I am supposed to use the delta method to find the limiting distribution for $$\sqrt{n}\left(\frac{\overline{X}_n}{1-\overline{X}_n} - \frac{E(X)}{1-E(X)}\right)$$ where $f(x, \theta)=\theta x^{\theta-1}I\{x\in (0,1)\}$. I'm utterly lost on how to do this, as this doesn't line up with how he briefly explained the Delta method in class.


1 Answer 1



let we know by central limit theorem $$\sqrt{n}(\bar{X}-\mu_{x})\rightarrow N(0,\sigma^2)$$

By using Taylor series we get $$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)=\sqrt{n} (\bar{X}-\mu_{x})h^{\prime}(\mu_{x}) +\sqrt{n} (\bar{X}-\mu_{x})^2\frac{h^{\prime \prime}(\mu_{x})}{2!}+\cdots$$

lets we use an approximation like (if $h^{\prime}(\mu_{x})\neq 0$)

$$\sqrt{n}\left(h(\bar{X})-h(\mu_{x})\right)\cong \color{red}{\sqrt{n} (\bar{X}-\mu_{x})}h^{\prime}(\mu_{x})\rightarrow N(0,?) $$

  • $\begingroup$ That would go to $N(0,(h'(\mu_x))^2\sigma^2)$ I believe $\endgroup$ Commented Apr 11, 2020 at 19:10
  • $\begingroup$ I think, its look good to me. $\endgroup$
    – Masoud
    Commented Apr 11, 2020 at 19:13
  • $\begingroup$ Okay, so how do I extend that to the entire Taylor expansion? $\endgroup$ Commented Apr 11, 2020 at 19:14
  • $\begingroup$ Usually you just use the first term of Taylor series. But some times $h^{\prime}(\mu_{x})= 0$ so you use the second term. It is depend on the situation. $\endgroup$
    – Masoud
    Commented Apr 11, 2020 at 19:18
  • $\begingroup$ Oh okay. Thank you! $\endgroup$ Commented Apr 11, 2020 at 19:25

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