Finding UMVUE for uniform distribution $U(alpha, beta)$












1












$begingroup$


Let $X = (X_1, X_2, ldots, X_n)$ be a sample from uniform distribution $U(alpha, beta): alpha, beta in mathbb{R}, alpha < beta$. I am to find UMVUE for the parameters $alpha, beta$.



Using factorization theorem I showed that $T(X) = (min{ X_1, X_2, ldots, X_2}, max{ X_1, X_2, ldots, X_n })$ is a sufficient statistics.



I think that I should use Lehmann–Scheffé theorem. Am I to solve this problem fixing $beta$ and then computing UMVUE for $alpha$? Then vive versa?



I tried to find an unbiased estimator for $alpha$ first. Thinking of $beta$ as fixed I managed to calculate that
$$mathbb{E}(2X_1 - beta) = alpha.$$
Thus using Lehmann–Scheffé theorem UMVUE would be
$$mathbb{E}(2X_1 - beta|T(X)).$$
How can I find conditional expected value when $T(X)$ is a vector?



On the other hand I fixed $beta$ in order to find my unbiased estimator so should I calculate $mathbb{E}(2X_1 - beta|X_{(1)})$?



I am a bit confused. I would appreciate any hints or tips.






statistics estimation







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $X = (X_1, X_2, ldots, X_n)$ be a sample from uniform distribution $U(alpha, beta): alpha, beta in mathbb{R}, alpha < beta$. I am to find UMVUE for the parameters $alpha, beta$.



    Using factorization theorem I showed that $T(X) = (min{ X_1, X_2, ldots, X_2}, max{ X_1, X_2, ldots, X_n })$ is a sufficient statistics.



    I think that I should use Lehmann–Scheffé theorem. Am I to solve this problem fixing $beta$ and then computing UMVUE for $alpha$? Then vive versa?



    I tried to find an unbiased estimator for $alpha$ first. Thinking of $beta$ as fixed I managed to calculate that
    $$mathbb{E}(2X_1 - beta) = alpha.$$
    Thus using Lehmann–Scheffé theorem UMVUE would be
    $$mathbb{E}(2X_1 - beta|T(X)).$$
    How can I find conditional expected value when $T(X)$ is a vector?



    On the other hand I fixed $beta$ in order to find my unbiased estimator so should I calculate $mathbb{E}(2X_1 - beta|X_{(1)})$?



    I am a bit confused. I would appreciate any hints or tips.






    statistics estimation







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $X = (X_1, X_2, ldots, X_n)$ be a sample from uniform distribution $U(alpha, beta): alpha, beta in mathbb{R}, alpha < beta$. I am to find UMVUE for the parameters $alpha, beta$.



      Using factorization theorem I showed that $T(X) = (min{ X_1, X_2, ldots, X_2}, max{ X_1, X_2, ldots, X_n })$ is a sufficient statistics.



      I think that I should use Lehmann–Scheffé theorem. Am I to solve this problem fixing $beta$ and then computing UMVUE for $alpha$? Then vive versa?



      I tried to find an unbiased estimator for $alpha$ first. Thinking of $beta$ as fixed I managed to calculate that
      $$mathbb{E}(2X_1 - beta) = alpha.$$
      Thus using Lehmann–Scheffé theorem UMVUE would be
      $$mathbb{E}(2X_1 - beta|T(X)).$$
      How can I find conditional expected value when $T(X)$ is a vector?



      On the other hand I fixed $beta$ in order to find my unbiased estimator so should I calculate $mathbb{E}(2X_1 - beta|X_{(1)})$?



      I am a bit confused. I would appreciate any hints or tips.






      statistics estimation







      share|cite|improve this question











      $endgroup$




      Let $X = (X_1, X_2, ldots, X_n)$ be a sample from uniform distribution $U(alpha, beta): alpha, beta in mathbb{R}, alpha < beta$. I am to find UMVUE for the parameters $alpha, beta$.



      Using factorization theorem I showed that $T(X) = (min{ X_1, X_2, ldots, X_2}, max{ X_1, X_2, ldots, X_n })$ is a sufficient statistics.



      I think that I should use Lehmann–Scheffé theorem. Am I to solve this problem fixing $beta$ and then computing UMVUE for $alpha$? Then vive versa?



      I tried to find an unbiased estimator for $alpha$ first. Thinking of $beta$ as fixed I managed to calculate that
      $$mathbb{E}(2X_1 - beta) = alpha.$$
      Thus using Lehmann–Scheffé theorem UMVUE would be
      $$mathbb{E}(2X_1 - beta|T(X)).$$
      How can I find conditional expected value when $T(X)$ is a vector?



      On the other hand I fixed $beta$ in order to find my unbiased estimator so should I calculate $mathbb{E}(2X_1 - beta|X_{(1)})$?



      I am a bit confused. I would appreciate any hints or tips.






      statistics estimation




      statistics estimation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 13 at 18:25







      Hendrra

















      asked Jan 13 at 18:17









      HendrraHendrra

      1,214516




      1,214516






















          1 Answer
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          1












          $begingroup$

          $T=(X_{(1)},X_{(n)})$ is not only sufficient but a complete sufficient statistic which is needed here, $X_{(k)}$ being the $k$th $,(1le kle n)$ order statistic.



          Since the $X_i$'s are i.i.d $U(a,b)$ variables, $Y_i=frac{X_i-a}{b-a}$ are i.i.d $U(0,1)$ variables, $, 1le ile n$.



          Now it is well-known that $Y_{(1)}sim text{Beta}(1,n)$ and $Y_{(n)}simtext{Beta}(n,1)$, implying $E(Y_{(1)})=frac{1}{n+1}$ and $E(Y_{(n)})=frac{n}{n+1}$. So all you have to do is solve for $a$ and $b$ from the equations



          $$E(X_{(1)})=frac{b-a}{n+1}+a\ E(X_{(n)})=frac{(b-a)n}{n+1}+a$$



          You would get $a$ and $b$ as unbiased estimators of some function of $T$, and those will be the corresponding UMVUEs by Lehmann-Scheffe theorem.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does that mean that my UMVUE will be a vector too?
            $endgroup$
            – Hendrra
            Jan 13 at 20:19






          • 1




            $begingroup$
            @Hendrra No, the UMVUEs are linear combinations of $X_{(1)}$ and $X_{(n)}$.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:21










          • $begingroup$
            One for b and one for a?
            $endgroup$
            – Hendrra
            Jan 13 at 20:24






          • 1




            $begingroup$
            @Hendrra If you have found an unbiased estimator of your parameter based on a complete sufficient statistic, then that is bound to be the UMVUE and it equals the conditional expectation ( the latter is sometimes difficult to compute). This follows from Lehmann-Scheffe.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:46






          • 1




            $begingroup$
            @Hendrra The linear combination part is irrelevant (it's only here that the UMVUEs are linear combinations, not in general).
            $endgroup$
            – StubbornAtom
            Jan 13 at 21:01












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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          $T=(X_{(1)},X_{(n)})$ is not only sufficient but a complete sufficient statistic which is needed here, $X_{(k)}$ being the $k$th $,(1le kle n)$ order statistic.



          Since the $X_i$'s are i.i.d $U(a,b)$ variables, $Y_i=frac{X_i-a}{b-a}$ are i.i.d $U(0,1)$ variables, $, 1le ile n$.



          Now it is well-known that $Y_{(1)}sim text{Beta}(1,n)$ and $Y_{(n)}simtext{Beta}(n,1)$, implying $E(Y_{(1)})=frac{1}{n+1}$ and $E(Y_{(n)})=frac{n}{n+1}$. So all you have to do is solve for $a$ and $b$ from the equations



          $$E(X_{(1)})=frac{b-a}{n+1}+a\ E(X_{(n)})=frac{(b-a)n}{n+1}+a$$



          You would get $a$ and $b$ as unbiased estimators of some function of $T$, and those will be the corresponding UMVUEs by Lehmann-Scheffe theorem.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does that mean that my UMVUE will be a vector too?
            $endgroup$
            – Hendrra
            Jan 13 at 20:19






          • 1




            $begingroup$
            @Hendrra No, the UMVUEs are linear combinations of $X_{(1)}$ and $X_{(n)}$.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:21










          • $begingroup$
            One for b and one for a?
            $endgroup$
            – Hendrra
            Jan 13 at 20:24






          • 1




            $begingroup$
            @Hendrra If you have found an unbiased estimator of your parameter based on a complete sufficient statistic, then that is bound to be the UMVUE and it equals the conditional expectation ( the latter is sometimes difficult to compute). This follows from Lehmann-Scheffe.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:46






          • 1




            $begingroup$
            @Hendrra The linear combination part is irrelevant (it's only here that the UMVUEs are linear combinations, not in general).
            $endgroup$
            – StubbornAtom
            Jan 13 at 21:01
















          1












          $begingroup$

          $T=(X_{(1)},X_{(n)})$ is not only sufficient but a complete sufficient statistic which is needed here, $X_{(k)}$ being the $k$th $,(1le kle n)$ order statistic.



          Since the $X_i$'s are i.i.d $U(a,b)$ variables, $Y_i=frac{X_i-a}{b-a}$ are i.i.d $U(0,1)$ variables, $, 1le ile n$.



          Now it is well-known that $Y_{(1)}sim text{Beta}(1,n)$ and $Y_{(n)}simtext{Beta}(n,1)$, implying $E(Y_{(1)})=frac{1}{n+1}$ and $E(Y_{(n)})=frac{n}{n+1}$. So all you have to do is solve for $a$ and $b$ from the equations



          $$E(X_{(1)})=frac{b-a}{n+1}+a\ E(X_{(n)})=frac{(b-a)n}{n+1}+a$$



          You would get $a$ and $b$ as unbiased estimators of some function of $T$, and those will be the corresponding UMVUEs by Lehmann-Scheffe theorem.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does that mean that my UMVUE will be a vector too?
            $endgroup$
            – Hendrra
            Jan 13 at 20:19






          • 1




            $begingroup$
            @Hendrra No, the UMVUEs are linear combinations of $X_{(1)}$ and $X_{(n)}$.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:21










          • $begingroup$
            One for b and one for a?
            $endgroup$
            – Hendrra
            Jan 13 at 20:24






          • 1




            $begingroup$
            @Hendrra If you have found an unbiased estimator of your parameter based on a complete sufficient statistic, then that is bound to be the UMVUE and it equals the conditional expectation ( the latter is sometimes difficult to compute). This follows from Lehmann-Scheffe.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:46






          • 1




            $begingroup$
            @Hendrra The linear combination part is irrelevant (it's only here that the UMVUEs are linear combinations, not in general).
            $endgroup$
            – StubbornAtom
            Jan 13 at 21:01














          1












          1








          1





          $begingroup$

          $T=(X_{(1)},X_{(n)})$ is not only sufficient but a complete sufficient statistic which is needed here, $X_{(k)}$ being the $k$th $,(1le kle n)$ order statistic.



          Since the $X_i$'s are i.i.d $U(a,b)$ variables, $Y_i=frac{X_i-a}{b-a}$ are i.i.d $U(0,1)$ variables, $, 1le ile n$.



          Now it is well-known that $Y_{(1)}sim text{Beta}(1,n)$ and $Y_{(n)}simtext{Beta}(n,1)$, implying $E(Y_{(1)})=frac{1}{n+1}$ and $E(Y_{(n)})=frac{n}{n+1}$. So all you have to do is solve for $a$ and $b$ from the equations



          $$E(X_{(1)})=frac{b-a}{n+1}+a\ E(X_{(n)})=frac{(b-a)n}{n+1}+a$$



          You would get $a$ and $b$ as unbiased estimators of some function of $T$, and those will be the corresponding UMVUEs by Lehmann-Scheffe theorem.






          share|cite|improve this answer









          $endgroup$



          $T=(X_{(1)},X_{(n)})$ is not only sufficient but a complete sufficient statistic which is needed here, $X_{(k)}$ being the $k$th $,(1le kle n)$ order statistic.



          Since the $X_i$'s are i.i.d $U(a,b)$ variables, $Y_i=frac{X_i-a}{b-a}$ are i.i.d $U(0,1)$ variables, $, 1le ile n$.



          Now it is well-known that $Y_{(1)}sim text{Beta}(1,n)$ and $Y_{(n)}simtext{Beta}(n,1)$, implying $E(Y_{(1)})=frac{1}{n+1}$ and $E(Y_{(n)})=frac{n}{n+1}$. So all you have to do is solve for $a$ and $b$ from the equations



          $$E(X_{(1)})=frac{b-a}{n+1}+a\ E(X_{(n)})=frac{(b-a)n}{n+1}+a$$



          You would get $a$ and $b$ as unbiased estimators of some function of $T$, and those will be the corresponding UMVUEs by Lehmann-Scheffe theorem.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 19:08









          StubbornAtomStubbornAtom

          6,45431440




          6,45431440












          • $begingroup$
            Does that mean that my UMVUE will be a vector too?
            $endgroup$
            – Hendrra
            Jan 13 at 20:19






          • 1




            $begingroup$
            @Hendrra No, the UMVUEs are linear combinations of $X_{(1)}$ and $X_{(n)}$.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:21










          • $begingroup$
            One for b and one for a?
            $endgroup$
            – Hendrra
            Jan 13 at 20:24






          • 1




            $begingroup$
            @Hendrra If you have found an unbiased estimator of your parameter based on a complete sufficient statistic, then that is bound to be the UMVUE and it equals the conditional expectation ( the latter is sometimes difficult to compute). This follows from Lehmann-Scheffe.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:46






          • 1




            $begingroup$
            @Hendrra The linear combination part is irrelevant (it's only here that the UMVUEs are linear combinations, not in general).
            $endgroup$
            – StubbornAtom
            Jan 13 at 21:01


















          • $begingroup$
            Does that mean that my UMVUE will be a vector too?
            $endgroup$
            – Hendrra
            Jan 13 at 20:19






          • 1




            $begingroup$
            @Hendrra No, the UMVUEs are linear combinations of $X_{(1)}$ and $X_{(n)}$.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:21










          • $begingroup$
            One for b and one for a?
            $endgroup$
            – Hendrra
            Jan 13 at 20:24






          • 1




            $begingroup$
            @Hendrra If you have found an unbiased estimator of your parameter based on a complete sufficient statistic, then that is bound to be the UMVUE and it equals the conditional expectation ( the latter is sometimes difficult to compute). This follows from Lehmann-Scheffe.
            $endgroup$
            – StubbornAtom
            Jan 13 at 20:46






          • 1




            $begingroup$
            @Hendrra The linear combination part is irrelevant (it's only here that the UMVUEs are linear combinations, not in general).
            $endgroup$
            – StubbornAtom
            Jan 13 at 21:01
















          $begingroup$
          Does that mean that my UMVUE will be a vector too?
          $endgroup$
          – Hendrra
          Jan 13 at 20:19




          $begingroup$
          Does that mean that my UMVUE will be a vector too?
          $endgroup$
          – Hendrra
          Jan 13 at 20:19




          1




          1




          $begingroup$
          @Hendrra No, the UMVUEs are linear combinations of $X_{(1)}$ and $X_{(n)}$.
          $endgroup$
          – StubbornAtom
          Jan 13 at 20:21




          $begingroup$
          @Hendrra No, the UMVUEs are linear combinations of $X_{(1)}$ and $X_{(n)}$.
          $endgroup$
          – StubbornAtom
          Jan 13 at 20:21












          $begingroup$
          One for b and one for a?
          $endgroup$
          – Hendrra
          Jan 13 at 20:24




          $begingroup$
          One for b and one for a?
          $endgroup$
          – Hendrra
          Jan 13 at 20:24




          1




          1




          $begingroup$
          @Hendrra If you have found an unbiased estimator of your parameter based on a complete sufficient statistic, then that is bound to be the UMVUE and it equals the conditional expectation ( the latter is sometimes difficult to compute). This follows from Lehmann-Scheffe.
          $endgroup$
          – StubbornAtom
          Jan 13 at 20:46




          $begingroup$
          @Hendrra If you have found an unbiased estimator of your parameter based on a complete sufficient statistic, then that is bound to be the UMVUE and it equals the conditional expectation ( the latter is sometimes difficult to compute). This follows from Lehmann-Scheffe.
          $endgroup$
          – StubbornAtom
          Jan 13 at 20:46




          1




          1




          $begingroup$
          @Hendrra The linear combination part is irrelevant (it's only here that the UMVUEs are linear combinations, not in general).
          $endgroup$
          – StubbornAtom
          Jan 13 at 21:01




          $begingroup$
          @Hendrra The linear combination part is irrelevant (it's only here that the UMVUEs are linear combinations, not in general).
          $endgroup$
          – StubbornAtom
          Jan 13 at 21:01


















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