Proving a statistic is ancillary












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Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|alpha)=alpha x^{alpha-1}e^{-x^alpha}$, $x>0$, $alpha>0$. Show that $frac{log X_1}{log X_2}$ is an ancillary statistic.



I guess I need to show that the distribution of this statistic is independent of $alpha$, i.e. it is the same as if $alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.










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  • So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
    – StubbornAtom
    Dec 28 '18 at 15:09
















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Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|alpha)=alpha x^{alpha-1}e^{-x^alpha}$, $x>0$, $alpha>0$. Show that $frac{log X_1}{log X_2}$ is an ancillary statistic.



I guess I need to show that the distribution of this statistic is independent of $alpha$, i.e. it is the same as if $alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.










share|cite|improve this question






















  • So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
    – StubbornAtom
    Dec 28 '18 at 15:09














1












1








1







Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|alpha)=alpha x^{alpha-1}e^{-x^alpha}$, $x>0$, $alpha>0$. Show that $frac{log X_1}{log X_2}$ is an ancillary statistic.



I guess I need to show that the distribution of this statistic is independent of $alpha$, i.e. it is the same as if $alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.










share|cite|improve this question













Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|alpha)=alpha x^{alpha-1}e^{-x^alpha}$, $x>0$, $alpha>0$. Show that $frac{log X_1}{log X_2}$ is an ancillary statistic.



I guess I need to show that the distribution of this statistic is independent of $alpha$, i.e. it is the same as if $alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.







statistics statistical-inference






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asked Dec 28 '18 at 14:49









Analysis801Analysis801

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  • So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
    – StubbornAtom
    Dec 28 '18 at 15:09


















  • So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
    – StubbornAtom
    Dec 28 '18 at 15:09
















So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
– StubbornAtom
Dec 28 '18 at 15:09




So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
– StubbornAtom
Dec 28 '18 at 15:09










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If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.





So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.



Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.






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    If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.





    So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.



    Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.






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      If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.





      So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.



      Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.






      share|cite|improve this answer


























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        If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.





        So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.



        Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.






        share|cite|improve this answer














        If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.





        So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.



        Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.







        share|cite|improve this answer














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        edited Dec 28 '18 at 15:23

























        answered Dec 28 '18 at 15:14









        drhabdrhab

        98.4k544129




        98.4k544129






























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