# Likelihood ratio test for two-parameter exponential distribution

I am considering a random sample $$X_1, \ldots, X_m; \ m \geq 2$$, from a 2-parameter exponential distribution with pdf $$f_X(x; \mu, \sigma) = \frac{1}{\sigma} \exp \left( -\frac{x-\mu}{\sigma} \right) I(x \geq \mu),$$

where $$I$$ is an indicator function, $$\mu \in \mathbb{R}$$ and $$\sigma > 0$$. I got that the unconstrained MLE of the parameters is $$\hat{\mu} = X_{(1)} = \min\{X_1, \ldots, X_m\}; \ \hat{\sigma} = \sum_{i=1}^m \left(x_i - X_{(1)}\right).$$

I am interested in testing the hypothesis $$H_0: \sigma = \sigma_0$$ vs. $$H_1: \sigma \neq \sigma_0$$ for some constant value $$\sigma_0$$ using a likelihood ratio test. In this case the constrained MLE over $$H_0$$ is $$\hat{\mu}_0 = X_{(1)}; \ \hat{\sigma}_0 = \sigma_0.$$

Then the likelihood ratio is $$\Lambda = \frac{(\sigma_0)^{-m}\exp\left( -\frac{1}{\sigma_0}\sum_{i=1}^m\left(X_i - X_{(1)}\right) \right)}{(\hat{\sigma})^{-m}\exp\left(-\frac{1}{\hat{\sigma}}\sum_{i=1}^m\left(X_i - X_{(1)}\right)\right)}$$ which I got to simplify to $$\Lambda = \left( \frac{\hat{\sigma}}{\sigma_0} \right)^m \exp \left( -\frac{1}{\sigma_0}\sum_{i=1}^m\left(X_i - X_{(1)}\right)-m \right).$$

However I am not sure that this is right because even if I set $$\Lambda \leq c$$

I have no idea what to do next or how to get a quantity with a known distribution. What do I do for the LRT? I cannot appeal to the asymptotic result as I have to use this test to get a confidence interval for $$\sigma$$.

Consider $$Z = \frac{1}{\sigma_0} \sum (X_i - X_{(1)})$$, then: $$\Lambda = e^m Z^{m} e^{-Z}$$
Note that $$g(z) = z^me^{-z} = e^{m \log z - z}$$ has the following shape: Therefore, $$\Lambda < c$$ for some $$c$$ if and only if $$Z < c_1$$ or $$Z > c_2$$ for some $$c_1$$ and $$c_2$$. Therefore, the problem is reduced to finding $$c_1$$ and $$c_2$$ such that: $$P(Z < c_1) + P(Z > c_2) = \alpha, \qquad \text{under }\sigma = \sigma_0$$ Now, under the null hypothesis: $$Z = \frac{1}{\sigma_0} \sum (X_i - X_{(1)}) = \frac{1}{\sigma_0} \sum[ (X_i - a) - (X_{(1)} - a)]$$ Therefore $$Z \sim \Gamma(n-1,1)$$, i.e. $$2Z \sim \chi^2_{2(n-1)}$$. Now all we need to do is to find appropriate quanitles $$a,b$$ of the chi squared distribution such that we reject the null hypothese if $$2Z < a$$ or $$2Z > b$$.