Aysmptotic variance of an estimator of the parameter of an exponential distribution

Question

---

---

There is a hint: you may use the expression for gamma function. But I don't know how to use it since it the sum of squares of exponential random variables. I computed the expression of the asymptotic variance in an extremely complicated way. The result is $\frac 1{2\theta^2}~$ but I don't know if its correct. Does anyone know how to use the hint? Thank you.

Answer 1

The canonical way is to use the Delta method. However, it is sometimes possible to apply tricks that make calculations easier. And this is just such a case if you are familiar with the Slutsky's theorem. Denote $\overline {X^2}=\frac{\sum_{i=1}^n X_i^2}{n}$ $$ \sqrt{n}\left(T_2-\frac1\theta\right)=\sqrt{n}\left(\sqrt{\frac12\overline {X^2}}-\frac1\theta\right) $$ Multiply and divide by the conjugate: $$ =\sqrt{n}\left(\frac12\overline{X^2}-\frac1{\theta^2}\right)\frac{1}{\sqrt{\frac12\overline {X^2}}+\frac1\theta} = \sqrt{n}\left(\overline{X^2}-\frac2{\theta^2}\right)\frac{1}{2\sqrt{\frac12\overline {X^2}}+\frac2\theta} $$ $$ =\sqrt{n}\left(\overline{X^2}-\mathbb E[X_1^2]\right) \cdot \frac{1}{2T_2+\frac2\theta} $$ Consider multipliers separately: $$ \sqrt{n}\left(\overline{X^2}-\mathbb E[X_1^2]\right) \xrightarrow{\mathcal D_\theta} \mathcal N(0, \text{Var}(X_1^2)). $$ $$ T_2 \xrightarrow{p} \frac1\theta, \quad\text{ then } \quad \frac{1}{2T_2+\frac2\theta}\xrightarrow{p}\frac\theta4. $$

Next, Slutsky's theorem implies that the product converges in distribution to $$ \mathcal N(0, \left(\frac\theta4\right)^2 \cdot \text{Var}(X_1^2)). $$

You need to calculate $\text{Var}(X_1^2)$, and for it you definitelyneed to use Gamma function.