Why Parameterization Matters
A probability model can describe the same distribution using different parameters. For example, an exponential distribution may be specified by its rate or by its mean, and a gamma distribution may be specified by shape and scale or by its mean and variance. A method-of-moments estimate should describe the same fitted model whichever parameterization is used—provided the same moment equations are matched.
The important qualification is “the same moment equations.” Method of moments does not select a unique estimator independently of that choice. If a candidate matches a different sample moment, that candidate can be a different method-of-moments estimate, even when it targets the same parameter. This distinction is useful when translating estimates, checking work, and interpreting apparently different answers.
Suppose a model is instead written using \(\phi=g(\theta)\), where \(g\) is a one-to-one transformation on the allowed parameter space. If the original moment equations give \(\hat\theta_{\text{MM}}\), then the estimate in the new parameterization, using those same equations, is \(g(\hat\theta_{\text{MM}})\). Rewriting the model does not itself create new data or new moment equations.
Why the Same Equations Give Transformed Estimates
Let the selected theoretical moments be collected in a vector \(M(\theta)\), and let the corresponding sample moments be \(m\). The method-of-moments equations are \(M(\theta)=m\). Under the new parameter \(\phi=g(\theta)\), the model’s moment vector is \(M(g^{-1}(\phi))\). Therefore the new equations are \(M(g^{-1}(\phi))=m\). If \(\hat\theta\) solves the original equations, then \(\hat\phi=g(\hat\theta)\) solves the new ones, since \(M(g^{-1}(g(\hat\theta)))=M(\hat\theta)=m\).
This argument requires the parameter transformation to be invertible over the relevant parameter space and the equations to have a solution. If a moment system has multiple solutions, a boundary solution, or no feasible solution, changing notation does not resolve that issue. A one-to-one transformation carries each valid solution to its corresponding transformed solution.
Worked Examples
Worked Example: Exponential Rate and Mean
A hypothetical device’s operating time \(X\), measured in hours, is modeled as exponential with rate \(\lambda>0\). A sample has observed times \(1,2,3\) hours. Estimate the rate and the mean operating time by matching the sample mean, then compare with an estimate obtained by matching the second raw moment.
Method: match the first raw moment. For an exponential distribution with rate \(\lambda\), \(E[X]=1/\lambda\). The sample mean is \(\bar X=(1+2+3)/3=6/3=2\) hours. Equating the theoretical mean to the sample mean gives \(1/\hat\lambda_{\text{MM}}=2\), so \(\hat\lambda_{\text{MM}}=1/2=0.5\) per hour.
The distribution’s mean parameter is \(\theta=1/\lambda\). Transform the rate estimate using this relation: \(\hat\theta_{\text{MM}}=1/\hat\lambda_{\text{MM}}=1/(1/2)=2\) hours. If the model is written directly in terms of \(\theta\), the first-moment equation is \(E[X]=\theta=\bar X\), which also gives \(2\) hours. The estimates agree because both calculations use the same moment.
Now match the second raw moment instead. For this exponential model, \(E[X^2]=2\theta^2\). The sample second raw moment is \(m'_2=(1^2+2^2+3^2)/3=(1+4+9)/3=14/3\) square hours. Equating moments gives \(2\tilde\theta^2=14/3\), hence \(\tilde\theta^2=7/3\) and \(\tilde\theta=\sqrt{7/3}\approx1.5275\) hours, rounded to four decimal places.
This second estimate differs from \(2\) because it matches a different sample moment. It is not the first-moment estimate transformed into a new parameterization. The calculations are valid for their stated procedures, and the values are positive, as required for an exponential mean.
Worked Example: Transforming Gamma Shape and Scale Estimates
A hypothetical laboratory records measurements \(1,3,5\), in a common measurement unit. Model \(X\) as gamma with shape \(\alpha>0\) and scale \(\beta>0\), so \(E[X]=\alpha\beta\) and \(\operatorname{Var}(X)=\alpha\beta^2\). Use the first two raw moments to estimate shape and scale, then transform the estimates to the model’s mean and variance.
The sample mean is \(\bar X=(1+3+5)/3=9/3=3\). The sample second raw moment is \(m'_2=(1^2+3^2+5^2)/3=(1+9+25)/3=35/3\). The sample variance moment used for matching is \(m'_2-\bar X^2=35/3-9=35/3-27/3=8/3\).
Match the population mean and variance to these sample quantities: \(\alpha\beta=3\) and \(\alpha\beta^2=8/3\). Dividing the second equation by the first gives \(\hat\beta_{\text{MM}}=(8/3)/3=8/9\). Then \(\hat\alpha_{\text{MM}}=3/(8/9)=27/8=3.375\).
Now express the model through its mean \(\mu=\alpha\beta\) and variance \(\tau^2=\alpha\beta^2\). Transform the estimates: \(\hat\mu_{\text{MM}}=\hat\alpha_{\text{MM}}\hat\beta_{\text{MM}}=(27/8)(8/9)=3\), and \(\widehat{\tau^2}_{\text{MM}}=\hat\alpha_{\text{MM}}\hat\beta_{\text{MM}}^2=(27/8)(64/81)=8/3\).
As a check, direct matching in the mean-variance parameterization gives \(\hat\mu=\bar X=3\) and \(\widehat{\tau^2}=m'_2-\bar X^2=8/3\). Both estimates are positive, and the two parameterizations describe the same fitted gamma model. No new moment was introduced by transforming the estimates.
Worked Example: A Different Moment for a Transformed Uniform Parameter
A hypothetical sensor reading \(X\) is modeled as uniform on \([0,\theta]\), with \(\theta>0\). The observations are \(2,4,6\). Estimate \(\phi=\theta^2\) first by matching the sample mean and transforming the result, then by matching the second raw moment directly to \(\phi\).
For the uniform model, \(E[X]=\theta/2\). The sample mean is \(\bar X=(2+4+6)/3=4\). Thus \(4=\hat\theta/2\), so \(\hat\theta=8\), and transforming gives \(\hat\phi=8^2=64\).
For a uniform random variable on \([0,\theta]\), \(E[X^2]=\theta^2/3=\phi/3\). The sample second raw moment is \(m'_2=(2^2+4^2+6^2)/3=(4+16+36)/3=56/3\). Matching this moment gives \(\tilde\phi/3=56/3\), hence \(\tilde\phi=56\).
The estimates \(64\) and \(56\) differ because the first is based on the sample mean, while the second is based on the sample second raw moment. Both are nonnegative, as required for \(\phi=\theta^2\). The discrepancy is not caused by reparameterizing the same moment equations.
Common Mistake: Transforming Each Observation Instead
For the exponential sample \(1,2,3\), a common incorrect rate calculation is to average the reciprocals of the observations: \((1/1+1/2+1/3)/3=(1+0.5+0.3333\ldots)/3\approx0.6111\) per hour. This is not the method-of-moments estimate obtained by matching \(E[X]=1/\lambda\). The correct calculation first averages the observations and then takes the reciprocal: \(\hat\lambda_{\text{MM}}=1/\bar X=1/2=0.5\) per hour.
In general, transforming an estimate means applying the parameter transformation to the estimate, such as \(g(\hat\theta)\). It does not mean applying that transformation separately to every data value and averaging. Nonlinear operations do not generally commute with averaging.
When to Check for a Different Moment System
When two solutions in different parameterizations disagree, write down the equations used in each calculation before deciding that one is wrong. If the equations are algebraically equivalent and the parameter transformation is one-to-one, their unique feasible solutions must correspond. If the equations use different sample moments, a difference is expected and does not by itself indicate an error.
- Check the target transformation. Write the relationship between the parameters, including its domain and inverse.
- Check the matched moments. Identify which population moments are equated to which sample moments.
- Check the algebra. Substitute the proposed transformed estimate into the rewritten moment equations.
- Check feasibility. Confirm that the estimate lies in the parameter space and that any required root is real.
A transformation of an estimator also does not, by itself, imply that the transformed estimator has any particular bias or variance. Those properties require separate analysis. For this topic, the key conclusion is narrower: with the same moment equations and an invertible parameter transformation, the estimates transform together; changing the matched moments can change the estimate.
Check Your Understanding
For each problem, identify whether the calculation is a transformation of the same moment estimate or uses a different moment equation.
- An exponential model has rate \(\lambda\) and mean \(\theta=1/\lambda\). For observed times \(2,2,5\) hours, find the first-moment estimate of \(\lambda\) and its transformed estimate of \(\theta\). (Answer: \(\hat\lambda=1/3\) per hour; \(\hat\theta=3\) hours.)
- A gamma model has shape \(\alpha\) and scale \(\beta\), with mean \(\alpha\beta\) and variance \(\alpha\beta^2\). For observations \(2,4,6\), use the first two moments and report the transformed mean and variance estimates. (Answer: \(\hat\mu=4\), \(\widehat{\tau^2}=8/3\).)
- For an exponential model with mean \(\theta\), observations are \(1,2,5\). Find the estimate of \(\theta\) from the first moment, then the estimate from matching \(E[X^2]=2\theta^2\). (Answer: first-moment estimate \(8/3\); second-moment estimate \(\sqrt{5}\approx2.2361\).)
- For \(X\sim\operatorname{Uniform}(0,\theta)\), let \(\phi=\theta^2\). Observations are \(2,4,6\). Find the estimate of \(\phi\) by transforming the first-moment estimate of \(\theta\), then by matching \(E[X^2]=\phi/3\). (Answer: \(64\) by transformation; \(56\) by second-moment matching.)