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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.

Michael Hardy
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Mariana
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1 Answers1

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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 $$C = 2 - \frac{1}{\sqrt{10}}.$$

heropup
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  • I don't know why $\Pr[X_1 + X_2 > 1] = \Pr[X_1 + X_2 < 1] $ – Mariana Mar 31 '21 at 13:45
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    @Yichuan If $X$ is uniform on $(0,1)$, then $1-X$ is also uniform on $(0,1)$. Draw a picture. If $(X_1, X_2)$ is a point in the unit square, what portion of the square satisfies $X_1 + X_2 < 1$ and what portion satisfies $X_1 + X_2 > 1$? – heropup Mar 31 '21 at 20:10
  • Yes, now I understand. TY! – Mariana Apr 01 '21 at 16:37