Testing certain estimators for consistency












1












$begingroup$


Let $X_1, X_2, ldots , X_n$ be a random sample from $U(0, θ)$, where $θ > 0$ is the unknown parameter. Let
$X_{(n)} = max{X_1,X_2, ldots , X_n }$. Then which of the following is (are) consistent estimator(s) of $θ^3$?



(A) $8X_n^3$



(B) $X_{(n)}^3$



(C) $(frac{2}{n}∑_{i=5}^n X_i)^3$



(D) $frac{nX_{(n)}^3 +1}{n+1}$




How do I test option (A) and (C)? I have been stuck for a while and some help would be appreciated.



My approach so far:



$$f(X_{(n)})=frac{nx^n}{theta^n}$$
Using the above density function I proved that



$E(X_{(n)})rightarrowtheta as nrightarrowinfty$



$Var(X_{(n)})rightarrow0 as nrightarrowinfty$



Thus, sufficient conditions for consistency have been met and $X_{(n)}$ is a consistent estimator for $theta$. Since $theta^3$ is a continuous function of $theta$, and $X_{(n)}$ is a consistent estimator for $theta$, it can be said that $X_{(n)}^3$ must be a consistent estimator for $theta^3$. Validity of (B) and (D) now follows from this fact.






statistics statistical-inference parameter-estimation







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$endgroup$

















    1












    $begingroup$


    Let $X_1, X_2, ldots , X_n$ be a random sample from $U(0, θ)$, where $θ > 0$ is the unknown parameter. Let
    $X_{(n)} = max{X_1,X_2, ldots , X_n }$. Then which of the following is (are) consistent estimator(s) of $θ^3$?



    (A) $8X_n^3$



    (B) $X_{(n)}^3$



    (C) $(frac{2}{n}∑_{i=5}^n X_i)^3$



    (D) $frac{nX_{(n)}^3 +1}{n+1}$




    How do I test option (A) and (C)? I have been stuck for a while and some help would be appreciated.



    My approach so far:



    $$f(X_{(n)})=frac{nx^n}{theta^n}$$
    Using the above density function I proved that



    $E(X_{(n)})rightarrowtheta as nrightarrowinfty$



    $Var(X_{(n)})rightarrow0 as nrightarrowinfty$



    Thus, sufficient conditions for consistency have been met and $X_{(n)}$ is a consistent estimator for $theta$. Since $theta^3$ is a continuous function of $theta$, and $X_{(n)}$ is a consistent estimator for $theta$, it can be said that $X_{(n)}^3$ must be a consistent estimator for $theta^3$. Validity of (B) and (D) now follows from this fact.






    statistics statistical-inference parameter-estimation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $X_1, X_2, ldots , X_n$ be a random sample from $U(0, θ)$, where $θ > 0$ is the unknown parameter. Let
      $X_{(n)} = max{X_1,X_2, ldots , X_n }$. Then which of the following is (are) consistent estimator(s) of $θ^3$?



      (A) $8X_n^3$



      (B) $X_{(n)}^3$



      (C) $(frac{2}{n}∑_{i=5}^n X_i)^3$



      (D) $frac{nX_{(n)}^3 +1}{n+1}$




      How do I test option (A) and (C)? I have been stuck for a while and some help would be appreciated.



      My approach so far:



      $$f(X_{(n)})=frac{nx^n}{theta^n}$$
      Using the above density function I proved that



      $E(X_{(n)})rightarrowtheta as nrightarrowinfty$



      $Var(X_{(n)})rightarrow0 as nrightarrowinfty$



      Thus, sufficient conditions for consistency have been met and $X_{(n)}$ is a consistent estimator for $theta$. Since $theta^3$ is a continuous function of $theta$, and $X_{(n)}$ is a consistent estimator for $theta$, it can be said that $X_{(n)}^3$ must be a consistent estimator for $theta^3$. Validity of (B) and (D) now follows from this fact.






      statistics statistical-inference parameter-estimation







      share|cite|improve this question









      $endgroup$




      Let $X_1, X_2, ldots , X_n$ be a random sample from $U(0, θ)$, where $θ > 0$ is the unknown parameter. Let
      $X_{(n)} = max{X_1,X_2, ldots , X_n }$. Then which of the following is (are) consistent estimator(s) of $θ^3$?



      (A) $8X_n^3$



      (B) $X_{(n)}^3$



      (C) $(frac{2}{n}∑_{i=5}^n X_i)^3$



      (D) $frac{nX_{(n)}^3 +1}{n+1}$




      How do I test option (A) and (C)? I have been stuck for a while and some help would be appreciated.



      My approach so far:



      $$f(X_{(n)})=frac{nx^n}{theta^n}$$
      Using the above density function I proved that



      $E(X_{(n)})rightarrowtheta as nrightarrowinfty$



      $Var(X_{(n)})rightarrow0 as nrightarrowinfty$



      Thus, sufficient conditions for consistency have been met and $X_{(n)}$ is a consistent estimator for $theta$. Since $theta^3$ is a continuous function of $theta$, and $X_{(n)}$ is a consistent estimator for $theta$, it can be said that $X_{(n)}^3$ must be a consistent estimator for $theta^3$. Validity of (B) and (D) now follows from this fact.






      statistics statistical-inference parameter-estimation




      statistics statistical-inference parameter-estimation






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 27 at 20:43









      s0ulr3aper07s0ulr3aper07

      580111




      580111






















          2 Answers
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          $begingroup$

          For (C), we can write
          $$dfrac{2}{n}sum_{i = 5}^{n}X_{i} = 2left(dfrac{1}{n}sum_{i = 1}^{n}X_{i}right) - 2dfrac{X_{1} + dots + X_{4}}{n}$$
          which using the law of large numbers, converges in probability to $2dfrac{theta}{2} = theta$. Then by applying the continuous mapping theorem, we can show that the estimator is consistent for $theta^{3}$.



          For (A), it is not consistent because it doesn't converge in probability to anything (it will have some scaled beta distribution).






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Consider the following:




            • If $T_n$ is consistent for $theta$, then $a_nT_n$ is also consistent for $theta$ provided $lim a_n=1$.


            • If $T_n$ is consistent for $theta$, then $T_n+frac{a}{psi(n)}$ is also consistent for $theta$ if $psi(n)$ is increasing and $a$ is independent of $n$.



            Once you know that $X_{(n)}$ is consistent for $theta$ (and hence $X_{(n)}^3$ is consistent for $theta^3$ ), it follows from the above facts that option (D) is correct; no need to check for the sufficient condition. The first bullet also rules out option (A).



            As for option (C), I suspect it was meant to be $left(frac{2}{n}sumlimits_{i=color{red}1}^n X_iright)^3=8overline X^3$, where $overline X$ is the sample mean.



            By the weak law of large numbers, $overline Xstackrel{P}longrightarrow frac{theta}{2}$ as $ntoinfty$.



            Then by the continuous mapping theorem it follows that $8overline X^3stackrel{P}longrightarrow theta^3$ as $ntoinfty$, so that (C) is correct.






            share|cite|improve this answer











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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

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              1












              $begingroup$

              For (C), we can write
              $$dfrac{2}{n}sum_{i = 5}^{n}X_{i} = 2left(dfrac{1}{n}sum_{i = 1}^{n}X_{i}right) - 2dfrac{X_{1} + dots + X_{4}}{n}$$
              which using the law of large numbers, converges in probability to $2dfrac{theta}{2} = theta$. Then by applying the continuous mapping theorem, we can show that the estimator is consistent for $theta^{3}$.



              For (A), it is not consistent because it doesn't converge in probability to anything (it will have some scaled beta distribution).






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                For (C), we can write
                $$dfrac{2}{n}sum_{i = 5}^{n}X_{i} = 2left(dfrac{1}{n}sum_{i = 1}^{n}X_{i}right) - 2dfrac{X_{1} + dots + X_{4}}{n}$$
                which using the law of large numbers, converges in probability to $2dfrac{theta}{2} = theta$. Then by applying the continuous mapping theorem, we can show that the estimator is consistent for $theta^{3}$.



                For (A), it is not consistent because it doesn't converge in probability to anything (it will have some scaled beta distribution).






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For (C), we can write
                  $$dfrac{2}{n}sum_{i = 5}^{n}X_{i} = 2left(dfrac{1}{n}sum_{i = 1}^{n}X_{i}right) - 2dfrac{X_{1} + dots + X_{4}}{n}$$
                  which using the law of large numbers, converges in probability to $2dfrac{theta}{2} = theta$. Then by applying the continuous mapping theorem, we can show that the estimator is consistent for $theta^{3}$.



                  For (A), it is not consistent because it doesn't converge in probability to anything (it will have some scaled beta distribution).






                  share|cite|improve this answer











                  $endgroup$



                  For (C), we can write
                  $$dfrac{2}{n}sum_{i = 5}^{n}X_{i} = 2left(dfrac{1}{n}sum_{i = 1}^{n}X_{i}right) - 2dfrac{X_{1} + dots + X_{4}}{n}$$
                  which using the law of large numbers, converges in probability to $2dfrac{theta}{2} = theta$. Then by applying the continuous mapping theorem, we can show that the estimator is consistent for $theta^{3}$.



                  For (A), it is not consistent because it doesn't converge in probability to anything (it will have some scaled beta distribution).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 27 at 21:48

























                  answered Jan 27 at 21:35









                  rzchrzch

                  1765




                  1765























                      1












                      $begingroup$

                      Consider the following:




                      • If $T_n$ is consistent for $theta$, then $a_nT_n$ is also consistent for $theta$ provided $lim a_n=1$.


                      • If $T_n$ is consistent for $theta$, then $T_n+frac{a}{psi(n)}$ is also consistent for $theta$ if $psi(n)$ is increasing and $a$ is independent of $n$.



                      Once you know that $X_{(n)}$ is consistent for $theta$ (and hence $X_{(n)}^3$ is consistent for $theta^3$ ), it follows from the above facts that option (D) is correct; no need to check for the sufficient condition. The first bullet also rules out option (A).



                      As for option (C), I suspect it was meant to be $left(frac{2}{n}sumlimits_{i=color{red}1}^n X_iright)^3=8overline X^3$, where $overline X$ is the sample mean.



                      By the weak law of large numbers, $overline Xstackrel{P}longrightarrow frac{theta}{2}$ as $ntoinfty$.



                      Then by the continuous mapping theorem it follows that $8overline X^3stackrel{P}longrightarrow theta^3$ as $ntoinfty$, so that (C) is correct.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Consider the following:




                        • If $T_n$ is consistent for $theta$, then $a_nT_n$ is also consistent for $theta$ provided $lim a_n=1$.


                        • If $T_n$ is consistent for $theta$, then $T_n+frac{a}{psi(n)}$ is also consistent for $theta$ if $psi(n)$ is increasing and $a$ is independent of $n$.



                        Once you know that $X_{(n)}$ is consistent for $theta$ (and hence $X_{(n)}^3$ is consistent for $theta^3$ ), it follows from the above facts that option (D) is correct; no need to check for the sufficient condition. The first bullet also rules out option (A).



                        As for option (C), I suspect it was meant to be $left(frac{2}{n}sumlimits_{i=color{red}1}^n X_iright)^3=8overline X^3$, where $overline X$ is the sample mean.



                        By the weak law of large numbers, $overline Xstackrel{P}longrightarrow frac{theta}{2}$ as $ntoinfty$.



                        Then by the continuous mapping theorem it follows that $8overline X^3stackrel{P}longrightarrow theta^3$ as $ntoinfty$, so that (C) is correct.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Consider the following:




                          • If $T_n$ is consistent for $theta$, then $a_nT_n$ is also consistent for $theta$ provided $lim a_n=1$.


                          • If $T_n$ is consistent for $theta$, then $T_n+frac{a}{psi(n)}$ is also consistent for $theta$ if $psi(n)$ is increasing and $a$ is independent of $n$.



                          Once you know that $X_{(n)}$ is consistent for $theta$ (and hence $X_{(n)}^3$ is consistent for $theta^3$ ), it follows from the above facts that option (D) is correct; no need to check for the sufficient condition. The first bullet also rules out option (A).



                          As for option (C), I suspect it was meant to be $left(frac{2}{n}sumlimits_{i=color{red}1}^n X_iright)^3=8overline X^3$, where $overline X$ is the sample mean.



                          By the weak law of large numbers, $overline Xstackrel{P}longrightarrow frac{theta}{2}$ as $ntoinfty$.



                          Then by the continuous mapping theorem it follows that $8overline X^3stackrel{P}longrightarrow theta^3$ as $ntoinfty$, so that (C) is correct.






                          share|cite|improve this answer











                          $endgroup$



                          Consider the following:




                          • If $T_n$ is consistent for $theta$, then $a_nT_n$ is also consistent for $theta$ provided $lim a_n=1$.


                          • If $T_n$ is consistent for $theta$, then $T_n+frac{a}{psi(n)}$ is also consistent for $theta$ if $psi(n)$ is increasing and $a$ is independent of $n$.



                          Once you know that $X_{(n)}$ is consistent for $theta$ (and hence $X_{(n)}^3$ is consistent for $theta^3$ ), it follows from the above facts that option (D) is correct; no need to check for the sufficient condition. The first bullet also rules out option (A).



                          As for option (C), I suspect it was meant to be $left(frac{2}{n}sumlimits_{i=color{red}1}^n X_iright)^3=8overline X^3$, where $overline X$ is the sample mean.



                          By the weak law of large numbers, $overline Xstackrel{P}longrightarrow frac{theta}{2}$ as $ntoinfty$.



                          Then by the continuous mapping theorem it follows that $8overline X^3stackrel{P}longrightarrow theta^3$ as $ntoinfty$, so that (C) is correct.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 27 at 21:48

























                          answered Jan 27 at 21:42









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