Why Compare Two Estimation Methods?
For a specified probability model and sample, method of moments and maximum likelihood provide two distinct ways to estimate an unknown parameter. Method of moments matches sample summaries to population moments. Maximum likelihood chooses the parameter value that makes the observed data most likely under the model. They can produce different estimates, but in some models they agree exactly.
A comparison should go beyond the numerical answers. The estimates can have different bias and sampling variability, and neither method is automatically preferable in every finite sample. In particular, an estimate’s closeness to a parameter for one observed sample does not tell us how variable that estimation procedure would be across repeated samples.
Both methods depend on the model being specified correctly. A MOM equation must use moments that exist under the assumed parameter values and identify the parameter. An MLE must maximize the likelihood over the full permitted parameter space, including its boundary when relevant. In either method, estimates from one particular sample are not the parameters themselves.
Different Constructions, Sometimes Different Answers
For a one-parameter model with finite mean \(E[X]=m_1(\theta)\), the usual first-moment MOM estimate solves \(\bar X=m_1(\theta)\), where \(\bar X=n^{-1}\sum_{i=1}^n X_i\). An MLE instead maximizes the joint likelihood. If the likelihood is positive and differentiable in the interior of the parameter space, one can often find an interior candidate by differentiating the log-likelihood and setting its derivative to zero. That candidate must still be checked against the parameter space and any boundary values.
The two approaches can agree because a likelihood equation sometimes reduces to a moment equation. Agreement is a property of a model and estimation problem, not a general rule. The following uniform model shows why a careful support check can matter.
Worked Example: A Uniform Endpoint Produces Different Estimates
A hypothetical sample of \(n=4\) losses, in thousands of dollars, is \(2,4,5,9\). Assume the observations are independent from a uniform distribution on \([0,\theta]\), where the upper endpoint \(\theta>0\) is unknown. Find the first-moment MOM estimate and the MLE, then compare their sampling uncertainty.
MOM construction. For \(X\sim\operatorname{Uniform}(0,\theta)\), the density is \(f(x;\theta)=1/\theta\) for \(0\leq x\leq\theta\), and zero otherwise. Its mean is \(E[X]=\theta/2\). The sample mean is
Equating the sample and population means gives \(5=\theta/2\), so \(\hat\theta_{\mathrm{MOM}}=10\) thousand dollars.
Likelihood construction. The largest observation is \(x_{(4)}=9\). The joint likelihood is
For every permitted \(\theta\geq9\), \(\theta^{-4}\) decreases as \(\theta\) increases. Therefore its maximum is attained at the smallest permitted value, \(\hat\theta_{\mathrm{MLE}}=9\) thousand dollars. The density convention includes both endpoints of its support; in particular, \(f(9;9)=1/9\). This makes the likelihood positive at \(\theta=9\).
Sampling uncertainty. Under this model, \(\operatorname{Var}(X)=\theta^2/12\), so
For the sample maximum \(M=X_{(n)}\), the exact variance is \(n\theta^2/((n+1)^2(n+2))\). With \(n=4\), the MOM variance is \(\theta^2/12\), whereas the MLE variance is \(2\theta^2/75\). Plugging in the corresponding estimates gives standard-error estimates \(10/\sqrt{12}=2.8868\) and \(9\sqrt{2/75}=1.4697\), respectively, in thousands of dollars. These are plug-in estimates of sampling standard deviations, not known standard errors because \(\theta\) is unknown.
The MLE is biased downward: \(E[M]=n\theta/(n+1)\), so its bias is \(-\theta/(n+1)\). Its mean squared error is \(2\theta^2/((n+1)(n+2))\), which for \(n=4\) is \(\theta^2/15\). The MOM estimate is unbiased and has mean squared error \(\theta^2/12\). Thus this example has a smaller MLE variance and a smaller MLE mean squared error, despite the MLE’s bias.
Method and trap avoided: The MOM estimate comes from the population mean; the MLE comes from the likelihood’s maximum over the support-constrained parameter space. Do not treat the sample mean and sample maximum as interchangeable estimates, or omit the support restriction.
Why the Uniform Sampling Formulas Hold
The uncertainty comparison above follows from the sampling distributions of the two estimators. The MOM estimator is \(2\bar X\). Independence gives \(\operatorname{Var}(\bar X)=\operatorname{Var}(X)/n\); multiplying a random variable by 2 multiplies its variance by \(2^2\). Hence \(\operatorname{Var}(2\bar X)=4(\theta^2/12)/n=\theta^2/(3n)\).
For the maximum \(M\), the event \(M\leq m\) means that every observation is at most \(m\). For \(0\leq m\leq\theta\), independence therefore gives
Integrating this density yields \(E[M]=n\theta/(n+1)\) and \(E[M^2]=n\theta^2/(n+2)\). Subtracting the square of the mean from the second moment gives
This derivation also shows why “uncertainty” needs a stated measure. Variance describes spread around the estimator’s own expectation; mean squared error describes expected squared distance from the target parameter and includes both variance and squared bias.
When the Methods Coincide
The methods need not disagree. In several familiar models, maximizing the likelihood produces the same equation as matching a sample moment to its population counterpart. The next example shows this directly for a Poisson rate and calculates the common estimator’s sampling uncertainty.
Worked Example: The Poisson Rate Has the Same MOM Estimate and MLE
A hypothetical set of five independent monthly claim counts is \(1,0,2,3,4\). Assume each count has a Poisson distribution with unknown mean \(\lambda>0\). Find both estimates and the standard error estimate for their common value.
MOM estimate. A Poisson random variable has mean \(\lambda\). The observed sample mean is \((1+0+2+3+4)/5=10/5=2\). Equating sample and population means gives \(\hat\lambda_{\mathrm{MOM}}=2\).
MLE. For observed counts \(x_1,\ldots,x_n\), the likelihood is proportional to \(e^{-n\lambda}\lambda^{\sum x_i}\); the factors \(1/x_i!\) do not depend on \(\lambda\). The log-likelihood, apart from those constant terms, is \(\ell(\lambda)=-n\lambda+(\sum x_i)\log\lambda\). Its derivative is \(-n+(\sum x_i)/\lambda\), so the stationary point is \(\hat\lambda=(\sum x_i)/n=10/5=2\). The second derivative is \(-(\sum x_i)/\lambda^2<0\), confirming a maximum. Thus \(\hat\lambda_{\mathrm{MLE}}=2\), the same as the MOM estimate.
Sampling uncertainty. Since a Poisson count has variance \(\lambda\), independence gives \(\operatorname{Var}(\bar X)=\lambda/n\). Here the common estimator has estimated variance \(2/5=0.4\), and estimated standard error \(\sqrt{0.4}=0.6325\) claims per month. This is obtained by substituting \(\hat\lambda=2\) into the exact variance formula \(\lambda/n\).
Method and trap avoided: The MOM estimate uses the sample mean and the MLE uses the likelihood, even though the algebra leads to the same answer. Do not mistake agreement in this model for a theorem that the two methods always agree.
When estimates coincide, their sampling variability as estimators is also the same because they are the same function of the sample. But the derivations can still be useful: a moment equation can be simple to form, while a likelihood gives a systematic way to handle the model’s full probability structure. In other models the two constructions lead to different functions of the data.
Worked Example: Bernoulli Estimates and an Exact Uncertainty Calculation
A hypothetical sample records whether each of \(n=8\) submitted claims is accepted under a particular rule. Four are accepted and four are not. Assume independent Bernoulli observations with unknown acceptance probability \(p\). Find the MOM estimate, the MLE, and an estimated standard error.
MOM estimate. A Bernoulli observation has mean \(p\). The sample mean is \(4/8=0.5\), so equating it to the population mean gives \(\hat p_{\mathrm{MOM}}=0.5\).
MLE. If \(k\) of the \(n\) observations equal 1, the likelihood is \(L(p)=p^k(1-p)^{n-k}\), for \(0\leq p\leq1\). Here \(k=4\) and \(n-k=4\). For \(0<p<1\), the log-likelihood derivative is \(k/p-(n-k)/(1-p)\). Setting it to zero gives \(k(1-p)=(n-k)p\), hence \(p=k/n=4/8=0.5\). The likelihood is maximized there, so \(\hat p_{\mathrm{MLE}}=0.5\).
Sampling uncertainty. The sample proportion has exact variance \(p(1-p)/n\). Substituting the common estimate gives estimated variance \(0.5(1-0.5)/8=0.25/8=0.03125\), and estimated standard error \(\sqrt{0.03125}=0.1768\). The two estimation methods agree because both yield the sample proportion.
Method and trap avoided: The number accepted is \(k=4\), while the denominator is the total number observed, \(n=8\). Using \(k\) rather than \(n\) as the denominator would not give the sample proportion or the MLE.
How to Make a Fair Comparison
A useful comparison separates how an estimator is constructed from how it behaves over repeated samples. The sample values determine the realized estimate; the model and sampling assumptions determine its bias, variance, and mean squared error. A single estimate cannot establish which method is more precise in general.
| Question | Method of moments | Maximum likelihood |
|---|---|---|
| How is the estimate constructed? | Match sample moments to population moments. | Maximize the likelihood over the parameter space. |
| What should be checked? | The moment exists, identifies the parameter, and gives a permitted value. | The candidate maximizes the likelihood, including boundary values. |
| Does it always agree with the other method? | No; agreement depends on the model and chosen moments. | No; agreement depends on the model and likelihood. |
| How is sampling uncertainty assessed? | Derive or calculate the estimator’s sampling variance under the model. | Derive or calculate the estimator’s sampling variance under the model. |
For either method, state what uncertainty measure is being reported. If an exact estimator variance is available, substituting an estimate for an unknown parameter produces an estimated standard error. That plug-in quantity is not the exact sampling standard deviation unless the substituted value happens to equal the true parameter.
Common Mistakes and Exam Traps
- Assuming the methods must agree. The uniform example gives \(2\bar x\) for MOM and \(x_{(n)}\) for the MLE. Derive each estimate from its own definition.
- Ignoring the parameter space. A likelihood may be maximized at a boundary. For the uniform model, the maximum occurs at the observed maximum, not at an interior solution of a derivative equation.
- Using an inconsistent endpoint convention. If the uniform density is defined as zero at \(x=\theta\), then observing \(x_{(n)}=\theta\) gives zero likelihood at that value. With that convention the likelihood’s supremum is approached as \(\theta\) decreases to \(x_{(n)}\), but is not attained. To state an MLE at the observed maximum, use a density convention that includes the endpoint, as in the worked example.
- Calling an estimate precise because it looks plausible. Precision concerns repeated-sample variability, which must be derived under the stated model. It cannot be read off from one estimate.
- Confusing variance and mean squared error. For a biased estimator, mean squared error equals variance plus squared bias; comparing variance alone does not compare mean squared error.
Check Your Understanding
For each problem, use the stated model and distinguish the construction of the estimate from its sampling uncertainty.
- A hypothetical sample from \(\operatorname{Uniform}(0,\theta)\) is \(1,4,5\), with the density defined to include its endpoints. Find the first-moment MOM estimate and the MLE. (Answer: \(\hat\theta_{\mathrm{MOM}}=20/3\), \(\hat\theta_{\mathrm{MLE}}=5\).)
- Four independent hypothetical counts are \(0,2,2,4\), each Poisson with unknown mean \(\lambda\). Find the MOM estimate and MLE, and the plug-in variance and standard error of their common estimator. (Answer: both estimates are \(2\); estimated variance \(0.5\), estimated standard error \(0.7071\).)
- In a hypothetical Bernoulli sample of \(n=10\), seven observations equal 1. Find the MOM estimate, the MLE, and the estimated standard error of the sample proportion. (Answer: both estimates are \(0.7\); estimated standard error \(\sqrt{0.7(0.3)/10}=0.1449\).)
- For \(n=2\) observations from \(\operatorname{Uniform}(0,\theta)\), the observed sample is \(2,6\). Find both estimates. If the true value is \(\theta=9\), calculate the MLE’s bias and variance. (Answer: \(\hat\theta_{\mathrm{MOM}}=8\), \(\hat\theta_{\mathrm{MLE}}=6\); MLE bias \(-3\), variance \(4.5\).)