Problem:
Let $(X,Y)$ be a random sample from the Poisson distribution with unknown mean $\theta$. In order to test $H_0:\theta=1$ against $H_1:\theta >1$, the following procedure is suggested:
If $X\leq 1$, then accept $H_0$.
If $X\geq 4$, then reject $H_0$.
If $X$ equals $2$ or $3$, then reject $H_0$ if $X+Y\geq 4$, otherwise accept $H_0$.
Find the size of the test.
My attempt:
$X$ and $Y$ follow $Poisson(1)$ distribution independently.
Size of the test
$=P_{H_0}(X\geq 4)+P_{H_0}(X=2, X+Y\geq 4)+P_{H_0}(X=3, X+Y\geq 4)$
$=P_{H_0}(X\geq 4)+P_{H_0}(X=2, Y\geq 2)+P_{H_0}(X=3, Y\geq 1)$
$=P_{H_0}(X\geq 4)+P_{H_0}(X=2)(Y\geq 2)+P_{H_0}(X=3)(Y\geq 1)$
$=[1-\sum_{x=0}^{3}e^{-1}\frac{1}{x!}]+\frac{e^{-1}}{2}[1-\sum_{y=0}^{1}e^{-1}\frac{1}{x!}]+\frac{e^{-1}}{6}[1-e^{-1}]$
$=0.10635$
I want to know if I have done it correctly, or otherwise, I want to know where the mistake is. Thanks in advance.
Yes, your answer is correct. The size of the test is the conditional probability of rejecting $H_0$ given that $H_0$ is true. Since $H_0$ is rejected in exactly two mutually exclusive cases:
the latter of which can be further broken apart into the two sub-cases you used, then the desired conditional probability is $$\Pr[X \ge 4 \mid H_0] + \Pr[(X = 2) \cap (Y \ge 2) \mid H_0] + \Pr[(X = 3) \cap (Y \ge 1) \mid H_0]$$ which is, apart from minor notational differences, is the same as what you wrote.
As an exercise, how would you calculate the power curve for this test? In other words, given $H_1 : \theta = \theta_1$ for some $\theta_1 \ne 1$, what is the power of this test to reject $H_0$ as a function of $\theta_1$?