Why Moment Choice Matters
Method of moments begins with a model and a set of population moments, then matches those moments to their sample counterparts. The choice of moments is consequential: equations that repeat the same information may leave parameters undetermined, while a suitable set can distinguish all parameters. Before calculating estimates, ask whether the chosen population equations can identify the model parameters at all.
A distribution may be identifiable from its full probability law, yet a particular selection of its moments may not identify its parameters. This tutorial focuses on that second question: whether the equations formed from the selected moments have a unique solution over the model’s allowed parameter space.
The usual raw moments take \(g_j(X)=X^j\), giving \(E[X], E[X^2],\ldots\). Other choices, such as indicators or functions of a transformed observation, can also be used when their expectations are known under the model. In method of moments, the selected population moment \(E_{\boldsymbol{\theta}}[g_j(X)]\) is matched to the corresponding sample average \(n^{-1}\sum_{i=1}^{n}g_j(X_i)\). The population equations determine whether the chosen moments can distinguish parameters; the sample averages supply the values to match.
How to Assess a Moment Set
A practical assessment has three parts. First, write each theoretical moment as a function of the unknown parameters. Second, check whether two different parameter values could give the same selected moments. Third, enforce the parameter constraints: a scale, for example, must be positive. Merely having as many equations as parameters is not enough. Two equations can be redundant, or they may determine a parameter only up to multiple possible values.
State what each parameter represents and the restrictions it must satisfy.
Write their theoretical expectations as functions of the model parameters.
Check whether equal values of all selected moments force the parameter vectors to be equal. Look for redundant equations, lost parameter dependence, or multiple solutions.
After matching sample and population moments, substitute the resulting parameters back into every selected equation and verify that they satisfy the model constraints.
For a differentiable moment map with the same number of moments as parameters, a nonsingular Jacobian at a parameter value is a useful local check. The Jacobian has entries \(\partial m_j/\partial\theta_k\). Full rank means the moment equations distinguish small parameter changes near that point under the usual regularity conditions. It does not, by itself, prove global uniqueness: a map can be locally one-to-one in separate regions yet produce the same moments at distinct parameter values.
Worked Moment Choices
The examples below use population equations to examine whether selected moments determine the parameters. The sample moment values are hypothetical summaries, not real company or industry data.
Worked Example: Two Moments Identify a Normal Model
A hypothetical study models a signed claim-adjustment measure \(X\) as \(N(\mu,\sigma^2)\), where \(\mu\) is unrestricted and \(\sigma^2>0\). The sample first raw moment is 4 and the sample second raw moment is 25. Use these moments to estimate the parameters, and check whether the selected population moments identify them uniquely.
Method. Use the first two raw moments. For a normal random variable, \(E[X]=\mu\) and \(E[X^2]=\mu^2+\sigma^2\). The second identity follows from \(\operatorname{Var}(X)=E[X^2]-(E[X])^2=\sigma^2\).
Match the first moment. The equation is \(\mu=4\), so \(\hat{\mu}=4\).
Match the second moment. Substitute \(\hat{\mu}=4\) into \(\mu^2+\sigma^2=25\):
Thus the moment estimates are \(\hat{\mu}=4\) and \(\hat{\sigma}^2=9\), with \(\hat{\sigma}=3\). The parameter restriction is satisfied because \(9>0\). Substitution checks the fit: \(E[X]=4\), and \(E[X^2]=4^2+9=25\).
Identifiability check. The moment map is \(m_1=\mu\), \(m_2=\mu^2+\sigma^2\). Equality of first moments forces equal values of \(\mu\); equality of second moments then forces equal values of \(\sigma^2\). The pair therefore identifies both parameters over \(\sigma^2>0\). The common mistake is to match the second raw moment directly to the variance, which would incorrectly give \(\sigma^2=25\).
Worked Example: First and Third Normal Moments Fail at Zero Mean
In a separate hypothetical study, a modeler again assumes \(X\sim N(\mu,\sigma^2)\), with \(\sigma^2>0\), but selects the first and third raw moments. The corresponding sample moments are both zero. Determine whether these equations identify \(\mu\) and \(\sigma^2\).
Write the theoretical moments. The first is \(E[X]=\mu\). To obtain the third, write \(X=\mu+Z\), where \(E[Z]=0\), \(E[Z^2]=\sigma^2\), and \(E[Z^3]=0\) for a centered normal variable. Expanding the cube gives
Match the first moment. The equation \(\mu=0\) forces \(\mu=0\).
Match the third moment. With \(\mu=0\), the theoretical third moment is \(0^3+3(0)\sigma^2=0\) for every \(\sigma^2>0\). Thus the second equation is satisfied by every allowed variance. For example, both \((\mu,\sigma^2)=(0,1)\) and \((0,16)\) produce the selected moments \((0,0)\).
The selected moments do not identify the variance, so there is no unique method-of-moments estimate from these equations. A sample can have both first and third raw moments equal to zero without having zero spread; for instance, the hypothetical observations \(-3,-1,1,3\) have both moments equal to zero. The trap is to treat a zero third moment as evidence of zero variance. Here it contains no scale information once the first-moment equation sets \(\mu=0\).
Use Model Constraints and Moment Structure
Choosing moments is not only a matter of counting equations. Their dependence on the parameters matters, as do the parameter-space restrictions. A useful check is to ask what each equation would tell you if the others had already fixed some parameters. If a remaining parameter disappears from every remaining equation, it cannot be identified from that selection.
Worked Example: Identifying Both Endpoints of a Uniform Model
A hypothetical bounded-loss model assumes \(X\) is uniform on \([a,b]\), with \(a<b\). The selected population moments are a mean of 10 and a variance of 12. Determine the endpoints and verify that the moment choice identifies them.
Derive the moment relationships. Let \(w=b-a>0\) be the interval width. Symmetry gives \(E[X]=(a+b)/2\). For the variance, center the uniform distribution at its midpoint. Its centered variable is uniform on \([-w/2,w/2]\), so
Use the variance equation. Since \(w>0\), the equation \(w^2/12=12\) gives \(w^2=144\) and \(w=12\).
Use the mean equation. The midpoint is 10, so the endpoints are 6 units on either side of 10: \(a=10-6=4\) and \(b=10+6=16\).
Check the result. The midpoint is \((4+16)/2=10\), and the variance is \((16-4)^2/12=144/12=12\). The restriction \(a<b\) selects the positive width, so these moments uniquely identify the endpoints. A common mistake is to use the negative square root \(w=-12\); it contradicts \(w=b-a>0\).
Jacobian Checks and Common Traps
The Jacobian can quickly reveal when a proposed set of moments loses a parameter locally. For the normal first and second raw moments, the Jacobian with respect to \((\mu,\sigma^2)\) is
Its full rank is consistent with the global uniqueness established in the first example. For the first and third raw moments, the Jacobian is
At \(\mu=0\), the second column is zero, so the moments have no local sensitivity to changes in \(\sigma^2\) there. This agrees with the failure in the second example. A full-rank Jacobian is a local diagnostic, not a substitute for checking whether two separated parameter values could produce the same selected moments.
The distinction between raw and central moments also matters. A central moment is calculated about the distribution’s own mean; it is not automatically a separate source of information from a raw moment. For example, once \(E[X]\) is known, \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\), so using the mean and variance is equivalent to using the first and second raw moments. Do not count equivalent ways of writing the same information as independent equations.
- Counting equations instead of checking them. Two equations do not guarantee two independent pieces of parameter information.
- Ignoring constraints. A mathematical solution outside the allowed parameter space is not an admissible model fit.
- Assuming uniqueness from one solution. Finding a solution does not rule out another one; test the equations over the full parameter space.
- Confusing fit with identification. A parameter choice may satisfy the selected equations even though many other choices satisfy them too.
Check Your Understanding
For each hypothetical model, examine what the selected moments determine and check relevant parameter restrictions.
- A normal model has unknown \(\mu\) and \(\sigma^2>0\). The sample first and second raw moments are 2 and 13. Find the moment estimates and verify them. (Answer: \(\hat{\mu}=2,\ \hat{\sigma}^2=9\); the fitted moments are 2 and 13.)
- A uniform model on \([a,b]\), with \(a<b\), has selected mean 8 and variance 3. Find the endpoints, showing how the width restriction affects the result. (Answer: \(b-a=6,\ a=5,\ b=11\).)
- A model has \(X=cB\), where \(B\) is Bernoulli with success probability \(p\), \(c>0\), and \(0<p<1\). Its sample first and second raw moments are 4 and 24. Use \(E[X]=cp\) and \(E[X^2]=c^2p\) to find \(c\) and \(p\), then check both moments. (Answer: \(p=2/3,\ c=6\); \(cp=4\) and \(c^2p=24\).)
- For a normal model, the selected sample first and third raw moments are both zero. State whether they uniquely determine the variance, and give two allowed parameter pairs that fit. (Answer: No; both \((\mu,\sigma^2)=(0,1)\) and \((0,4)\) fit.)
- A normal model uses the first and third raw moments, with sample values 1 and 7. First match the first moment, then use \(E[X^3]=\mu^3+3\mu\sigma^2\) to determine whether the parameters are unique. (Answer: \(\hat{\mu}=1,\ \hat{\sigma}^2=2\); these equations uniquely determine the two parameters under \(\sigma^2>0\).)