If $X_isim U(0,theta)$, then there exists UMP test for $H_0:theta=theta_0$ vs $H_1:theta>theta_0$












0















Let $X_{1},dots,X_{n} sim Unif(0,theta)$ independently. Show there exists a UMP size $alpha$ test for testing $H_{0}:theta=theta_{0}$ vs $H_{1}:theta>theta_{0}$.




My attempt:



I attempted to show that the distribution is of increasing monotone likelihood ratio by computing $$frac{f(X_{1},dots, X_{n}midtheta_{1})}{f(X_{1}dots X_{n}midtheta_{2})}$$



This computation gave me: $$frac{f(X_{1},dots, X_{n}midtheta_{1})}{f(X_{1},dots ,X_{n}midtheta_{2})}=frac{theta_{2}^{n}I(X_{(n)}<theta_{1})}{theta_{1}^{n}I(X_{(n)}<theta_{2})}$$ This already confuses me because I'm not familiar with what happens when dividing indicator functions. Moreover, this does not seem to be a non-decreasing function of our test-statistic $t(x)=X_{(n)}$, which, according to my notes, means we cannot apply the Karlin-Rubin theorem for finding a UMP test. Since I get stuck here I cannot prove that there exists a UMP size $alpha$ test and I can also not find its form.



Question: What is going wrong in my approach above and how can I solve this question?



Thanks!






statistics probability-distributions statistical-inference uniform-distribution hypothesis-testing







share|cite|improve this question
























  • See math.stackexchange.com/questions/1736322/….
    – StubbornAtom
    Dec 3 '18 at 18:41
















0















Let $X_{1},dots,X_{n} sim Unif(0,theta)$ independently. Show there exists a UMP size $alpha$ test for testing $H_{0}:theta=theta_{0}$ vs $H_{1}:theta>theta_{0}$.




My attempt:



I attempted to show that the distribution is of increasing monotone likelihood ratio by computing $$frac{f(X_{1},dots, X_{n}midtheta_{1})}{f(X_{1}dots X_{n}midtheta_{2})}$$



This computation gave me: $$frac{f(X_{1},dots, X_{n}midtheta_{1})}{f(X_{1},dots ,X_{n}midtheta_{2})}=frac{theta_{2}^{n}I(X_{(n)}<theta_{1})}{theta_{1}^{n}I(X_{(n)}<theta_{2})}$$ This already confuses me because I'm not familiar with what happens when dividing indicator functions. Moreover, this does not seem to be a non-decreasing function of our test-statistic $t(x)=X_{(n)}$, which, according to my notes, means we cannot apply the Karlin-Rubin theorem for finding a UMP test. Since I get stuck here I cannot prove that there exists a UMP size $alpha$ test and I can also not find its form.



Question: What is going wrong in my approach above and how can I solve this question?



Thanks!






statistics probability-distributions statistical-inference uniform-distribution hypothesis-testing







share|cite|improve this question
























  • See math.stackexchange.com/questions/1736322/….
    – StubbornAtom
    Dec 3 '18 at 18:41














0












0








0


0






Let $X_{1},dots,X_{n} sim Unif(0,theta)$ independently. Show there exists a UMP size $alpha$ test for testing $H_{0}:theta=theta_{0}$ vs $H_{1}:theta>theta_{0}$.




My attempt:



I attempted to show that the distribution is of increasing monotone likelihood ratio by computing $$frac{f(X_{1},dots, X_{n}midtheta_{1})}{f(X_{1}dots X_{n}midtheta_{2})}$$



This computation gave me: $$frac{f(X_{1},dots, X_{n}midtheta_{1})}{f(X_{1},dots ,X_{n}midtheta_{2})}=frac{theta_{2}^{n}I(X_{(n)}<theta_{1})}{theta_{1}^{n}I(X_{(n)}<theta_{2})}$$ This already confuses me because I'm not familiar with what happens when dividing indicator functions. Moreover, this does not seem to be a non-decreasing function of our test-statistic $t(x)=X_{(n)}$, which, according to my notes, means we cannot apply the Karlin-Rubin theorem for finding a UMP test. Since I get stuck here I cannot prove that there exists a UMP size $alpha$ test and I can also not find its form.



Question: What is going wrong in my approach above and how can I solve this question?



Thanks!






statistics probability-distributions statistical-inference uniform-distribution hypothesis-testing







share|cite|improve this question
















Let $X_{1},dots,X_{n} sim Unif(0,theta)$ independently. Show there exists a UMP size $alpha$ test for testing $H_{0}:theta=theta_{0}$ vs $H_{1}:theta>theta_{0}$.




My attempt:



I attempted to show that the distribution is of increasing monotone likelihood ratio by computing $$frac{f(X_{1},dots, X_{n}midtheta_{1})}{f(X_{1}dots X_{n}midtheta_{2})}$$



This computation gave me: $$frac{f(X_{1},dots, X_{n}midtheta_{1})}{f(X_{1},dots ,X_{n}midtheta_{2})}=frac{theta_{2}^{n}I(X_{(n)}<theta_{1})}{theta_{1}^{n}I(X_{(n)}<theta_{2})}$$ This already confuses me because I'm not familiar with what happens when dividing indicator functions. Moreover, this does not seem to be a non-decreasing function of our test-statistic $t(x)=X_{(n)}$, which, according to my notes, means we cannot apply the Karlin-Rubin theorem for finding a UMP test. Since I get stuck here I cannot prove that there exists a UMP size $alpha$ test and I can also not find its form.



Question: What is going wrong in my approach above and how can I solve this question?



Thanks!






statistics probability-distributions statistical-inference uniform-distribution hypothesis-testing




statistics probability-distributions statistical-inference uniform-distribution hypothesis-testing






share|cite|improve this question















share|cite|improve this question













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edited Dec 4 '18 at 15:09









StubbornAtom

5,36411138




5,36411138










asked Dec 3 '18 at 15:47









S. Crim

14212




14212












  • See math.stackexchange.com/questions/1736322/….
    – StubbornAtom
    Dec 3 '18 at 18:41


















  • See math.stackexchange.com/questions/1736322/….
    – StubbornAtom
    Dec 3 '18 at 18:41
















See math.stackexchange.com/questions/1736322/….
– StubbornAtom
Dec 3 '18 at 18:41




See math.stackexchange.com/questions/1736322/….
– StubbornAtom
Dec 3 '18 at 18:41










1 Answer
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Just to elaborate on the division of indicator functions:



To test $H_0:theta=theta_0$ against $H_1:theta=theta_1(>theta_0)$ according to Neyman-Pearson lemma, note that
the likelihood ratio $lambda$ is of the form



begin{align}
lambda(x_1ldots,x_n)&=frac{f_{H_1}(x_1,ldots,x_n)}{f_{H_0}(x_1,ldots,x_n)}
\\&=left(frac{theta_0}{theta_1}right)^nfrac{mathbf1_{{x_{(n)}<theta_1}}}{mathbf1_{{x_{(n)}<theta_0}}}
\\&=begin{cases}left(frac{theta_0}{theta_1}right)^n&,text{ if }0<x_{(n)}<theta_0\,,, infty&,text{ if }theta_0< x_{(n)}<theta_1end{cases}
end{align}



So I think it follows from here that $lambda$ is a monotone non-decreasing function of $x_{(n)}$.



And by N-P lemma we know that an MP test of size $alpha$ is given by $$varphi(x_1,ldots,x_n)=mathbf1_{lambda(x_1,ldots,x_n)>k}$$, where $k$ is so chosen that $$E_{H_0}varphi(X_1,ldots,X_n)=alpha$$



Now you can proceed with your proof.






share|cite|improve this answer





















  • Could you go into more detail on why we can conside the function $lambda (x_{1}, dots, x_{n})$ a monotone non-decreasing function of $x_{n}$? I know the definition of such a function but am kind of lost as to how it follows directly from this representation of $lambda$. Thanks!
    – S. Crim
    Dec 5 '18 at 7:12






  • 1




    @S.Crim Roughly, plotting $lambda$ against $x_{(n)}$, we see that it takes a constant value when $0<x_{(n)}<theta_0$ and increases without bound when $theta_0<x_{(n)}<theta_1$.
    – StubbornAtom
    Dec 5 '18 at 12:56











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1 Answer
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Just to elaborate on the division of indicator functions:



To test $H_0:theta=theta_0$ against $H_1:theta=theta_1(>theta_0)$ according to Neyman-Pearson lemma, note that
the likelihood ratio $lambda$ is of the form



begin{align}
lambda(x_1ldots,x_n)&=frac{f_{H_1}(x_1,ldots,x_n)}{f_{H_0}(x_1,ldots,x_n)}
\\&=left(frac{theta_0}{theta_1}right)^nfrac{mathbf1_{{x_{(n)}<theta_1}}}{mathbf1_{{x_{(n)}<theta_0}}}
\\&=begin{cases}left(frac{theta_0}{theta_1}right)^n&,text{ if }0<x_{(n)}<theta_0\,,, infty&,text{ if }theta_0< x_{(n)}<theta_1end{cases}
end{align}



So I think it follows from here that $lambda$ is a monotone non-decreasing function of $x_{(n)}$.



And by N-P lemma we know that an MP test of size $alpha$ is given by $$varphi(x_1,ldots,x_n)=mathbf1_{lambda(x_1,ldots,x_n)>k}$$, where $k$ is so chosen that $$E_{H_0}varphi(X_1,ldots,X_n)=alpha$$



Now you can proceed with your proof.






share|cite|improve this answer





















  • Could you go into more detail on why we can conside the function $lambda (x_{1}, dots, x_{n})$ a monotone non-decreasing function of $x_{n}$? I know the definition of such a function but am kind of lost as to how it follows directly from this representation of $lambda$. Thanks!
    – S. Crim
    Dec 5 '18 at 7:12






  • 1




    @S.Crim Roughly, plotting $lambda$ against $x_{(n)}$, we see that it takes a constant value when $0<x_{(n)}<theta_0$ and increases without bound when $theta_0<x_{(n)}<theta_1$.
    – StubbornAtom
    Dec 5 '18 at 12:56
















1














Just to elaborate on the division of indicator functions:



To test $H_0:theta=theta_0$ against $H_1:theta=theta_1(>theta_0)$ according to Neyman-Pearson lemma, note that
the likelihood ratio $lambda$ is of the form



begin{align}
lambda(x_1ldots,x_n)&=frac{f_{H_1}(x_1,ldots,x_n)}{f_{H_0}(x_1,ldots,x_n)}
\\&=left(frac{theta_0}{theta_1}right)^nfrac{mathbf1_{{x_{(n)}<theta_1}}}{mathbf1_{{x_{(n)}<theta_0}}}
\\&=begin{cases}left(frac{theta_0}{theta_1}right)^n&,text{ if }0<x_{(n)}<theta_0\,,, infty&,text{ if }theta_0< x_{(n)}<theta_1end{cases}
end{align}



So I think it follows from here that $lambda$ is a monotone non-decreasing function of $x_{(n)}$.



And by N-P lemma we know that an MP test of size $alpha$ is given by $$varphi(x_1,ldots,x_n)=mathbf1_{lambda(x_1,ldots,x_n)>k}$$, where $k$ is so chosen that $$E_{H_0}varphi(X_1,ldots,X_n)=alpha$$



Now you can proceed with your proof.






share|cite|improve this answer





















  • Could you go into more detail on why we can conside the function $lambda (x_{1}, dots, x_{n})$ a monotone non-decreasing function of $x_{n}$? I know the definition of such a function but am kind of lost as to how it follows directly from this representation of $lambda$. Thanks!
    – S. Crim
    Dec 5 '18 at 7:12






  • 1




    @S.Crim Roughly, plotting $lambda$ against $x_{(n)}$, we see that it takes a constant value when $0<x_{(n)}<theta_0$ and increases without bound when $theta_0<x_{(n)}<theta_1$.
    – StubbornAtom
    Dec 5 '18 at 12:56














1












1








1






Just to elaborate on the division of indicator functions:



To test $H_0:theta=theta_0$ against $H_1:theta=theta_1(>theta_0)$ according to Neyman-Pearson lemma, note that
the likelihood ratio $lambda$ is of the form



begin{align}
lambda(x_1ldots,x_n)&=frac{f_{H_1}(x_1,ldots,x_n)}{f_{H_0}(x_1,ldots,x_n)}
\\&=left(frac{theta_0}{theta_1}right)^nfrac{mathbf1_{{x_{(n)}<theta_1}}}{mathbf1_{{x_{(n)}<theta_0}}}
\\&=begin{cases}left(frac{theta_0}{theta_1}right)^n&,text{ if }0<x_{(n)}<theta_0\,,, infty&,text{ if }theta_0< x_{(n)}<theta_1end{cases}
end{align}



So I think it follows from here that $lambda$ is a monotone non-decreasing function of $x_{(n)}$.



And by N-P lemma we know that an MP test of size $alpha$ is given by $$varphi(x_1,ldots,x_n)=mathbf1_{lambda(x_1,ldots,x_n)>k}$$, where $k$ is so chosen that $$E_{H_0}varphi(X_1,ldots,X_n)=alpha$$



Now you can proceed with your proof.






share|cite|improve this answer












Just to elaborate on the division of indicator functions:



To test $H_0:theta=theta_0$ against $H_1:theta=theta_1(>theta_0)$ according to Neyman-Pearson lemma, note that
the likelihood ratio $lambda$ is of the form



begin{align}
lambda(x_1ldots,x_n)&=frac{f_{H_1}(x_1,ldots,x_n)}{f_{H_0}(x_1,ldots,x_n)}
\\&=left(frac{theta_0}{theta_1}right)^nfrac{mathbf1_{{x_{(n)}<theta_1}}}{mathbf1_{{x_{(n)}<theta_0}}}
\\&=begin{cases}left(frac{theta_0}{theta_1}right)^n&,text{ if }0<x_{(n)}<theta_0\,,, infty&,text{ if }theta_0< x_{(n)}<theta_1end{cases}
end{align}



So I think it follows from here that $lambda$ is a monotone non-decreasing function of $x_{(n)}$.



And by N-P lemma we know that an MP test of size $alpha$ is given by $$varphi(x_1,ldots,x_n)=mathbf1_{lambda(x_1,ldots,x_n)>k}$$, where $k$ is so chosen that $$E_{H_0}varphi(X_1,ldots,X_n)=alpha$$



Now you can proceed with your proof.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 3 '18 at 19:39









StubbornAtom

5,36411138




5,36411138












  • Could you go into more detail on why we can conside the function $lambda (x_{1}, dots, x_{n})$ a monotone non-decreasing function of $x_{n}$? I know the definition of such a function but am kind of lost as to how it follows directly from this representation of $lambda$. Thanks!
    – S. Crim
    Dec 5 '18 at 7:12






  • 1




    @S.Crim Roughly, plotting $lambda$ against $x_{(n)}$, we see that it takes a constant value when $0<x_{(n)}<theta_0$ and increases without bound when $theta_0<x_{(n)}<theta_1$.
    – StubbornAtom
    Dec 5 '18 at 12:56


















  • Could you go into more detail on why we can conside the function $lambda (x_{1}, dots, x_{n})$ a monotone non-decreasing function of $x_{n}$? I know the definition of such a function but am kind of lost as to how it follows directly from this representation of $lambda$. Thanks!
    – S. Crim
    Dec 5 '18 at 7:12






  • 1




    @S.Crim Roughly, plotting $lambda$ against $x_{(n)}$, we see that it takes a constant value when $0<x_{(n)}<theta_0$ and increases without bound when $theta_0<x_{(n)}<theta_1$.
    – StubbornAtom
    Dec 5 '18 at 12:56
















Could you go into more detail on why we can conside the function $lambda (x_{1}, dots, x_{n})$ a monotone non-decreasing function of $x_{n}$? I know the definition of such a function but am kind of lost as to how it follows directly from this representation of $lambda$. Thanks!
– S. Crim
Dec 5 '18 at 7:12




Could you go into more detail on why we can conside the function $lambda (x_{1}, dots, x_{n})$ a monotone non-decreasing function of $x_{n}$? I know the definition of such a function but am kind of lost as to how it follows directly from this representation of $lambda$. Thanks!
– S. Crim
Dec 5 '18 at 7:12




1




1




@S.Crim Roughly, plotting $lambda$ against $x_{(n)}$, we see that it takes a constant value when $0<x_{(n)}<theta_0$ and increases without bound when $theta_0<x_{(n)}<theta_1$.
– StubbornAtom
Dec 5 '18 at 12:56




@S.Crim Roughly, plotting $lambda$ against $x_{(n)}$, we see that it takes a constant value when $0<x_{(n)}<theta_0$ and increases without bound when $theta_0<x_{(n)}<theta_1$.
– StubbornAtom
Dec 5 '18 at 12:56


















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