Uniformly Most Powerful test (UMP) (for Poisson random variables)

Question

Consider testing $H_0:\theta \leq \theta_0$ against $H_1:\theta>\theta_0$. Let $0<\alpha<1$ be given. Suppose $X_1,...,X_n$ are i.i.d. Poisson distributed with parameter $\theta>0$: for all $i\in\{ 1,...,n\}$

\begin{align*} P_\theta(X_i=x)=e^{-\theta}\frac{\theta^x}{x!}, \; x\in\{ 0,1,... \}. \end{align*}

I want to find the UMP (Uniformly Most Powerful) test $\phi$ at level $\alpha$.

My attempt: I know the following thm.: Suppose $\mathcal{P}$ is a 1-d exponential family, i.e. $p_\theta(x)=\exp(c(\theta)T(x)-d(\theta))h(x)$ and assume $c(\theta)$ is trictly increasing, then a UMP test is \begin{align*} \Phi(T(x)) : \begin{cases} 1 \quad ,T(x)>c \\ q \quad, T(x)=c\\ 0 \quad, T(x)<c, \end{cases} \end{align*} where $q$ and $c$ are chosen such that $\mathbb{E}_{\theta_0}[\phi(T)]=\alpha$.

So, since $p_\theta=e^{-\theta}\frac{\theta^x}{x!}=\frac{1}{x!}\exp(x\ln(\theta)-\theta)$ is an exp-familiy with $c(\theta)=ln(\theta)$ increasing, the thm. gives a UMP test an we need to find $q$ and $c$:

\begin{align*} \alpha=\mathbb{E}_{\theta_0}[\Phi(T)]=\mathbb{P}_{\theta_0}[X>c]+q\mathbb{P}_{\theta_0}[X=c] \quad \text{with} \quad X\sim\text{Poisson}(\theta). \end{align*}

Is this correct so far? And how do i get the values $c$ and $q$ from here?

Answer 1

I learned how to do this in a different way and idk if it helps you but,

you can form the likelihood ratio, $$\Lambda = \frac{l(\theta_1)}{l(\theta)}$$

We take, $\theta$ to be the largest value which satisfies $H_0$, and $\theta_1$ to be some value that satisfies $H_1$ and define the test,

$$\Lambda > k$$ for some constant k

$$\to \Pi_{i =0}^n e^{-\theta_1+\theta} (\frac{\theta_1}{\theta})^{x_i} > k$$

$$\to e^{-n\theta_1+n\theta} (\frac{\theta_1}{\theta})^{\sum_{i =0}^n x_i} > k$$

then we can take the ln of both sides since the log is order preserving, $$\to -n\theta_1+n\theta + ln(\frac{\theta_1}{\theta}) \sum_{i =0}^n x_i > k_2$$ $$\to ln(\frac{\theta_1}{\theta}) \sum_{i =0}^n x_i > k_3$$ $$\to \sum_{i =0}^n x_i > k_4$$ now we just need $$p_{\theta}(\sum_{i =0}^n x_i > k_4) =\alpha $$