510)\approx[0.0260,2.5780]\). The moment estimator’s usual asymptotic variance calculation requires a finite severity variance; that condition fails here. Reporting the moment standard error from the finite-variance formula would not be justified. The trap is to confuse existence of the mean, which requires \(\alpha>1\), with existence of the variance, which requires \(\alpha>2\).

Worked Example: Comparing Uncertainty From Sample Summaries

A hypothetical sample of \(n=25\) fully observed Pareto losses has known scale \(\theta=1\), sample mean \(\bar X=1.50\), and log-ratio sum \(\sum\ln(X_j/\theta)=10\). Find the two estimates and their approximate standard errors and 95% intervals.

Estimates. The moment estimate is \(1.50/(1.50-1)=3\). The likelihood estimate is \(25/10=2.5\). The supplied summaries are compatible: the geometric mean is \(e^{10/25}=e^{0.4}\approx1.4918\), below the arithmetic mean 1.50, as required.

Uncertainty estimates. For the moment estimator, \(\widehat{\operatorname{Var}}=3(3-1)^2/[(3-2)25]=12/25=0.48\), so the standard error is \(\sqrt{0.48}\approx0.6928\). For the MLE, the estimated variance is \(2.5^2/25=0.25\), so its standard error is \(0.5\).

Intervals and check. The moment interval is \(3\pm1.96(0.6928)\approx[1.6421,4.3579]\), and the likelihood interval is \(2.5\pm1.96(0.5)=[1.5200,3.4800]\). Both estimates exceed 2, so the moment variance calculation is permitted under their fitted values. The intervals are approximate, not exact finite-sample intervals.

Interpreting the Comparison

A comparison should not be reduced to choosing whichever standard error is smaller. The moment estimator is directly tied to matching the sample mean, while the likelihood estimator uses the entire sample through the logarithms of scale ratios. Their point estimates can therefore differ even under the same model. In addition, the moment-based variance depends on the existence of the severity variance; the likelihood information formula does not.

The intervals above use the large-sample normal approximation for the estimator. A standard error describes estimated sampling variability, not a guarantee that the interval has exact coverage in a small sample. In particular, an interval for a positive shape parameter can extend below zero under a normal approximation. Such an endpoint signals a limitation of that approximation, not a permissible negative Pareto shape. The intervals in the worked examples remain positive, but they should still be interpreted as approximate.

Exam trap: Do not insert a method-of-moments estimate below 2 into the usual variance formula and report the resulting number as a standard error. That formula comes from the finite variance of \(X\), which does not exist for \(\alpha\leq2\). Check the parameter condition before calculating uncertainty.

Check Your Understanding

For each problem, use the stated Pareto Type I model and treat the supplied losses or summaries as fully observed.

With known \(\theta=2\), \(n=5\), and sample mean \(\bar X=5\), calculate the method-of-moments estimate. If the log-ratio sum is 3, calculate the MLE as well. (Answer: \(\widehat{\alpha}_{\mathrm{MM}}=5/3\approx1.6667;\ \widehat{\alpha}_{\mathrm{ML}}=5/3\approx1.6667\).)
A sample has \(n=20\) and \(\sum\ln(X_j/\theta)=40/7\). Find the MLE and its information-based estimated variance and standard error. (Answer: \(\widehat{\alpha}_{\mathrm{ML}}=3.5;\ \widehat{\operatorname{Var}}=0.6125;\ \mathrm{SE}\approx0.7826\).)
For \(n=16\), known \(\theta=1\), and \(\bar X=1.4\), calculate the moment estimate and its plug-in asymptotic variance and standard error. (Answer: \(\widehat{\alpha}_{\mathrm{MM}}=3.5;\ \widehat{\operatorname{Var}}\approx0.9115;\ \mathrm{SE}\approx0.9547\).)
An MLE is \(\widehat{\alpha}_{\mathrm{ML}}=2\) from \(n=100\) observations. Give its information-based standard error and approximate 95% interval. (Answer: \(\mathrm{SE}=0.2;\ [1.6080,2.3920]\).)
For \(n=10\), known \(\theta=1\), the sample mean is 2.2 and the log-ratio sum is 8. Find the moment estimate and MLE. Is the usual moment-estimator variance available? (Answer: \(\widehat{\alpha}_{\mathrm{MM}}=11/6\approx1.8333;\ \widehat{\alpha}_{\mathrm{ML}}=1.25;\) no, because the fitted moment estimate is at most 2 and the fitted severity variance is infinite.)

What You'll Learn