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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.

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

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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 $H_1$.

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