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Exercise: Let $X$ be a random variable that takes values in $\{-1,0,1\}$. We want to test $$H_0: Pr(X = -1) = Pr(X = 1) = Pr(X = 0) = 1/3$$ versus $$H_1: Pr(X = -1) = Pr(X = 1) = 1/4, \, Pr(X = 0) = 1/2.$$

Show that the uniformly most powerful (UMP) test of size $\alpha = 1/3$ based on $X$ leads to rejecting the null hypothesis when $X = 0$, and compute the power of the test.

What I've tried: I first need to find the critical region $C$ for the UMP test based on the test statistic $T(X) = X$. This can be done in two steps:

1) Choose $C$ such that under $H_0$, $P_\theta(T(X)\in C)\leq \alpha$ (this ensures that the probability of a type I error is no greater than $\alpha$). I choose $C = \{0\}$.

2) Conditional on 1), maximise $P(T(X)\in C)$ under $H_1$ (this ensures minimising the probability a type II error). With my choice of $C = \{0\}$, $P(T(X)\in C)$ is already maximised.

So I have a test $(T,C)$ with $T(X) = X$ and $C = \{0\}$. Since $X = 0$, $X\in C$ and $H_0$ is rejected. The power of the test is equal to the probability of rejecting $H_0$ when $H_1$ is correct. Hence the power of the test is equal to $1/2$.

Question: Is my answer correct? If not; what am I doing wrong? The thing that I'm most uncertain about in my approach is whether or not the region $C$ that I derived is the critical region of a UMP test.

Thanks in advance!

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  • $\begingroup$ @AlexFrancisco Thanks edited. $\endgroup$
    – titusAdam
    Jan 19, 2018 at 11:49

1 Answer 1

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Looks right. I'm not $100\%$ percent about the technical details. But you can check the answer. In order to keep the $\alpha$, i.e., the probability of erroneously rejecting $H_0$ to be equal $1/3$, you must choose one of the values $\{-1,0,1\}$ as a rejection region. Now, you can simply check what would be the power for every value, i.e., the probability of getting the value under $H_1$, that is $\{1/4, 1/2, 1/4\}$ respectively, so the maximal power $(1/2)$ is attained for $C=\{0\}$.

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  • $\begingroup$ I hope I can revive this question a bit because something is unclear to me. It's about the original question but the poster is not active anymore on the website so maybe you can help me. I'm confused about the line after 2) ... . He states that with this choise of $C=\{0\}$, the power function is already maximised. Is this true though? What if you were to pick $C=\{0,1\}$ or $C=\{-1,0,1\}$? Then your power function would take an even higher value, right? There must be something that I'm missing here. Any help is appreciated. $\endgroup$ Dec 3, 2018 at 13:07
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    $\begingroup$ @S.Crim You are right, but in this case the $\alpha$ will be $2/3$ and $1$, respectively. $\endgroup$
    – V. Vancak
    Dec 3, 2018 at 18:32

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