Assume we have an exponential distribution $$ p_{\theta}(x) = \theta \exp(-\theta x), \quad x>0 $$
and the estimator $$T = \sqrt\frac{{\sum_{i=1}^{n} X^2_i}}{2n}$$ and as parameter of interest $$\gamma =1/\theta $$ How can I find the limiting distribution of it ? I understand finding the limiting distribution in case of the arithmetic mean. Then I can apply the CLT by
$\sqrt{n}(\bar{X}-\frac{1}{\theta}) {\rightarrow}N(0, \frac{1}{\theta^2})$. But I have no idea how I can derive the distribution of
$$\sqrt{n}(T-\frac{1}{\theta}) \rightarrow N(0,V_\theta )$$ with $V$ as asymptotic variance and with $n \rightarrow \infty$.
Best Answer
Verify that for every $\theta>0$, $$E_{\theta}(X_i^2)=\operatorname{Var}_{\theta}(X_i)+[E_{\theta}(X_i)]^2=\frac2{\theta^2}=\mu_\theta\,,\,\text{say}$$
And $$\operatorname{Var}_{\theta}(X_i^2)=E_{\theta}(X_i^4)-[E_{\theta}(X_i^2)]^2=\frac{20}{\theta^4}=\sigma^2_\theta\,,\,\text{say}$$
Taking $Y_i=X_i^2$, you have by classical CLT $$\sqrt n\left(\overline Y-\mu_\theta\right) \stackrel{L}\longrightarrow N\left(0,\sigma^2_\theta\right)$$
Now use Delta-method with the transformation $g(x)=\sqrt\frac{x}{2}$ for $x\ge 0$, so that
$$\sqrt n\left(g(\overline Y)-g(\mu_\theta)\right)\stackrel{L}\longrightarrow N\left(0,[g'(\theta)]^2\sigma^2_{\theta}\right)\tag{*}$$
This $(*)$ would give you the limiting distribution of $g(\overline Y)=\sqrt{\frac{1}{2n}\sum\limits_{i=1}^n X_i^2}$.