For which $alpha$-sizes is there a MP level-$alpha$ test?












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Let $X_1,...,X_n$ be idd and uniformly distributed over $(0,theta)$. Choosing between the null hypothesis which states that $theta=theta_0$ and the alternate hypothesis which states that $theta=theta_1$, with $theta_1<theta_0$, suppose we reject the null if and only if $x_{n:n}<c$ for some fixed, positive and known number $c$.



My question asks: for which values of $alpha$ does there exist a MP level-$alpha$ test?



By the Neyman-Pearson Lemma, I will reject the null for sufficiently large values of $L(x,theta_0)/L(x,theta_1)=(frac{theta_0}{theta_1})^n*I(x_{n:n}<theta_1)/I(x_{n:}<theta_0)$. However, this expression can only take two values for a particular sample $x$: either the likelihood ratio is $1$, or the likelihood ratio is $0$.



My question is, does this mean that there are only two values of $alpha$ for which this test can be MP level-$alpha$? I would find this strange, since I'm pretty sure that for any $alpha$ between $0$ and $1$, we can find a value $c$ to make the test level-$alpha$.






statistics statistical-inference hypothesis-testing







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    Let $X_1,...,X_n$ be idd and uniformly distributed over $(0,theta)$. Choosing between the null hypothesis which states that $theta=theta_0$ and the alternate hypothesis which states that $theta=theta_1$, with $theta_1<theta_0$, suppose we reject the null if and only if $x_{n:n}<c$ for some fixed, positive and known number $c$.



    My question asks: for which values of $alpha$ does there exist a MP level-$alpha$ test?



    By the Neyman-Pearson Lemma, I will reject the null for sufficiently large values of $L(x,theta_0)/L(x,theta_1)=(frac{theta_0}{theta_1})^n*I(x_{n:n}<theta_1)/I(x_{n:}<theta_0)$. However, this expression can only take two values for a particular sample $x$: either the likelihood ratio is $1$, or the likelihood ratio is $0$.



    My question is, does this mean that there are only two values of $alpha$ for which this test can be MP level-$alpha$? I would find this strange, since I'm pretty sure that for any $alpha$ between $0$ and $1$, we can find a value $c$ to make the test level-$alpha$.






    statistics statistical-inference hypothesis-testing







    share|cite|improve this question



























      0












      0








      0


      1





      Let $X_1,...,X_n$ be idd and uniformly distributed over $(0,theta)$. Choosing between the null hypothesis which states that $theta=theta_0$ and the alternate hypothesis which states that $theta=theta_1$, with $theta_1<theta_0$, suppose we reject the null if and only if $x_{n:n}<c$ for some fixed, positive and known number $c$.



      My question asks: for which values of $alpha$ does there exist a MP level-$alpha$ test?



      By the Neyman-Pearson Lemma, I will reject the null for sufficiently large values of $L(x,theta_0)/L(x,theta_1)=(frac{theta_0}{theta_1})^n*I(x_{n:n}<theta_1)/I(x_{n:}<theta_0)$. However, this expression can only take two values for a particular sample $x$: either the likelihood ratio is $1$, or the likelihood ratio is $0$.



      My question is, does this mean that there are only two values of $alpha$ for which this test can be MP level-$alpha$? I would find this strange, since I'm pretty sure that for any $alpha$ between $0$ and $1$, we can find a value $c$ to make the test level-$alpha$.






      statistics statistical-inference hypothesis-testing







      share|cite|improve this question















      Let $X_1,...,X_n$ be idd and uniformly distributed over $(0,theta)$. Choosing between the null hypothesis which states that $theta=theta_0$ and the alternate hypothesis which states that $theta=theta_1$, with $theta_1<theta_0$, suppose we reject the null if and only if $x_{n:n}<c$ for some fixed, positive and known number $c$.



      My question asks: for which values of $alpha$ does there exist a MP level-$alpha$ test?



      By the Neyman-Pearson Lemma, I will reject the null for sufficiently large values of $L(x,theta_0)/L(x,theta_1)=(frac{theta_0}{theta_1})^n*I(x_{n:n}<theta_1)/I(x_{n:}<theta_0)$. However, this expression can only take two values for a particular sample $x$: either the likelihood ratio is $1$, or the likelihood ratio is $0$.



      My question is, does this mean that there are only two values of $alpha$ for which this test can be MP level-$alpha$? I would find this strange, since I'm pretty sure that for any $alpha$ between $0$ and $1$, we can find a value $c$ to make the test level-$alpha$.






      statistics statistical-inference hypothesis-testing




      statistics statistical-inference hypothesis-testing






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      edited Nov 27 at 15:48

























      asked Nov 27 at 12:17









      DavidS

      337111




      337111






















          1 Answer
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          The test should be a function of the minimal sufficient statistic, namely $X_{(n)} = max{X_1,...,X_n}$, such that if $X_{(n)} > theta_1$ then you reject $H_0$ with $alpha = 0$ and where $X_{(n)} le theta_1$ then you should set a rejection region $c$.
          $$
          alpha = mathbb{E}_{theta_0}Psi (x) = mathbb{P}(Psi (x) = 1) = mathbb{P}( X_{(n)} le c)=F_X(c)=left(frac{c}{theta_0}right)^n,
          $$
          namely, for $X_{(n)} le theta_0$, reject $H_0$ if
          $$
          X_{(n)} le a^{1/n} theta_0.
          $$






          share|cite|improve this answer





















          • Thanks for the response. So is there any alpha for which no MP level alpha test exists? And why?
            – DavidS
            Dec 1 at 12:16










          • This test holds for any $alpha in (0,1)$.
            – V. Vancak
            Dec 1 at 12:21










          • I would find that very strange given the question, not to say that you aren’t correct.
            – DavidS
            Dec 1 at 12:43











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          The test should be a function of the minimal sufficient statistic, namely $X_{(n)} = max{X_1,...,X_n}$, such that if $X_{(n)} > theta_1$ then you reject $H_0$ with $alpha = 0$ and where $X_{(n)} le theta_1$ then you should set a rejection region $c$.
          $$
          alpha = mathbb{E}_{theta_0}Psi (x) = mathbb{P}(Psi (x) = 1) = mathbb{P}( X_{(n)} le c)=F_X(c)=left(frac{c}{theta_0}right)^n,
          $$
          namely, for $X_{(n)} le theta_0$, reject $H_0$ if
          $$
          X_{(n)} le a^{1/n} theta_0.
          $$






          share|cite|improve this answer





















          • Thanks for the response. So is there any alpha for which no MP level alpha test exists? And why?
            – DavidS
            Dec 1 at 12:16










          • This test holds for any $alpha in (0,1)$.
            – V. Vancak
            Dec 1 at 12:21










          • I would find that very strange given the question, not to say that you aren’t correct.
            – DavidS
            Dec 1 at 12:43
















          1














          The test should be a function of the minimal sufficient statistic, namely $X_{(n)} = max{X_1,...,X_n}$, such that if $X_{(n)} > theta_1$ then you reject $H_0$ with $alpha = 0$ and where $X_{(n)} le theta_1$ then you should set a rejection region $c$.
          $$
          alpha = mathbb{E}_{theta_0}Psi (x) = mathbb{P}(Psi (x) = 1) = mathbb{P}( X_{(n)} le c)=F_X(c)=left(frac{c}{theta_0}right)^n,
          $$
          namely, for $X_{(n)} le theta_0$, reject $H_0$ if
          $$
          X_{(n)} le a^{1/n} theta_0.
          $$






          share|cite|improve this answer





















          • Thanks for the response. So is there any alpha for which no MP level alpha test exists? And why?
            – DavidS
            Dec 1 at 12:16










          • This test holds for any $alpha in (0,1)$.
            – V. Vancak
            Dec 1 at 12:21










          • I would find that very strange given the question, not to say that you aren’t correct.
            – DavidS
            Dec 1 at 12:43














          1












          1








          1






          The test should be a function of the minimal sufficient statistic, namely $X_{(n)} = max{X_1,...,X_n}$, such that if $X_{(n)} > theta_1$ then you reject $H_0$ with $alpha = 0$ and where $X_{(n)} le theta_1$ then you should set a rejection region $c$.
          $$
          alpha = mathbb{E}_{theta_0}Psi (x) = mathbb{P}(Psi (x) = 1) = mathbb{P}( X_{(n)} le c)=F_X(c)=left(frac{c}{theta_0}right)^n,
          $$
          namely, for $X_{(n)} le theta_0$, reject $H_0$ if
          $$
          X_{(n)} le a^{1/n} theta_0.
          $$






          share|cite|improve this answer












          The test should be a function of the minimal sufficient statistic, namely $X_{(n)} = max{X_1,...,X_n}$, such that if $X_{(n)} > theta_1$ then you reject $H_0$ with $alpha = 0$ and where $X_{(n)} le theta_1$ then you should set a rejection region $c$.
          $$
          alpha = mathbb{E}_{theta_0}Psi (x) = mathbb{P}(Psi (x) = 1) = mathbb{P}( X_{(n)} le c)=F_X(c)=left(frac{c}{theta_0}right)^n,
          $$
          namely, for $X_{(n)} le theta_0$, reject $H_0$ if
          $$
          X_{(n)} le a^{1/n} theta_0.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 at 23:40









          V. Vancak

          10.8k2926




          10.8k2926












          • Thanks for the response. So is there any alpha for which no MP level alpha test exists? And why?
            – DavidS
            Dec 1 at 12:16










          • This test holds for any $alpha in (0,1)$.
            – V. Vancak
            Dec 1 at 12:21










          • I would find that very strange given the question, not to say that you aren’t correct.
            – DavidS
            Dec 1 at 12:43


















          • Thanks for the response. So is there any alpha for which no MP level alpha test exists? And why?
            – DavidS
            Dec 1 at 12:16










          • This test holds for any $alpha in (0,1)$.
            – V. Vancak
            Dec 1 at 12:21










          • I would find that very strange given the question, not to say that you aren’t correct.
            – DavidS
            Dec 1 at 12:43
















          Thanks for the response. So is there any alpha for which no MP level alpha test exists? And why?
          – DavidS
          Dec 1 at 12:16




          Thanks for the response. So is there any alpha for which no MP level alpha test exists? And why?
          – DavidS
          Dec 1 at 12:16












          This test holds for any $alpha in (0,1)$.
          – V. Vancak
          Dec 1 at 12:21




          This test holds for any $alpha in (0,1)$.
          – V. Vancak
          Dec 1 at 12:21












          I would find that very strange given the question, not to say that you aren’t correct.
          – DavidS
          Dec 1 at 12:43




          I would find that very strange given the question, not to say that you aren’t correct.
          – DavidS
          Dec 1 at 12:43


















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