A transformation $g$ such that $g(S^2)$ has asymptotic distribution depending on $\beta_2$ only

Question

Let $X_1, X_2,\ldots,X_n$ be i.i.d. RVs with $E|X_1|^4 < \infty$. Let $\operatorname{var}(X_1) = \sigma^2$, $\beta_2 = \mu_4/\sigma^4$.

(a) Using CLT for i.i.d. RVs, show that $\sqrt{n}(S^2-\sigma^2)\stackrel{L}\rightarrow N(0, \mu_4-\sigma^4)$.

(b) Find a transformation $g$ such that $g(S^2)$ has asymptotic distribution that depends on $\beta_2$ alone, not on $\sigma^2$.

I have completed part (a). But got stuck on finding $g$. I know I have to use the theorem:

$Y_n$ is $AN(\mu, \sigma^2_n)$, with $\sigma^2_n\rightarrow 0$ and $\mu$ fixed real. $g$ be differentiable at $\mu$ with $g'(\mu)\neq 0$, then $g(Y_n)$ is $AN(g(\mu), [g'(\mu)]^2\sigma^2_n)$

Any help appreciated. Thanks.

($AN(\cdot)$ means asymptotically normal distribution.)

Answer 1

You can apply a variance stabilising transformation on $S^2$.

Let $\sigma_T^2=\mu_4-\sigma^4$, so that $\sigma_T^2=\beta_2\sigma^4-\sigma^4=\sigma^4(\beta_2-1)$.

By delta method, for a function $g$ that is differentiable at $\sigma^2$ with $g'(\sigma^2)\ne 0$, we have

$$\sqrt{n}\left(g(S^2)-g(\sigma^2)\right)\stackrel{a}\sim N\left(0,\sigma_T^2([g'(\sigma^2)]^2\right)\tag{*}$$

Set $\sigma_T\,g'(\sigma^2)=c$, a non-zero constant.

Therefore,

$$\sigma^2\sqrt{\beta_2-1}\,g'(\sigma^2)=c \implies \int g'(\sigma^2)\,d\sigma^2=\frac{c}{\sqrt{\beta_2-1}}\int \frac{1}{\sigma^2}\,d\sigma^2$$

Choosing $c=1/2$ (and ignoring constant of integration), you get

$$\color{blue}{g(\sigma^2)=\frac{\ln\sigma}{\sqrt{\beta_2-1}}}$$

So $(*)$ gives

$$\frac{\sqrt{n}}{\sqrt{\beta_2-1}}(\ln S-\ln \sigma)\stackrel{a}\sim N\left(0,\frac{1}{4}\right)$$

That is,

$$\sqrt{n}(\ln S-\ln \sigma)\stackrel{a}\sim N\left(0,\frac{1}{4}(\beta_2-1)\right)$$

Answer 2

Rearranging the asymptotic distribution in (a) gives:

$$\frac{S_n^2}{\sigma^2} \overset{\text{Asymp}}{\sim} \text{N} \Big( 1, \frac{\beta_2-1}{n} \Big).$$

So, taking the parameter $\beta_2$ as fixed, this gives you an asymptotically pivotal quantity (where the distribution does not depend on $\sigma^2$ but that parameter appears in the quantity). If you are allowed to use the true variance parameter in your function $g$ then you are done, but if not, then it is unlikely there is a solution. Because it gives the scale of the sample variance, the only way you will be able to remove the variance parameter from this quantity (without having the function degenerate down to a form whose distribution no longer depends on $\beta_2$) is if you replace the variance parameter with an alternative estimator of the variance.