Assume that $Y$ has a Pareto distribution with parameters ($\theta, t$ = 1). An estimator of $\theta$ is $\tilde{\theta}$ where $\bar{Y} = \frac{\tilde{\theta}t}{\tilde{\theta} - 1}$. Solve for $\tilde{\theta}$ and then use the delta method to derive the asymptotic distribution of $\sqrt{n}(\tilde{\theta} - \theta)$, assuming $\theta$ > 2.

A good place to begin, if I am correct, is to assume $\bar{Y}$ = $\frac{\theta t}{\theta - 1}$. From there, we can solve for $\tilde{\theta}$. Except, this doesn't necessarily get rid of our random variable. Additionally, I'm confused about how this relates to the delta method. As someone utterly lost, can I get any help for this? Thanks!


Since $t=1$,

$$\bar{Y} = \frac{\tilde{\theta}}{\tilde{\theta} - 1} \implies \tilde \theta = \frac {\bar Y}{\bar Y-1}$$

We know the limiting distribution of

$$\sqrt{n}\left(\bar Y - \frac{\theta}{\theta-1}\right)$$

since it is the sample mean. Then we need the Delta method to find the limiting distribution of $\sqrt{n}\left(\tilde \theta - \theta\right)$, since $\theta$ is a non-linear function of $\bar Y$. The $g$ function here is $g(z) = z/(z-1)$.

I guess the rest are up to the OP.

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  • $\begingroup$ Limiting distribution of a sample mean being a N(0,variance) distribution? $\endgroup$ – Brendan G Mar 28 '18 at 22:35
  • $\begingroup$ @BrendanG Yes. The "i.i.d. sample" assumption is usually implied when not mentioned. $\endgroup$ – Alecos Papadopoulos Mar 28 '18 at 23:13

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