How to determine the estimator of the asymptotic variance of the MLE estimator of the Pareto distribution?

Question

For my statistics class I have to determine for a pareto distribution with shape parameter 1 and scale parameter $\theta > 0$ the asymptotic distribution of $\sqrt{n}(\hat{\theta}_n - \theta)$, where $\hat{\theta}_n$ is the MLE estimator, the estimator $\widehat{Avar}(\hat{\theta}_n)$ of the asymptotic variance and what does $\frac{\sqrt{n}(\hat{\theta}_n - \theta)}{\sqrt{Avar(\hat{\theta_n})}}$ converge to and how. I don't know if what I did, however it makes intuitively sense to me, is correct, so could you please check it?

I know that by the asymptotic normality of the MLE, $\sqrt{n}(\hat{\theta}_n - \theta)$ $\xrightarrow{\mathcal{D}}$ $\mathcal{N}(0, (\frac{1}{n}i(\theta))^{-1})$, i.e. $\sqrt{n}(\hat{\theta}_n - \theta)$ $\xrightarrow{\mathcal{D}}$ $\mathcal{N}(0, \theta^2)$, given $i(\theta)$ = $\frac{n}{\theta^2}$.

I also know that a consistent estimator of this distribution is given by $\widehat{Avar}(\hat{\theta}_n)$ = $[\frac{1}{n}i(\hat{\theta}_n)]^{-1} = \hat{\theta}_n^2$.

Now, I thought that $\frac{\sqrt{n}(\hat{\theta}_n - \theta)}{\sqrt{Avar(\hat{\theta_n})}}$ would converge in distribution to $\mathcal{N}(0, 1)$. Is this correct?.

Answer 1

(I am going to fully answer this home study, because it appears that the OP is rather away from any path that could be steered in the right direction by simple hints).

It can be shown that the MLE for $\theta$ is the minimum-order statistic,

$$\hat \theta_n = X_{1:n}$$

In fact, one can derive that $\hat \theta_n$ follows itself a Pareto distribution with scale parameter $\theta$ and (since here the original shape parameter is $1$), with shape parameter $n$, i.e. with distribution function

$$F(\hat \theta_n) = 1- \left(\frac {\hat \theta_n}{\theta}\right)^{-n}$$

So in this case we have available the finite sample distribution of the ML estimator.

Still, the OP is asked to consider asymptotics for the function $\sqrt {n} (\hat \theta _n - \theta)$.

Subtracting $\theta$ does not make the distribution centered at zero or anything like that, since we always have

$$\hat \theta_n = X_{1:n} \geq \min {X}= \theta$$

By this observation alone we know that even if $\sqrt {n} (\hat \theta _n - \theta)$ has a limiting distribution, it won't be the normal.

Now, $(\hat \theta_n - \theta)$ has an exact Lomax distribution, with the same parameters as the distribution of $\hat \theta_n$. The Variance of the Lomax distribution is the same as the variance of the Pareto distribution so in our case

$$\text{Var}[(\hat \theta_n - \theta)] = \frac {\theta^2 n}{(n-1)^2(n-2)}$$

Note that the denominator has leading term $n^3$. So to stabilize this variance as $n$ goes to infinity, so that it neither explodes nor goes to zero, one needs to multiply the variable by $n$, not $\sqrt {n}$.

$$\text{Var}[n(\hat \theta_n - \theta)] = \frac {\theta^2 n^3}{(n-1)^2(n-2)} \to \theta^2$$

So $\sqrt {n} (\hat \theta _n - \theta)$ goes to the constant zero (convergence of the estimator is faster than usual and multiplying by $\sqrt {n}$ "is not enough" to maintain a distribution, we need to multiply by $n$). Standardized by its variance (which goes to zero), it explodes.

Can we obtain the limiting distribution of $Z_n = n(\hat \theta_n - \theta)$?

We have for the distribution function

$$F_{Z_n}(z) = \text{Prob}(Z_n \leq z) = \text{Prob}(n(\hat \theta_n - \theta) \leq z) = \text{Prob}\left(\hat \theta_n \leq \theta + \frac {z}{n} \right)$$

$$ \implies F_{Z_n}(z) = 1- \left(\frac {\theta + \frac {z}{n}}{\theta}\right)^{-n} = 1- \left(1+ \frac {(z/\theta)}{n}\right)^{-n}$$

$$\implies \lim_{n \to \infty}F_{Z_n}(z) = 1 - \exp\{-z/\theta\}$$

which is the CDF of the Exponential distribution, with expected value $\theta$ and variance $\theta^2$.