Let $X_1,X_2$ be iid uniform ($\theta,\theta+1$). For testing $H_0:\theta=0$ versus $H_1:\theta>0$, we have two competing tests:
$$\phi_1(X_1): \text{Reject H}_0 \text{ if }X_1>.95,$$
$$\phi_2(X_1,X_2): \text{Reject H}_0 \text{ if }X_1+X_2>C.$$
Find the value of $C$ so that $\phi_2$ has the same size as $\phi_1$.
Attempt:
Size for $\phi_1$ is $\alpha_1 = P(X_1 > .95\mid\theta=0)=.05$. This is easy, since under $H_0$, we have a uniform (0,1) distribution.
Size for $\phi_2$ is $\alpha_2 = P(X_1 +X_2 > C\mid\theta=0)=.05$ I don't know how to solve this probability (integral). I will appreciate detailed steps.
Best Answer
If $\theta = 0$, then $X_1$, $X_2$ are iid Uniform on $(0,1)$. We require $$0.05 = \Pr[X_1 + X_2 > C].$$ Note that $1 < C < 2$. This is because if $C = 1$, by symmetry, $$\Pr[X_1 + X_2 > 1] = \Pr[(1 - X_1) + (1 - X_2) > 1] = \Pr[X_1 + X_2 < 1] = 1/2 > 0.05,$$ and if $C = 2$, obviously this probability is $0$, which is too small.
So if we imagine $(X_1, X_2)$ being plotted on a coordinate plane, then the support is the unit square $[0,1]^2$. The set of points in this square that satisfy the inequality $X_1 + X_2 > C$ is given by a triangle whose area is $0.05$ and has vertices at $$(X_1, X_2) \in \{ (C - 1, 1), (1,1), (1, C - 1) \}.$$ Therefore, we require $$0.05 = \frac{(1 - (C-1))^2}{2} = \frac{(2-C)^2}{2},$$ or $$C = 2 - \frac{1}{\sqrt{10}}.$$