Let $X_i\sim \text{Exp}(\theta)$ for $i=1,\dots,n$ i.i.d with density \begin{equation*} f(x,\theta) = \begin{cases} \theta \exp(-x\theta ),& x\geq 0\\ 0,& x< 0. \end{cases} \end{equation*} I want to test the hypothesis \begin{equation*} H_0: \theta = \theta_0 \quad \text{vs} \quad H_1: \theta = \theta_1 \quad \text{with}\quad \theta_1 > \theta_0 \end{equation*} The Neyman-Pearson lemma provides a UMP test at level $\alpha$ of the form \begin{equation} \varphi^*(X) = \begin{cases} 1, & \hspace{0.25cm} f_1(X) > k f_0(X),\\ 0, & \hspace{0.25cm} f_1(X) < k f_0(X). \end{cases} \end{equation}
For some critical value $k$. The densities are absolutely continious, therefore we can look at the likelihood ratio. Since they are i.i.d this is just the product of the respective densities and we get \begin{equation*} \frac{f(X,\theta_1)}{f(X,\theta_0)}=\frac{\theta_1^{n}\exp(-\theta_1\sum_{i=1}^n x_i )}{\theta_0\exp(-\theta_0\sum_{i=1}^n x_i )} = \frac{\theta_1^n}{\theta_0^n}\exp\left(-(\theta_1-\theta_0)\sum_{i=1}^n x_i\right)>k. \end{equation*} My problem arises, when I rearrange this term to get an equivalent condition. \begin{align*} \frac{\theta_1^n}{\theta_0^n}\exp\left(-(\theta_1-\theta_0)\sum_{i=1}^n x_i\right)&>k \\ n\log(\theta_1/\theta_0)-(\theta_1-\theta_0)\sum_{i=1}^n x_i&>k'\\ -(\theta_1-\theta_0)\sum_{i=1}^n x_i&>k'-n\log(\theta_1/\theta_0)\\ -\frac{(\theta_1-\theta_0)}{n}\sum_{i=1}^n x_i&>k'/n-\log(\theta_1/\theta_0)\\ \frac{1}{n}\sum_{i=1}^n x_i&<\frac{k'/n-\log(\theta_1/\theta_0)}{(\theta_1-\theta_0)}=:k''.\\ \end{align*} The inequality switches in the last step since we assume $\theta_1>\theta_0$. However this seems odd. Using some probability theory we calculate $k''$ for $\alpha=0.05$. Since the sum of i.i.d exponential r.v. is $Gamma(n,\theta)$ and $Gamma(n,1/2) = \chi^2(2n)$ this results in the condition \begin{equation*}\alpha = \mathcal{P}_{\theta_0}(\overline{X}_n <k'') = \mathcal{P}_{\theta_0}(2\theta_0n\overline{X}_n <2\theta_0nk''). \end{equation*} Thus we get $2\theta_0nk''= \chi^2_{1-\alpha}(2n)$, where the subscript denotes the $1-\alpha$ quantile. Therefore, $$ k''=\frac{\chi^2_{1-\alpha}(2n)}{2\theta_0n}.$$
The reason why the condition $\overline{X}_n <k''$ seems odd is the following simulation for a sample of 15 exponential r.v. with parameter $\theta = 1$, where i test $\theta_0 = 1$ vs $\theta_1 = 2$. The likehood ratio process goes very nicely to $0$, as one would expect with how i chose the values. However the condition for the sample mean would reject the null in favor of the paramter $\theta_1$, since the sample mean is lower than the critical value $k''$ (in the image discription it says k'). This cannot be right, therefore I assume an error when calculating $k''.$
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