Suppose X has distribution in $\mathcal{P}=\{B(10, \frac{1}{2}), P(1)\}$ (Binomial and Poisson respectively). Let $H_0 := X \stackrel{D}{=} B(10, \frac{1}{2})$ and $H_1 := X \stackrel{D}{=}P(1)$. Derive an UMP test for $H_0$ with $\alpha = 0.1$
There is also a hint to use tables of Poisson or Binomial distribution.
I know $\alpha$ = $\mathbb{P}_0(f_1(x) > cf_0(x)) + \gamma \mathbb{P}_0(f_1(x)=cf_0(x)$.
I calculated $f_1(x) > c f_0(x)$ as $$ \frac{1}{e x!} > c \frac{10!}{(10-x)!x!}2^{-10} \\ \frac{2^{10}}{e} > c 10^{\underline{x}} \\ \frac{2^{10}}{e 10^{\underline{x}}} > c $$
When I plug that in $$ \alpha = \mathbb{P}_0(f_1(x) > cf_0(x)) + \gamma \mathbb{P}_0(f_1(x)=cf_0(x) = \\ = \mathbb{P}_0(\frac{2^{10}}{e 10^{\underline{x}}} > c) + \gamma \mathbb{P}_0(\frac{2^{10}}{e 10^{\underline{x}}} = c) $$ Here I got stuck. I know that I need to assume the $H_0$ is true so $X \stackrel{D}{=} B(10, \frac{1}{2})$ but I have $10^{\underline{X}}$ here. Surely I must be missing something. I usually used CLT here but I have no idea how to do it here. Thank you for any insights.
You are going to want to use the Neyman–Pearson lemma
It should be obvious that if $x \gt 10$ then $\mathbb P(X=x \mid H_0)=0$ and $\mathbb P(X=x \mid H_1)>0$, so the likelihood ratio is then zero
For $0 \le x \le 10$, the likelihood ratio is $\dfrac{\frac{10!}{(10-x)!x!}2^{-10} }{e^{-1}\frac{1^x}{x!}} = \dfrac{10!2^{-10}e}{(10-x)!}$, which is an increasing function of $x$.
So for a UMP test, you are going to reject $H_0$ when $X$ is strictly greater than $10$ or strictly smaller than some value $k$ when $\mathbb P(k \le X=x \le 10 \mid H_0)$ is at least $1-\alpha=0.9$, where $k$ is the largest possible value subject to that constraint. I suspect $k=3$; depending on what you have been taught, you might or might not want to do something special when $X=3$