How to compare asymptotic efficiency of MOM estimator with MLE? Compare the asymptotic efficiency of MOM estimator of parameter $\alpha$ of the pareto distributions with MLE (assume $X_m$ known)
$$f(x;\alpha;X_m) = \alpha X_m^{\alpha} x^{-(\alpha + 1)}.$$
I computed the MLE of the pareto distribution equating to $0$ the first derivative of the log-likelihood getting $$\frac{n}{n\log(X_m)+\sum_{i=0}^n \log(X_i)}.$$
And for MOM I get $$\frac{\bar X}{\bar X- X_m}.$$
My question is: In order to compare the asymptotic efficiency should I compute the variance of the Mom and the variance of MLE?
If so someone can direct me in the right direction?
 A: Some simplification is possible.  Note that if $X_m$ is known, then consider the transformation $Y = X/X_m$, hence $Y$ is Pareto with shape $\alpha$ and location $1$ with density $f_Y(y) = \alpha y^{-(\alpha+1)} \mathbb 1 (y \ge 1)$.  Then the transformed sample is simply $$\boldsymbol Y = (y_1, y_2, \ldots, y_n) = \boldsymbol X/X_m = (x_1/X_m, x_2/X_m, \ldots, x_n/X_m).$$  So for the purposes of estimation and efficiency, we can work with $\boldsymbol Y$, or equivalently, assume $X_m = 1$, because this transformation is one-to-one.
That said, the efficiency of an estimator $w(\theta)$ of $\theta$ is defined as $$\mathcal E(w(\theta)) = \frac{1/\mathcal I(\theta)}{\operatorname{Var}[w(\theta)]},$$ where $\mathcal I(\theta)$ is the Fisher information.  So you need to compute the variance of each estimator.
I leave it as an exercise to show that for your distribution, the Fisher information is:
$$\mathcal I(\alpha) = \frac{n}{\alpha^2}. \tag{1}$$
The exact variance of the MLE is:
$$\operatorname{Var}[\hat \alpha] = \frac{(n \alpha)^2}{(n-1)^2 (n-2)}, \quad n > 2, \alpha > 1. \tag{2}$$
The asymptotic variance of the method of moments estimator via the CLT and the delta method is:
$$\operatorname{Var}[\tilde \alpha] = \frac{\alpha (\alpha-1)^2 ((n-1)\alpha - 2n)}{n^2 (\alpha-2)^2}. \tag{3}$$
