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.
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ What do you know about the delta method? $\endgroup$– StubbornAtomCommented Apr 11, 2020 at 19:05
-
$\begingroup$ I know that $\sqrt{n}(h(\overline{x})-h(\theta)) \to N(0, h'(x)^2\theta)$ or something like that $\endgroup$– Michael Devin SmithCommented Apr 11, 2020 at 19:09
-
1$\begingroup$ en.wikipedia.org/wiki/Delta_method#Univariate_delta_method $\endgroup$– StubbornAtomCommented Apr 11, 2020 at 19:43
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
6
Hint:
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
-
-
$\begingroup$ 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. $\endgroup$– MasoudCommented Apr 11, 2020 at 19:18
-