Why Nonlinear Moment Systems Need a Plan
When a model has two unknown parameters, the method of moments typically gives two equations. Those equations need not be linear. Attempting to solve both at once can obscure the useful structure: one equation may let us express a parameter in terms of the other, leaving a single nonlinear equation to solve. The resulting estimate should then be substituted back into the original equations as a check.
This tutorial develops that approach for a two-parameter Pareto Type I model. The example illustrates both algebraic substitution and numerical root-finding. The model and data below are hypothetical; they are used only to demonstrate the calculation.
A Two-Parameter Pareto Moment System
Suppose \(X\) follows a Pareto Type I distribution with scale parameter \(\theta>0\) and shape parameter \(\alpha>0\). Its density is
The raw moment of order \(r\) exists when \(\alpha>r\). For that range, integrate the density:
In particular, using the first and second raw moments requires \(\alpha>2\). Equating them to the sample raw moments gives
The first equation can be rearranged to express the scale in terms of the shape:
Substituting this expression into the second equation eliminates \(\theta\). Defining \(Q=m'_2/(m'_1)^2\), the two-equation system reduces to
Thus the nonlinear part of the problem is to solve one equation for \(\alpha\); the scale parameter follows by substitution. This reduction is useful even when the remaining equation cannot be solved conveniently by hand.
Solving the Reduced Equation
If \(Q\) is known, rearranging the reduced equation gives
A root can be found by algebra when the resulting equation is simple. Otherwise, a numerical method such as Newton's method can approximate it. Newton's method starts from a value \(\alpha_0\) and updates according to
This update comes from replacing \(g\), near the current estimate \(\alpha_k\), by its tangent line \(g(\alpha_k)+g'(\alpha_k)(\alpha-\alpha_k)\), and setting that line equal to zero. Each iterate is an approximation, not an exact solution. The method can fail or converge to an unsuitable value if its starting point is poorly chosen; for this model, keep the search in the required range \(\alpha>2\) and verify the final estimates in the original moment equations.
Use sample raw moments and the corresponding population raw moments for the stated model.
Rearrange one equation to express one unknown in terms of the other.
Use algebra if practical; otherwise apply a numerical root-finding method while respecting the model's parameter restrictions.
Recover the remaining parameter and evaluate both model moments to check that they match the sample moments.
Worked Example: Using Newton's Method for Pareto Moments
A hypothetical sample of losses is summarized by \(m'_1=6\) and \(m'_2=54\), in consistent loss units. Estimate the shape and scale of a Pareto Type I model by the method of moments. Use Newton's method for the nonlinear equation.
Method and reduction. Equate the first two population raw moments to 6 and 54. First calculate \(Q=54/6^2=54/36=1.5\). The reduced shape equation is therefore \(\alpha(\alpha-2)=1/(1.5-1)=2\), or \(g(\alpha)=\alpha^2-2\alpha-2=0\), with \(\alpha>2\).
Newton iterations. Here \(g'(\alpha)=2\alpha-2\). Starting at \(\alpha_0=3\), which is in the required range:
At the next iterate, \(g(2.75)=2.75^2-2(2.75)-2=7.5625-5.5-2=0.0625\), and \(g'(2.75)=5.5-2=3.5\). Thus \(\alpha_2=2.75-0.0625/3.5=2.7321429\), to seven decimal places.
One further update gives approximately \(\alpha_3=2.7320508\). The scale estimate is \(\theta=6(\alpha-1)/\alpha\), so using \(\alpha_3\), \(\hat\theta=6(1.7320508)/2.7320508\approx3.8038\). Round only the final estimates: \(\hat\alpha=2.7321\) and \(\hat\theta=3.8038\), to four decimal places.
Check and trap avoided. Substitution gives a fitted mean of approximately \(2.7320508(3.8038)/(2.7320508-1)\approx6.0000\) and a fitted second raw moment of approximately \(2.7320508(3.8038)^2/(2.7320508-2)\approx54.0000\). A common mistake is to use \(m'_2\) as though it were the sample variance; here the supplied value 54 is the second raw moment. The two fitted moments confirm the estimates, up to rounding.
Substitution Can Also Give an Exact Solution
Numerical root-finding is not always necessary. Sometimes the reduced equation has a simple exact solution. The same substitution remains the key step; after solving, check both original moment equations rather than relying only on the reduced equation.
Worked Example: Solving from a Mean and Variance
A hypothetical sample from a Pareto Type I loss model has sample mean 12 and sample variance 48. Find the method-of-moments estimates of \(\alpha\) and \(\theta\), using the first two raw moments.
Method and conversion. The second raw moment is not the variance. The identity \(m'_2=s^2+(m'_1)^2\), for these supplied moment summaries, gives \(m'_2=48+12^2=48+144=192\). Therefore \(Q=192/144=4/3\).
The reduced equation becomes \(\alpha(\alpha-2)=1/(4/3-1)=1/(1/3)=3\). Rearranging gives \(\alpha^2-2\alpha-3=0\), which factors as \((\alpha-3)(\alpha+1)=0\). The solution \(\alpha=-1\) is outside the required range \(\alpha>2\), so \(\hat\alpha=3\).
Recover the scale. From the first moment, \(\hat\theta=12(3-1)/3=12(2/3)=8\).
Check and trap avoided. The fitted mean is \(3(8)/(3-1)=24/2=12\). The fitted second raw moment is \(3(8^2)/(3-2)=192\), so the implied variance is \(192-12^2=192-144=48\). A frequent error is to put 48 directly into the second raw moment equation; converting from variance first avoids that error.
Calculating Moments from Observations
A moment system is only as reliable as the summaries entered into it. When observations are supplied rather than sample moments, calculate each raw moment explicitly: square each observation for the second raw moment, add those squares, and divide by the sample size. This is not the same operation as calculating a sample variance about the sample mean.
Worked Example: Building and Solving the System from Data
Four hypothetical loss observations are \(5,6,7,8\). Fit a Pareto Type I model by matching the first two sample raw moments. Give the estimates to four decimal places.
Method and sample moments. The sample size is \(n=4\). The first raw moment is \(m'_1=(5+6+7+8)/4=26/4=6.5\). The second raw moment is \(m'_2=(5^2+6^2+7^2+8^2)/4=(25+36+49+64)/4=174/4=43.5\).
Thus \((m'_1)^2=6.5^2=42.25\), and \(Q=43.5/42.25=174/169\). Hence \(Q-1=5/169\), so the reduced equation is \(\alpha(\alpha-2)=169/5=33.8\). Rearranging gives \(\alpha^2-2\alpha-33.8=0\). Its solution in the required range is \(\alpha=1+\sqrt{34.8}\approx6.9833\); the other root is less than 2.
Recover the scale. Using the unrounded shape value, \(\hat\theta=6.5(\hat\alpha-1)/\hat\alpha\approx5.5692\). The estimates, rounded to four decimal places, are \(\hat\alpha=6.9833\) and \(\hat\theta=5.5692\).
Check and trap avoided. The fitted mean is approximately \(6.9833(5.5692)/(6.9833-1)\approx6.5000\). The fitted second raw moment is approximately \(6.9833(5.5692^2)/(6.9833-2)\approx43.5000\). A mistake would be to square the mean and call \(42.25\) the second raw moment; \(42.25\) is the square of the first raw moment, while the sample second raw moment is \(43.5\).
Checks That Make a Numerical Solution More Reliable
A numerical answer should be treated as a candidate until it has been checked in the system that generated it. For the Pareto example, a practical check has three parts:
- Confirm that the estimated shape is in the range required for the moments used. Matching the second raw moment requires \(\hat\alpha>2\).
- Substitute the unrounded shape into the equation used to find it and check that the residual is small. For Newton's method, this means checking that \(g(\hat\alpha)\) is close to zero.
- Substitute both parameter estimates into the original mean and second raw moment formulas. Those fitted moments should agree with the sample moments, allowing for the rounding displayed.
These checks catch common arithmetic and setup errors, including switching the scale and shape roles, losing a square when substituting, or using a variance where a raw moment is required. They do not turn an approximation into an exact value: retain extra digits during the calculation and round only the reported estimates.
Check Your Understanding
For each problem, use the stated Pareto Type I model and the first two raw moments unless the question says otherwise.
- A hypothetical sample has \(m'_1=9\) and \(m'_2=108\). Find the method-of-moments estimates of \(\alpha\) and \(\theta\). (Answer: \(\hat\alpha=3,\ \hat\theta=6\).)
- A hypothetical sample has mean 7 and variance 21. Convert to the second raw moment, then find the two method-of-moments estimates. (Answer: \(m'_2=70,\ \hat\alpha=1+\sqrt{10/3}\approx2.8257,\ \hat\theta\approx4.5243\).)
- Four hypothetical observations are \(3,5,7,9\). Calculate the first two raw sample moments and use them to estimate the Pareto parameters. (Answer: \(m'_1=6,\ m'_2=41,\ \hat\alpha=1+\sqrt{46/5}\approx4.0332,\ \hat\theta\approx4.5114\).)
- A candidate uses the supplied mean 4 and variance 4 as the first and second raw moments. Identify the error and give the corrected estimates. (Answer: the second raw moment is \(4+4^2=20\), not 4; \(\hat\alpha=1+\sqrt5,\ \hat\theta=4\sqrt5/(1+\sqrt5)\approx2.4721\).)