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Suppose that we are given an exponential distribution model with a pdf $f(x,\theta) = \theta^{-1}\exp(-x/\theta)$ with an iid sample $X_1, \ldots, X_n$, and we would like to test hypothesis $H_0 : \theta = 1$ and $H_1 : \theta > 1$. We shall investigate the limiting distribution of the likelihood test statistic $2n(l_n (\hat\theta) - l_n (1))$ where $l_n(\theta)$ is the log-likelihood function $\sum \log f(x,\theta$) and $\hat \theta$ is the MLE of $\theta$ in the parameter space. We shall investigate the asymptotic behavior of the test statistic.

Note that the parameter space is not the whole real line, so it does not create an open neighborhood around $H_0: \theta = 1$). Here is what I have discovered so far: we know that the MLE of exponential distribution is $\overline X_n$, the sample mean if we are allowed to take the whole parameter space. Using a similar argument, if the sample mean is greater than one, the same argument works. However, if it is smaller than or equal to one, the MLE is 1 (we can show it using a monotonicity argument). Hence, the test statistic becomes

$$\begin{cases} 2n(\ln \overline X_n - (\overline X_n - 1)) && \text{if $\overline X_n > 1$} \\ 0 && \text{if $\overline X_n \leq 1$} \end{cases}$$

we can use then a taylor expansion for the first case to show that it is indeed approximately equal to $n(\overline X_n - 1)^2$ that has a $\chi^2$-distribution asymptotically. However, how can we incorporate the case where $\overline X_n \leq 1$? If $H_1$ were $\theta \neq 1$ it would have been safe and sound. I guess the test becomes $\sum \overline X_n > k$ for some $k$, due to the zero part for the second case?

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In the case where you observe $\bar{x}_n \leqslant 1$ you have test statistic $t = 0$ and the corresponding p-value for your test is $p=1$. This means that there is no evidence to reject the null (i.e., it is not rejected at any significance level), which is as it should be.

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