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?

  • $\begingroup$ Note $\frac{1}{n}\sum_{i=1}^n\log(X_i)$ is approximately $\mathcal{N}\left(\mathbb{E}\left(\log(X)\right),\frac{\mathbb{Var}\left(\log(X)\right)}{n}\right)$ with $X\sim f$ from CLT. I assume $f$ is supported on $[X_m,\infty)$. Now apply delta with with $g(x)=\frac{1}{\log(X_m)+x}$ to obtain the asymptotic distribution of you MLE. Can you proceed in this manner to find the asymptotic distribution of your MOM? $\endgroup$
    – user429040
    Jan 27, 2022 at 1:18

1 Answer 1


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: $$\operatorname{Var}[\tilde \alpha] = \frac{\alpha (\alpha-1)^2 ((n-1)\alpha - 2n)}{n^2 (\alpha-2)^2}. \tag{3}$$

  • $\begingroup$ to compute the exactly variance of the MLE have you computed the $E[Y]$ = $\int_{1}^{\infty} \alpha y^{-\alpha} dy$ and the $E[Y^2]$ = $\int_{1}^{\infty} \alpha y^{-\alpha+1} dy$ And then $VAR = E[Y^2]-E[Y]^2$ because i get a similar result but not exactly the same variance I Know why fisher info is equal to $\frac{n}{\alpha^2}$ I miss only the variance of MLE step $\endgroup$
    – V013
    Jan 27, 2022 at 11:23
  • $\begingroup$ @lollo $Y$ is Pareto, so $\log Y$ is exponential with rate $\alpha$. Then the sample mean of the log-transformed sample $\frac{1}{n} \sum \log Y_i$ is Gamma with shape $n$ and rate $n\alpha$. So the MLE $\hat \alpha$ is inverse gamma distributed. You cannot use $\operatorname{Var}[Y]$ to compute $\operatorname{Var}[\hat \alpha]$. $\endgroup$
    – heropup
    Jan 27, 2022 at 18:32

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