Let $X$ be a random variable having probability mass function
$f(x) = \begin{cases} \dfrac{2+4a_1+a_2}{6}, & \text{if $x=1$} \\ \dfrac{2-2a_1+a_2}{6}, & \text{if $x=2$} \\ \dfrac{1-a_1-a_2}{3}, & \text{if $x=3$} \end{cases}$
where $a_1>0$ and $a_2>0$ are unknown parameters such that $a_1+a_2\leq 1$. For testing the null hypothesis $H_0:a_1+a_2=1$, against the alternative hypothesis $H_1:a_1=a_2=0$, suppose that the critical region is $C=\{2,3\}$. Then calculate the size and power for the critical region.
Power=$1-\beta$ where $\beta=P(\text{accept }H_0\mid H_1\text{ is true})=\dfrac{2+4a_1+a_2}{6}=\dfrac{1}{3}$
So Power = 2/3
Now, size $= P(\text{reject }H_0\mid H_0\text{ is true})=\text{critical region} = \dfrac{2-2a_1+a_2}{6}+\dfrac{1-a_1-a_2}{3}$.
So, how do I calculate a numerical value here ? Please advise.
The size of the test is the upper bound of the probability of rejecting $H_0$ when $H_0$ is true, over the set of all parameter values satisfying $H_0$. That is to say, the size is $$\sup_{a_1+a_2 = 1} 1 - \frac{2 + 4a_1 + a_2}{6}.$$ It is not difficult to show that for $a_1, a_2 > 0$, this is simply $1/2$.