I have two simple hypotheses:
$$ H_0: \ \theta = \theta_0 \\ H_1: \ \theta = \theta_1 $$
Let $p_0$ be the density under $H_0$ and $p_1$ the density under $H_1$.
Then we defined a Neyman-Pearson test to be of the form:
$$ \phi_{NP}(X):=\begin{cases} 1, \quad \frac{p_1(X)}{p_0(X)}>c_0 \\ q, \quad \frac{p_1(X)}{p_0(X)}=c_0 \\ 0, \quad \frac{p_1(X)}{p_0(X)}<c_0 \end{cases} $$ Where $c_0 \geq 0, q \in [0,1]$. Then our version of the Neyman-Pearson lemma tells us that for any given level $\alpha \in (0,1)$ if $\mathbb{E}_{\theta_0}[\phi_{NP}]=\alpha$ then for all randomised tests $\tilde{\phi}$ with $\mathbb{E}_{\theta_0}[\tilde\phi] \leq \alpha$ it holds that $\mathbb{E}_{\theta_1}[\tilde\phi] \leq \mathbb{E}_{\theta_1}[\phi_{NP}]$
In my examples I‘m now asked to use this to construct the NP-test for two poisson distributions with unknown instensity $\lambda$, i.e. $p_0(x)=\frac{1}{x!}e^{-\lambda_0}\lambda_0^x, p_1(x)=\frac{1}{x!}e^{-\lambda_1}\lambda_1^x, x=0,1,2,\dots $ Now my solution manual states that we can always choose $c_0$ to be the $(1-\alpha)$-quantile of the distribution of $\frac{p_1(X)}{p_0(X)}$ under $H_0$ and then choose $q$ s.t.: $$ \mathbb{P}(\frac{p_1(X)}{p_0(X)}>c_0)+q \mathbb{P}(\frac{p_1(X)}{p_0(X)}=c_0)=\alpha $$ The latter of course makes sense, but why choose $c_0$ as this quantile?
Then they calculate explicitly: $$ \frac{p_1(X)}{p_0(X)}= \frac{p_1(x_1,\dots,x_n)}{p_1(x_1,\dots,x_n)}=e^{n(\lambda_1-\lambda_0)}(\frac{\lambda_1}{\lambda_0})^{\sum_{i=1}^n x_i}=g(\sum_{i=1}^n x_i) $$
Now because $g$ is strictly increasing the test can also be given as:
$$ \phi_{NP}(X):=\begin{cases} 1, \quad \sum_{i=1}^n X_i >t_0 \\ q, \quad \sum_{i=1}^n X_i =t_0 \\ 0, \quad \sum_{i=1}^n X_i <t_0 \end{cases} $$
where $t_0$ now simply is the $(1-\alpha)$-quantile of the distribution $\sum_{i=1}^n X_i $ under $H_0$. But I dont get why we would have $c_0=(1-\alpha)$-quantile of $\frac{p_1(X)}{p_0(X)} \ \Leftrightarrow \ t_0=(1-\alpha)$-quantile of $\sum_{i=1}^n X_i $?
So in summary my questions are:
- Why choose $c_0$ as this quantile, couldn’t we achieve the desired mean value in the lemma by choosing $c_0$ anything and then just choosing $q$ appropriately?
- I get that the $>c_0$ condition can be replaced by the $>t_0$ condition because of the monotonicity of $g$ but why would $c_0=(1-\alpha)$-quantile of $\frac{p_1(X)}{p_0(X)} \ \Leftrightarrow \ t_0=(1-\alpha)$-quantile of $\sum_{i=1}^n X_i $? (If this even is the case)
- Does the strategy above work in all simple hypotheses cases?