Asymptotic test statistic for exponentially distributed data

Question

I need an idea for tackling the following problem: Let $X_1,...,X_n\sim\mathrm{Exp}(\lambda)$ be an iid sample. We want to test $H_0:\lambda=\lambda_0$ vs. $H_1:\lambda\not=\lambda_0$ for some prescribed $\lambda_0>0$ using the statistic $T_n:=g(\lambda_0\overline{X}_n)$, where $g(u):=u\exp(1-u)$. Therefore, one wants to find - depending on some $\alpha\in(0,1)$ - $C_n$ and $D_n$, s.t. under $H_0$ we have: $\mathbb{P}(T_n<C_n)\to\frac{\alpha}{2}$ and $\mathbb{P}(T_n>D_n)\to\frac{\alpha}{2}$ as $n\to\infty$. Does someone have an idea how to choose these $C_n$ and $D_n$? Thank you in advance!

Answer 1

Since $\sqrt{n}\left(\lambda_0\bar X_n-1\right)\xrightarrow{d}\mathcal{N}\left(0,1\right)$, you could try the Delta Method:

$$\sqrt{n}\left[g(\lambda_0\bar X_n)-g(1) \right]\xrightarrow{d}\mathcal{N}\left(0,[g'(1)]^2\right)$$

Now, $g'(u)= \exp(1-u)-u\exp(1-u)=(1-u)\exp(1-u) \implies g'(1)=0$, which means that this method will not work for $\lambda_0 \bar X_n$, since $T_n\xrightarrow{p} 0$.

However, we can use the next higher order of the delta method to get:

$$\frac{n\left[g(\lambda_0\bar X_n)-g(1) \right]}{g''(1)} \xrightarrow{d} \chi^2_1 $$

Where $g''(u)=\exp(1-u)(u-2)\implies g''(1)=-1$ (which makes sense, since $g(1)$ is the local maximum, so $\left[g(\lambda_0\bar X_n)-g(1) \right]\leq 0$ and we keep the ratio positive). Simplifying a little, we get:

$$n\left[1-T_n\right]\xrightarrow{d} \chi^2_1$$

For a given $\alpha \in (0,1)$, let $U_n=\chi^2_{1,1-\frac{\alpha}{2}},L_n=\chi^2_{1,\frac{\alpha}{2}}$, then

$$P(n\left[1-T_n\right]<U_n)\text{ and }P(n\left[1-T_n\right]>L_n) \to \frac{\alpha}{2}$$

Isolating $T_n$, we get:

$$P\left(T_n>1-\frac{U_n}{n}\right)\text{ and }P\left(T_n<1-\frac{L_n}{n}\right) \to \frac{\alpha}{2}$$

Thus: $C_n=1-\frac{L_n}{n}, \text{ and } D_n=1-\frac{U_n}{n}$