Check if the estimator is unbiased












0












$begingroup$


For $X_isim U[0,a]$ where $i=1,2,dots,n$ so, $E(X_i)=dfrac a2$.



Is $a'=max{X_1,X_2,dots,X_n}$ an unbiased estimator of $a$?



This is what I thought.



Since $a'=max{X_1,X_2,dots,X_n}=X_k$ such that $X_kge X_h$ for any $h$,



$E(a')=E(X_k)=dfrac a2$.



Therefore $a'$ is not an unbiased estimator.



However, I don't think it is properly solved...






estimation parameter-estimation mean-square-error







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    For $X_isim U[0,a]$ where $i=1,2,dots,n$ so, $E(X_i)=dfrac a2$.



    Is $a'=max{X_1,X_2,dots,X_n}$ an unbiased estimator of $a$?



    This is what I thought.



    Since $a'=max{X_1,X_2,dots,X_n}=X_k$ such that $X_kge X_h$ for any $h$,



    $E(a')=E(X_k)=dfrac a2$.



    Therefore $a'$ is not an unbiased estimator.



    However, I don't think it is properly solved...






    estimation parameter-estimation mean-square-error







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      For $X_isim U[0,a]$ where $i=1,2,dots,n$ so, $E(X_i)=dfrac a2$.



      Is $a'=max{X_1,X_2,dots,X_n}$ an unbiased estimator of $a$?



      This is what I thought.



      Since $a'=max{X_1,X_2,dots,X_n}=X_k$ such that $X_kge X_h$ for any $h$,



      $E(a')=E(X_k)=dfrac a2$.



      Therefore $a'$ is not an unbiased estimator.



      However, I don't think it is properly solved...






      estimation parameter-estimation mean-square-error







      share|cite|improve this question











      $endgroup$




      For $X_isim U[0,a]$ where $i=1,2,dots,n$ so, $E(X_i)=dfrac a2$.



      Is $a'=max{X_1,X_2,dots,X_n}$ an unbiased estimator of $a$?



      This is what I thought.



      Since $a'=max{X_1,X_2,dots,X_n}=X_k$ such that $X_kge X_h$ for any $h$,



      $E(a')=E(X_k)=dfrac a2$.



      Therefore $a'$ is not an unbiased estimator.



      However, I don't think it is properly solved...






      estimation parameter-estimation mean-square-error




      estimation parameter-estimation mean-square-error






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 11:19









      Tianlalu

      3,08621038




      3,08621038










      asked Nov 27 '18 at 10:27









      NewtNewt

      207




      207






















          2 Answers
          2






          active

          oldest

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          0












          $begingroup$

          The $text{cdf}$ of the maximum of $n$ iid random variables is the $n^{th}$ power of the single $text{cdf}$. In this case,
          $$text{cdf}(X)=left(frac xaright)^n.$$



          The expectation is



          $$int_0^a xfrac naleft(frac xaright)^{n-1}dx=frac{na^{n+1}}{(n+1)a^n},$$ which is smaller than $a$ (but asymptotically unbiaised for large $n$).




          Your reasoning is wrong in that the distribution of $max{X_k}$ is not the same as that of a single $X_k$ (obviously the $max$ favors the larger values), and the expectations do not match.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't pdf of the X then becomes (n/a)(x/a)^(n-1)??
            $endgroup$
            – Newt
            Nov 27 '18 at 11:57










          • $begingroup$
            The link between cdf and bias is not clear to me. Is there a reference available?
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 12:06










          • $begingroup$
            @JanRothkegel: the link is between the expectation and the bias.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 12:49










          • $begingroup$
            @Newt: right, fixed.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 14:27



















          0












          $begingroup$

          By definition, $B_a(a')$ is the bias of the estimator $a'$ for $a$. That is,
          $$B_a(a')=mathbb E left [a'-a right]$$
          Always, $a' < a$, because $P(X=a)=0$. Therefore $B_a(a')<0$ and (more important) $B_a(a') ne 0$. It follows that $a'$ is biased.



          By the way: $a'$ is the MLE of $a$. Asymptotically (or approximately for large samples) it is unbiased.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But why does a′<a hold?
            $endgroup$
            – Newt
            Nov 27 '18 at 11:40












          • $begingroup$
            Because $max{X_1,X_2,...,X_n}<a$ or as described in my post: $P(X=a)=0$. You never hit the parameter.
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 11:46













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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

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          active

          oldest

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          0












          $begingroup$

          The $text{cdf}$ of the maximum of $n$ iid random variables is the $n^{th}$ power of the single $text{cdf}$. In this case,
          $$text{cdf}(X)=left(frac xaright)^n.$$



          The expectation is



          $$int_0^a xfrac naleft(frac xaright)^{n-1}dx=frac{na^{n+1}}{(n+1)a^n},$$ which is smaller than $a$ (but asymptotically unbiaised for large $n$).




          Your reasoning is wrong in that the distribution of $max{X_k}$ is not the same as that of a single $X_k$ (obviously the $max$ favors the larger values), and the expectations do not match.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't pdf of the X then becomes (n/a)(x/a)^(n-1)??
            $endgroup$
            – Newt
            Nov 27 '18 at 11:57










          • $begingroup$
            The link between cdf and bias is not clear to me. Is there a reference available?
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 12:06










          • $begingroup$
            @JanRothkegel: the link is between the expectation and the bias.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 12:49










          • $begingroup$
            @Newt: right, fixed.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 14:27
















          0












          $begingroup$

          The $text{cdf}$ of the maximum of $n$ iid random variables is the $n^{th}$ power of the single $text{cdf}$. In this case,
          $$text{cdf}(X)=left(frac xaright)^n.$$



          The expectation is



          $$int_0^a xfrac naleft(frac xaright)^{n-1}dx=frac{na^{n+1}}{(n+1)a^n},$$ which is smaller than $a$ (but asymptotically unbiaised for large $n$).




          Your reasoning is wrong in that the distribution of $max{X_k}$ is not the same as that of a single $X_k$ (obviously the $max$ favors the larger values), and the expectations do not match.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't pdf of the X then becomes (n/a)(x/a)^(n-1)??
            $endgroup$
            – Newt
            Nov 27 '18 at 11:57










          • $begingroup$
            The link between cdf and bias is not clear to me. Is there a reference available?
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 12:06










          • $begingroup$
            @JanRothkegel: the link is between the expectation and the bias.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 12:49










          • $begingroup$
            @Newt: right, fixed.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 14:27














          0












          0








          0





          $begingroup$

          The $text{cdf}$ of the maximum of $n$ iid random variables is the $n^{th}$ power of the single $text{cdf}$. In this case,
          $$text{cdf}(X)=left(frac xaright)^n.$$



          The expectation is



          $$int_0^a xfrac naleft(frac xaright)^{n-1}dx=frac{na^{n+1}}{(n+1)a^n},$$ which is smaller than $a$ (but asymptotically unbiaised for large $n$).




          Your reasoning is wrong in that the distribution of $max{X_k}$ is not the same as that of a single $X_k$ (obviously the $max$ favors the larger values), and the expectations do not match.






          share|cite|improve this answer











          $endgroup$



          The $text{cdf}$ of the maximum of $n$ iid random variables is the $n^{th}$ power of the single $text{cdf}$. In this case,
          $$text{cdf}(X)=left(frac xaright)^n.$$



          The expectation is



          $$int_0^a xfrac naleft(frac xaright)^{n-1}dx=frac{na^{n+1}}{(n+1)a^n},$$ which is smaller than $a$ (but asymptotically unbiaised for large $n$).




          Your reasoning is wrong in that the distribution of $max{X_k}$ is not the same as that of a single $X_k$ (obviously the $max$ favors the larger values), and the expectations do not match.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 27 '18 at 14:31

























          answered Nov 27 '18 at 11:15









          Yves DaoustYves Daoust

          125k671223




          125k671223












          • $begingroup$
            Isn't pdf of the X then becomes (n/a)(x/a)^(n-1)??
            $endgroup$
            – Newt
            Nov 27 '18 at 11:57










          • $begingroup$
            The link between cdf and bias is not clear to me. Is there a reference available?
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 12:06










          • $begingroup$
            @JanRothkegel: the link is between the expectation and the bias.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 12:49










          • $begingroup$
            @Newt: right, fixed.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 14:27


















          • $begingroup$
            Isn't pdf of the X then becomes (n/a)(x/a)^(n-1)??
            $endgroup$
            – Newt
            Nov 27 '18 at 11:57










          • $begingroup$
            The link between cdf and bias is not clear to me. Is there a reference available?
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 12:06










          • $begingroup$
            @JanRothkegel: the link is between the expectation and the bias.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 12:49










          • $begingroup$
            @Newt: right, fixed.
            $endgroup$
            – Yves Daoust
            Nov 27 '18 at 14:27
















          $begingroup$
          Isn't pdf of the X then becomes (n/a)(x/a)^(n-1)??
          $endgroup$
          – Newt
          Nov 27 '18 at 11:57




          $begingroup$
          Isn't pdf of the X then becomes (n/a)(x/a)^(n-1)??
          $endgroup$
          – Newt
          Nov 27 '18 at 11:57












          $begingroup$
          The link between cdf and bias is not clear to me. Is there a reference available?
          $endgroup$
          – Jan Rothkegel
          Nov 27 '18 at 12:06




          $begingroup$
          The link between cdf and bias is not clear to me. Is there a reference available?
          $endgroup$
          – Jan Rothkegel
          Nov 27 '18 at 12:06












          $begingroup$
          @JanRothkegel: the link is between the expectation and the bias.
          $endgroup$
          – Yves Daoust
          Nov 27 '18 at 12:49




          $begingroup$
          @JanRothkegel: the link is between the expectation and the bias.
          $endgroup$
          – Yves Daoust
          Nov 27 '18 at 12:49












          $begingroup$
          @Newt: right, fixed.
          $endgroup$
          – Yves Daoust
          Nov 27 '18 at 14:27




          $begingroup$
          @Newt: right, fixed.
          $endgroup$
          – Yves Daoust
          Nov 27 '18 at 14:27











          0












          $begingroup$

          By definition, $B_a(a')$ is the bias of the estimator $a'$ for $a$. That is,
          $$B_a(a')=mathbb E left [a'-a right]$$
          Always, $a' < a$, because $P(X=a)=0$. Therefore $B_a(a')<0$ and (more important) $B_a(a') ne 0$. It follows that $a'$ is biased.



          By the way: $a'$ is the MLE of $a$. Asymptotically (or approximately for large samples) it is unbiased.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But why does a′<a hold?
            $endgroup$
            – Newt
            Nov 27 '18 at 11:40












          • $begingroup$
            Because $max{X_1,X_2,...,X_n}<a$ or as described in my post: $P(X=a)=0$. You never hit the parameter.
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 11:46


















          0












          $begingroup$

          By definition, $B_a(a')$ is the bias of the estimator $a'$ for $a$. That is,
          $$B_a(a')=mathbb E left [a'-a right]$$
          Always, $a' < a$, because $P(X=a)=0$. Therefore $B_a(a')<0$ and (more important) $B_a(a') ne 0$. It follows that $a'$ is biased.



          By the way: $a'$ is the MLE of $a$. Asymptotically (or approximately for large samples) it is unbiased.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But why does a′<a hold?
            $endgroup$
            – Newt
            Nov 27 '18 at 11:40












          • $begingroup$
            Because $max{X_1,X_2,...,X_n}<a$ or as described in my post: $P(X=a)=0$. You never hit the parameter.
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 11:46
















          0












          0








          0





          $begingroup$

          By definition, $B_a(a')$ is the bias of the estimator $a'$ for $a$. That is,
          $$B_a(a')=mathbb E left [a'-a right]$$
          Always, $a' < a$, because $P(X=a)=0$. Therefore $B_a(a')<0$ and (more important) $B_a(a') ne 0$. It follows that $a'$ is biased.



          By the way: $a'$ is the MLE of $a$. Asymptotically (or approximately for large samples) it is unbiased.






          share|cite|improve this answer











          $endgroup$



          By definition, $B_a(a')$ is the bias of the estimator $a'$ for $a$. That is,
          $$B_a(a')=mathbb E left [a'-a right]$$
          Always, $a' < a$, because $P(X=a)=0$. Therefore $B_a(a')<0$ and (more important) $B_a(a') ne 0$. It follows that $a'$ is biased.



          By the way: $a'$ is the MLE of $a$. Asymptotically (or approximately for large samples) it is unbiased.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 27 '18 at 12:19

























          answered Nov 27 '18 at 11:37









          Jan RothkegelJan Rothkegel

          1235




          1235












          • $begingroup$
            But why does a′<a hold?
            $endgroup$
            – Newt
            Nov 27 '18 at 11:40












          • $begingroup$
            Because $max{X_1,X_2,...,X_n}<a$ or as described in my post: $P(X=a)=0$. You never hit the parameter.
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 11:46




















          • $begingroup$
            But why does a′<a hold?
            $endgroup$
            – Newt
            Nov 27 '18 at 11:40












          • $begingroup$
            Because $max{X_1,X_2,...,X_n}<a$ or as described in my post: $P(X=a)=0$. You never hit the parameter.
            $endgroup$
            – Jan Rothkegel
            Nov 27 '18 at 11:46


















          $begingroup$
          But why does a′<a hold?
          $endgroup$
          – Newt
          Nov 27 '18 at 11:40






          $begingroup$
          But why does a′<a hold?
          $endgroup$
          – Newt
          Nov 27 '18 at 11:40














          $begingroup$
          Because $max{X_1,X_2,...,X_n}<a$ or as described in my post: $P(X=a)=0$. You never hit the parameter.
          $endgroup$
          – Jan Rothkegel
          Nov 27 '18 at 11:46






          $begingroup$
          Because $max{X_1,X_2,...,X_n}<a$ or as described in my post: $P(X=a)=0$. You never hit the parameter.
          $endgroup$
          – Jan Rothkegel
          Nov 27 '18 at 11:46




















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