Suppose that $X_{1}, X_{2}$, and $X_{3}$ are independent identically distributed Poisson random variables with expectation $\theta$, so that

$$\mathbb{P}\left(X_{i}=x\right)=\frac{e^{-\theta} \theta^{x}}{x !} \quad x=0,1, \ldots$$

and consider testing $H_{0}: \theta=1$ against $H_{1}: \theta=\theta_{1}$, where $\theta_{1}$ is a known value greater than 1. Show that the test with critical region $\left\{\left(x_{1}, x_{2}, x_{3}\right): \sum_{i=1}^{3} x_{i}>5\right\}$ is a likelihood ratio test of $H_{0}$ against $H_{1}$. What is the size of this test? Write down an expression for its power.

A scientist counts the number of bird territories in $n$ randomly selected sections of a large park. Let $Y_{i}$ be the number of bird territories in the $i$ th section, and suppose that $Y_{1}, \ldots, Y_{n}$ are independent Poisson random variables with expectations $\theta_{1}, \ldots, \theta_{n}$ respectively. Let $a_{i}$ be the area of the $i$ th section. Suppose that $n=2 m$, $a_{1}=\cdots=a_{m}=a(>0)$ and $a_{m+1}=\cdots=a_{2 m}=2 a$. Derive the generalised likelihood ratio $\Lambda$ for testing

$$H_{0}: \theta_{i}=\lambda a_{i} \text { against } H_{1}: \theta_{i}= \begin{cases}\lambda_{1} & i=1, \ldots, m \\ \lambda_{2} & i=m+1, \ldots, 2 m\end{cases}$$

What should the scientist conclude about the number of bird territories if $2 \log _{e}(\Lambda)$ is $15.67 ?$

[Hint: Let $F_{\theta}(x)$ be $\mathbb{P}(W \leqslant x)$ where $W$ has a Poisson distribution with expectation $\theta$. Then

$$\left.F_{1}(3)=0.998, \quad F_{3}(5)=0.916, \quad F_{3}(6)=0.966, \quad F_{5}(3)=0.433 .\right]$$