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!


1 Answer 1


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.

  • $\begingroup$ Limiting distribution of a sample mean being a N(0,variance) distribution? $\endgroup$
    – Brendan G
    Mar 28, 2018 at 22:35
  • $\begingroup$ @BrendanG Yes. The "i.i.d. sample" assumption is usually implied when not mentioned. $\endgroup$ Mar 28, 2018 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.