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

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?.

• Thanks for showing us what you've done! Please add the [self-study] tag (you can probably replace the "distributions" tag as that is not so relevant to your question) & read its wiki so you know how we deal with self-study questions. – Silverfish Oct 16 '16 at 9:36
• @user130264 Do you think the MLE should have an asymptotically normal distribution? One of the regularity conditions (support depends on parameter space) does not hold here. – Greenparker Oct 16 '16 at 13:39
• Just to make sure we're all on the same page, you're talking about the density $f_X(x;\theta) = (\frac{x^2}{\theta})^{-1}\,,\:\theta>0\,,\: x>\theta$...? – Glen_b -Reinstate Monica Oct 16 '16 at 22:57
• @Alecos Yes I screwed that up. I should have looked more carefully at what I was writing. To save additional comments I have ninja edited what it said. – Glen_b -Reinstate Monica Oct 19 '16 at 1:42

(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 $\theta^2$.

• I was searching for an answer to a similar question, and interestingly enough I am quite certain that the question asked here is exactly the same question as that I am now being asked in a homework assignment. The reason why your answer states that the requested expression, namely, $\sqrt{n}(\hat{\theta}_n - \theta)$, converges to constant zero, which would make little sense given that the exercise asks for an asymptotic distribution, is that in the exercise the words 'shape' and 'scale' are in the wrong order. My teacher mentioned this to us by e-mail today. – Anon Oct 17 '19 at 20:11
• This would mean that the scale parameter is $1$, and the shape parameter is $\theta > 0$. In this case, I believe OP's answer is correct. It is worth noting that in this exercise, it is also provided that a certain information identity holds, which is an indirect hint to the fact that it should be used (which the OP has done). – Anon Oct 17 '19 at 20:15
• That said I think your answer should obtain some more information as to why the MLE for the scale parameter is equal to the smallest order statistic and why the MLE follows a Pareto distribution, even though both of these observations might be somewhat trivial. – Anon Oct 17 '19 at 20:18