# Why here we choose $T<k$ as the rejection region in the $\alpha$-level uniformly most powerful test?

Let $$X_i$$ be iid random sample from $$exp(\lambda)$$ with $$f(x;\lambda)=\lambda e^{-\lambda x}$$ for $$x>0$$ and $$\lambda>0$$. Find the $$\alpha$$-level uniformly most powerful test for $$H_0: \lambda\le \lambda_0$$ v.s. $$H_1: \lambda> \lambda_0$$.

I try to use the Karlin-Rubin theorem as follows. Under $$H_0$$, we take $$T=2\lambda \sum_i X_i\sim Gamma(n ,2)=\chi^2(2n)$$. So I take the test function of UMPT: $$$$\Phi(x)= \begin{cases} 1 & \text{if } T>\chi^2_{1-\alpha}(2n)\\ 0 & \text{if } T\le \chi^2_{1-\alpha}(2n) \end{cases}$$$$

But the solution used $$$$\Phi(x)= \begin{cases} 1 & \text{if } T\le \chi^2_{1-\alpha}(2n)\\ 0 & \text{if } T> \chi^2_{1-\alpha}(2n) \end{cases}$$$$

I am confused about why here we choose $$T as the rejection region.

So you agree with the solution on what the two sets are, but you disagree on which one goes with $$H_0$$ and which one goes with $$H_1$$.
• Under $$H_1$$, the rate parameter is larger than it is under $$H_0$$. A larger rate means a smaller mean, so $$X$$ tends to be smaller under $$H_1$$ than under $$H_0$$. -Your statistic $$T$$ is be larger when $$X$$ is larger and smaller when $$X$$ is smaller, so it will tend to be smaller under $$H_1$$ than under $$H_0$$.
• So the rejection region is the one with small $$T$$ and the non-rejection region is the one with larger $$T$$.
The loglikelihood is $$-\lambda X$$ (up to constants), and I think you probably dropped the minus sign in your derivation.