uniformly most powerful test: binomial distribution Let $X$ $\sim \mathrm{Bin}(2,p)$ a binomial distributed random variable. Give a uniformly most powerful test with size $\alpha \in (0,1)$ for the hypotheses $H_0:p=\frac{1}{2}$ vs. $H_1:p=\frac{3}{4}$. Especially determine the critical value and the randomization constant.
Since "great" values of $X$ imply that the null hypotheses is not true, the test is constructed as a right tailed test.
So we have:
$\Theta=\{\frac{1}{2},\frac{3}{4}\}, \Theta_0=\{\frac{1}{2}\}, \Theta_1=\{\frac{3}{4}\}$
I know that
$\mathbb P(X\ge2)=\mathbb P(X=2)=\frac{1}{4}$
$\mathbb P(X\ge1)=1-\mathbb P(X=0)=\frac{3}{4}$
$\mathbb P(X\ge 0)=1$
Since there is no specific $\alpha$ given here, I have no idea what is actually to do here. I think that if f.e. $\alpha=0.3$, the critical value would be $2$ and if $\alpha=0.8$, the critical value would be $1$. Is that true?
Anyway, I don't know how to solve this task. Can anyone help me with that?
 A: For the test with reject area $\{X\geq C\}$, if the critical value is $2$, then test size is $\mathbb P_{H_0}(X\geq 2)=\frac14\neq 0.3$. And for critical value $1$, the size is $\mathbb P_{H_0}(X\geq 1)=\frac34\neq 0.8$. 
Note that if we consider the test with reject area  $\{X\geq C\}$, then the size of this test can only be $0$, $1/4$, $3/4$ and $1$. If we want to construct the test with size $\alpha$ for any $\alpha\in(0,1)$, we should consider randomized test: 
Always reject $H_0$ when $X>c$, always accept $H_0$ when $X <c$, and reject $H_0$ with a certain probability $p$
when $X=c$, where this probability is chosen to make the size of the test exactly $\alpha$. 
The values of critical value $c$ and randomization probability $p$ are uniquelly determined by the equality
$$
\alpha=\mathbb P_{H_0}(H_0 \textit{ is rejected}\,)=\mathbb P_{H_0}(X>c)+p\mathbb P_{H_0}(X=c).
$$
This equation shows that we can restrict ourselves with $c\in\{0,1,2\}$ only. Then the first summand in this equation can be either $3/4$ or $1/4$ or $0$: 
$$\mathbb P_{H_0}(X>0)=3/4, \quad \mathbb P_{H_0}(X>1)=1/4, \quad \mathbb P_{H_0}(X>2)=0.$$
Say, let $\alpha=0.3\in(1/4,\,3/4)$. Then $c=1$ and randomization probability $p$ can be found as: 
$$
0.3=\mathbb P_{H_0}(X>1)+p\mathbb P_{H_0}(X=1)=\frac14+p\cdot\frac12,\quad p=\frac{1}{10}.
$$
More generally, for $\alpha\in(1/4,3/4)$ one need to take $c=1$ and then find that $p=2\alpha-\frac12$:
$$
\alpha=\mathbb P_{H_0}(X>1)+p\mathbb P_{H_0}(X=1)=\frac14+p\cdot\frac12.
$$
For $\alpha\in(0,1/4)$ we should take $c=2$ and the first summand $\mathbb P_{H_0}(X>2)$ equals zero. Then 
$$
\alpha=p \mathbb P_{H_0}(X=2) = p\cdot \frac14, \quad p=4\alpha.
$$
Consider the last case  $\alpha\in(3/4,1)$ and find $c$ and $p$. 
Also look at $\alpha=1/4$ or $\alpha=3/4$. For this values the test does not need to be randomized. Really, the test with reject area $\{X\geq 2\}$ has the size exactly $\alpha=1/4$, and the test with reject area $\{X\geq 1\}$ has the size exactly $\alpha=3/4$. But both these tests are the variants of randomized tests with some $c$ and $p=0\vee 1$. 
