Under appropriate regularity conditions it is well-known that Maximum Likelihood Estimation (MLE) produces asymptotically efficient estimators in the sense that their asymptotic covariance is given by the inverse of Fisher information, i.e.
$$
\sqrt n(\theta-\tilde\theta)\overset{d}{\to}\mathcal N(0,I^{-1}(\theta))
$$
as $n\to\infty$.


With the Method of Moments (MoM) this is not necessarily true. There are cases where MLE and MoM produce the same estimators; however, in general, there is no guarantee of asymptotic efficiency in the MoM's.


My question is why? Intuitively, MLE requires specifying the exact distribution from which the observed data is realized. On the other hand, MoM only requires specifying the first $m$-moments ($\theta\in\Bbb R^m$) of the data's distribution. So I suspect this lack of specificity is why we do not have any guarantees of asymptotic efficiency. Is this intuition correct? Can someone spell this out with a more compelling theoretical argument?



Why MLEs are asymptotically efficient whereas method of moment estimators are not?
The comment by [kjetil b halvorsen](https://mathoverflow.net/questions/451895/why-mles-are-asymptotically-efficient-whereas-method-of-moment-estimators-are-no?noredirect=1#comment1168436_451895) seems to have a good point. In view of the ensuing discussion, it may be of use to detail, clarify, and complement some of the raised points, which will be done below.


The [likelihood function](https://en.wikipedia.org/wiki/Likelihood_function#Definition) is usually defined as the map $\Theta\ni\theta\mapsto L\_x(\theta):= f\_\theta(x)$, where $\Theta$ is the parameter space, $x$ is a realization (value) of the random sample $X$ taken from the distribution with density $f\_{\theta\_0}$, $\theta\_0\in\Theta$ is the "true" value of the parameter $\theta$, and $(f\_\theta)\_{\theta\in\Theta}$ is a family of probability densities (referred to as the statistical model).


So, for each realization $x$ of $X$ we have its own likelihood function $L\_x$. We may then consider the function $x\mapsto \mathcal L(x):=L\_x$, and we may even want to refer to this function $\mathcal L$ as the likelihood function as well. The function $\mathcal L$ will be measurable with respect to the cylindrical $\sigma$-algebra over $\mathbb R^\Theta$. So, the random function $L\_X:=\mathcal L(X):=\mathcal L\circ X$ is a statistic with values in $\mathbb R^\Theta$. Trivially, the statistic $L\_X$ is sufficient, since $f\_\theta(x)=L\_x(\theta)$ for all $\theta$ and $x$. (However, to focus on the essential ideas, let us not be concerned with measurability matters in what follows in this answer.)


So, trivially, the "maximum likelihood estimator (MLE)" $$\operatorname{argmax}\limits\_{\theta\in\Theta}L\_X(\theta)$$
with values in the set of all subsets of $\Theta$ is a function of the sufficient statistic $L\_X$. This fact is hardly of any significance, though -- because the (entire) sample $X$ is of course always sufficient, and any statistic is, by definition, a function of the sufficient statistic $X$.


What is important is that, by the [factorization criterion](https://mathoverflow.net/questions/451895/why-mles-are-asymptotically-efficient-whereas-method-of-moment-estimators-are-no?noredirect=1#comment1168436_451895), the MLE is a function of **any** sufficient statistic, including [minimal sufficient statistics](https://en.wikipedia.org/wiki/Sufficient_statistic#Minimal_sufficiency).


(However, the MLE by itself of course does not have to be sufficient. E.g., if $\Theta=(0,\infty)$, $X=(X\_1,\dots,X\_n)$, $n\ge2$, and $X\_1,\dots,X\_n$ are i.i.d. normal random variables each with mean $\theta$ and variance $\theta^2$, then the almost surely unique MLE of $\theta$ is
\begin{equation}
 \hat\theta:=\sqrt{\overline X^2/4+\overline{X^2}}-\overline X/2,
\end{equation}
where $\overline X:=\frac1n\,\sum\_1^n X\_i$ and $\overline{X^2}:=\frac1n\,\sum\_1^n X\_i^2$.
However, here $(\overline X,\overline{X^2})$ is a minimal sufficient statistic -- which is not a function of $\hat\theta$, and therefore the MLE $\hat\theta$ is not sufficient.)




---


On the other hand, estimators other than the MLE (including method-of-moment estimators) do not have to be functions of a minimal sufficient statistic, and are therefore "less likely" to have good statistical properties. One way to see this is that, by the [Rao–Blackwell theorem](https://en.wikipedia.org/wiki/Rao%E2%80%93Blackwell_theorem#Mean-squared-error_version), if $S(X)$ is any estimator of $q(\theta)$ for some function $q$ and $T(X)$ is any sufficient statistic, then (i) $S\_T(X):=E\_\theta(S(X)|T(X))$ is a statistic (as it does not depend on $\theta$); (ii) **$S\_T(X)$ is a function of the sufficient statistic $T(X)$** (even if $T(X)$ is a minimal sufficient statistic); (iii) the bias of $S\_T(X)$ for $q(\theta)$ is the same as the bias of $S(X)$ for $q(\theta)$, for all values of $\theta$; (iv) **the variance of $S\_T(X)$ is no greater than the variance of $S(X)$, for all values of $\theta$ (and the latter property [generalizes to any convex loss function](https://en.wikipedia.org/wiki/Rao%E2%80%93Blackwell_theorem#Convex_loss_generalization))**.


**So, we can take any estimator which is not a function of a minimal sufficient statistic $T(X)$ and improve it by the described above Rao–Blackwell conditioning on $T(X)$ -- whereas the MLE cannot be improved this way, since the MLE is already a function of any (minimal) sufficient statistic and hence the conditioning on any sufficient statistic does not change the MLE.**




---


Finally, about this:



> 
> Intuitively, MLE requires specifying the exact distribution from which the observed data is realized. On the other hand, MoM only requires specifying the first $m$-moments ($\theta\in\Bbb R^m$) of the data's distribution. So I suspect this lack of specificity is why we do not have any guarantees of asymptotic efficiency. Is this intuition correct?
> 
> 
> 


The answer to this is no. Indeed, if $\theta$ is an $m$-dimensional parameter and the method of moments is applicable, then the knowledge of $m$ moments uniquely determines $\theta$, so that you have the complete specificity. E.g., if $f\_\theta$ is the density of the gamma distribution with parameter $\theta:=(\alpha,\beta)\in\Theta=(0,\infty)\times(0,\infty)$, then the first two moments $\mu\_1(\theta)=\alpha\beta$ and $\mu\_2(\theta)=\alpha\beta^2$ uniquely determine $\theta=(\alpha,\beta)$. Another way to look at this is that, in a parametric model, knowing the value of the parameter $\theta$, you fully know the density $f\_\theta$ and thus you fully know the corresponding distribution.


