# Rejection region in hypothesis testing based on asymptotic distribution

## Hypothesis testing: finding a rejection region

Let $$X_1, \ldots, X_n$$ be i.i.d. distributed with $$f_\theta (x) = \begin{cases} e^{\theta-x}, & x\ge\theta, \\ 0, & \text{elsewhere} \end{cases}$$

1. We have found the maximum likelihood estimator $$\hat{\theta}_n = \min_{i=1}^n\{X_i\}$$.

2. The cdf of $$\hat{\theta}_n$$ is $$1-e^{n(\theta - x)} \textbf{1}_{x \geq \theta}$$

3. We have shown that $$\hat{θ}_n$$ is consistent.

4. The asymptotic distribution of $$n(\hat{θ}_n -\theta)$$ is Exp(1).

For some fixed $$n$$, it is decided to use $$T = \hat{θ}_n$$ as the test statistic for testing

$$H_0 : θ ≤ θ_0,~~~~H_1 : θ > θ_0$$

Determine the rejection region $$R = [r, ∞)$$ for $$T$$ based on the asymptotic distribution of $$n(\hat{θ}_n -\theta)$$, Exp(1) with significance level $$α$$.

I was thinking about doing something with confidence intervals, but I'm not sure if it will help. Just feeling very lost.

• In fact Exp(1) is the exact distribution of $n(\hat\theta_n-\theta)$. Oct 21, 2019 at 18:05

With $$T(X)=\min\limits_{1\le i\le n} X_i$$, we have the exact distribution $$n(T-\theta)\sim \mathsf{Exp}(1)$$, which is the same as$$2n(T-\theta)\sim \chi^2_2\tag{*}$$

There is more than one way to derive a test for testing $$H_0:\theta\le\theta_0$$ against $$H_1:\theta>\theta_0$$.

One elementary way is to find the test corresponding to a given confidence interval for $$\theta$$ based on $$T$$.

Using the pivot in $$(*)$$, we have $$P_{\theta}\left(2n(T-\theta)< \chi^2_{2,\alpha}\right)=1-\alpha\quad\forall\,\theta\in\mathbb R\,,$$

where $$\chi^2_{2,\alpha}$$ denotes the $$(1-\alpha)$$th quantile of a $$\chi^2_2$$ distribution.

Or, $$P_{\theta}\left(\theta> T-\frac{\chi^2_{2,\alpha}}{2n}\right)=1-\alpha\quad,\forall\,\theta\tag{**}$$

This says that a (one-sided) $$100(1-\alpha)\%$$ confidence interval for $$\theta$$ is $$I(X)=\left(T-\frac{\chi^2_{2,\alpha}}{2n},\infty\right)$$

From $$(**)$$, we can say that for some $$\theta_0$$,

$$P_{\theta_0}\left(\theta_0< T-\frac{\chi^2_{2,\alpha}}{2n}\right)=P_{\theta_0}\left(T>\theta_0+\frac{\chi^2_{2,\alpha}}{2n}\right)=\alpha$$

So the size $$\alpha$$ test obtained by 'inverting' the interval $$I$$ rejects $$H_0$$ whenever $$T>\theta_0+\frac{\chi^2_{2,\alpha}}{2n}$$.

It can be shown that this test is in fact the likelihood ratio test and also the uniformly most powerful test (via this result) for testing $$H_0$$ against $$H_1$$.