# Method of Moments - What's the logic?

Method of Moments - What's the logic? We have a random vector $X=(X_1,X_2,...,X_n)$ that generates a sample. The hypothesis is that our components of the random vector $X_i$ are i.i.d.

The method of moments, for point estimation, in statistical inference, suggests a way to find an estimate of a parameter $\theta$. If you need to estimate only one parameter, the procedure seems to force you to equate the first theoretical moment $E(X_i)$ to the corresponding sample moment.

Question:

If I have only one parameter to estimate, can I equate the second theoretical moment to the second sample moment in order to find $\theta$? Does the fact that I have only one parameter to estimate force me to use just the first moments in my equation? Am I free to use every moment in my equation?

• For a Poisson distribution, you can take the average value of the observations as a method of moments estimator of the parameter, or the variance (using $\frac 1n$) or even some weighted average of these two. All of these would be methods of moments estimators, For an exponential distribution, you could take the reciprocal of the average of the observations or the average of the reciprocals of the observations as estimators of the rate parameter (the later uses a negative moment) Jun 4, 2022 at 1:44

You don't even have to use moments. Any function $$g(X)$$ whose expectation with respect to $$P_{\theta}$$ uniquely determines $$\theta$$ will do. By Law of Large Numbers, the average of $$g(X_i)$$ will converge to that expectation, and you estimate $$\theta$$ to be the $$\theta$$ giving this expectation (this only works if the inverse of $$\mathbb{E}_{\theta} g(X)$$ is continuous in $$\theta$$). One often uses $$g(X)=X^k$$, in which case this gives estimates of the k-th mooment of the underlying distribution (from which one deduces the parameters), hence the name.