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

Question

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. $H_1: \lambda> \lambda_0$.

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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: \begin{equation} \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} \end{equation}

But the solution used \begin{equation} \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} \end{equation}

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

Answer 1

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.