Correlation between a random variable and its rank











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Let $X_1,ldots,X_n$ be a random sample from $U(0,1)$ and $X_{(1)}<ldots<X_{(n)}$ be the corresponding order statistics.



Define,
$$
R(X_1) = rquad text{if}quad X_{(r)} = X_1;quad r = 1(1)n
$$
i.e. $R(X_1)$ is the rank of $X_1$ in the ordered sample. Then what will be correlation between $X_1$ and $R(X_1)$ ?






probability correlation symmetry order-statistics







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  • A more general version of this question: stats.stackexchange.com/questions/375989/…
    – kjetil b halvorsen
    Nov 16 at 16:32















up vote
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down vote

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Let $X_1,ldots,X_n$ be a random sample from $U(0,1)$ and $X_{(1)}<ldots<X_{(n)}$ be the corresponding order statistics.



Define,
$$
R(X_1) = rquad text{if}quad X_{(r)} = X_1;quad r = 1(1)n
$$
i.e. $R(X_1)$ is the rank of $X_1$ in the ordered sample. Then what will be correlation between $X_1$ and $R(X_1)$ ?






probability correlation symmetry order-statistics







share|cite|improve this question
























  • A more general version of this question: stats.stackexchange.com/questions/375989/…
    – kjetil b halvorsen
    Nov 16 at 16:32













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





Let $X_1,ldots,X_n$ be a random sample from $U(0,1)$ and $X_{(1)}<ldots<X_{(n)}$ be the corresponding order statistics.



Define,
$$
R(X_1) = rquad text{if}quad X_{(r)} = X_1;quad r = 1(1)n
$$
i.e. $R(X_1)$ is the rank of $X_1$ in the ordered sample. Then what will be correlation between $X_1$ and $R(X_1)$ ?






probability correlation symmetry order-statistics







share|cite|improve this question















Let $X_1,ldots,X_n$ be a random sample from $U(0,1)$ and $X_{(1)}<ldots<X_{(n)}$ be the corresponding order statistics.



Define,
$$
R(X_1) = rquad text{if}quad X_{(r)} = X_1;quad r = 1(1)n
$$
i.e. $R(X_1)$ is the rank of $X_1$ in the ordered sample. Then what will be correlation between $X_1$ and $R(X_1)$ ?






probability correlation symmetry order-statistics




probability correlation symmetry order-statistics






share|cite|improve this question















share|cite|improve this question













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edited Nov 23 at 15:30









kjetil b halvorsen

4,72742638




4,72742638










asked Mar 9 '16 at 9:13









Koushik

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  • A more general version of this question: stats.stackexchange.com/questions/375989/…
    – kjetil b halvorsen
    Nov 16 at 16:32


















  • A more general version of this question: stats.stackexchange.com/questions/375989/…
    – kjetil b halvorsen
    Nov 16 at 16:32
















A more general version of this question: stats.stackexchange.com/questions/375989/…
– kjetil b halvorsen
Nov 16 at 16:32




A more general version of this question: stats.stackexchange.com/questions/375989/…
– kjetil b halvorsen
Nov 16 at 16:32










2 Answers
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A generalization of the problem statement is that $X_1, dotsc,X_n$ is exchangeable. Then we will also get $R_1, dotsc, R_n$ is exchangeable, and even the bivariate vectors
$$
left( (begin{smallmatrix} X_1\R_1end{smallmatrix}), dotsc, (begin{smallmatrix} X_n\R_nend{smallmatrix}) right)
$$ is exchangeable. Let $mu$ be the common mean of the $X_j$. The common mean of the $R_j$ is $(n+1)/2$. By exchangeability we have that $DeclareMathOperator{E}{mathbb{E}} E X_1 R(X_1) =E X_j R(X_j)$ for $j=1, dotsc,n$. Then we find
$$
E X_1 R(X_1) =frac1{n}sum_{j=1}^n E X_j R(X_j) = \
frac1{n}sum_j X_{(j)} R(X_{(j)}) =frac1{n}sum_j E X_{(j)} j =\
frac1{n} E sum_j j X_j=frac1{n} mu sum_j j =\
frac{n(n+1)mu}{2 n}=frac{mu (n+1)}{2}
$$
and then it follows fast that the correlation is zero, under much greater generality than in the question.






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    begin{align}
    E[X_1R(X_1)]=frac1nsum_{r=1}^nrfrac{int_0^1xx^{r-1}(1-x)^{n-r}mathrm dx}{int_0^1x^{r-1}(1-x)^{n-r}mathrm dx}=frac1nsum_{r=1}^nfrac{r^2}{n+1}=frac{2n+1}6;,
    end{align}



    so with $E[X_1]=frac12$ and $E[R(X_1)]=frac{n+1}2$ we have



    $$
    operatorname{Cov}(X_1,R(X_1))=frac{2n+1}6-frac12cdotfrac{n+1}2=frac{n-1}{12};.
    $$



    With $operatorname{Var}(X_1)=frac1{12}$ and $operatorname{Var}(R(X_1))=frac{n^2-1}{12}$ the correlation coefficient is



    $$
    frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
    $$






    share|cite|improve this answer





















    • This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
      – kjetil b halvorsen
      Nov 16 at 16:30











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    2 Answers
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    2 Answers
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    active

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    up vote
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    down vote













    A generalization of the problem statement is that $X_1, dotsc,X_n$ is exchangeable. Then we will also get $R_1, dotsc, R_n$ is exchangeable, and even the bivariate vectors
    $$
    left( (begin{smallmatrix} X_1\R_1end{smallmatrix}), dotsc, (begin{smallmatrix} X_n\R_nend{smallmatrix}) right)
    $$     is exchangeable. Let $mu$ be the common mean of the $X_j$. The common mean of the $R_j$ is $(n+1)/2$. By exchangeability we have that $DeclareMathOperator{E}{mathbb{E}} E X_1 R(X_1) =E X_j R(X_j)$ for $j=1, dotsc,n$. Then we find
    $$
    E X_1 R(X_1) =frac1{n}sum_{j=1}^n E X_j R(X_j) = \
    frac1{n}sum_j X_{(j)} R(X_{(j)}) =frac1{n}sum_j E X_{(j)} j =\
    frac1{n} E sum_j j X_j=frac1{n} mu sum_j j =\
    frac{n(n+1)mu}{2 n}=frac{mu (n+1)}{2}
    $$
    and then it follows fast that the correlation is zero, under much greater generality than in the question.






    share|cite|improve this answer

























      up vote
      1
      down vote













      A generalization of the problem statement is that $X_1, dotsc,X_n$ is exchangeable. Then we will also get $R_1, dotsc, R_n$ is exchangeable, and even the bivariate vectors
      $$
      left( (begin{smallmatrix} X_1\R_1end{smallmatrix}), dotsc, (begin{smallmatrix} X_n\R_nend{smallmatrix}) right)
      $$       is exchangeable. Let $mu$ be the common mean of the $X_j$. The common mean of the $R_j$ is $(n+1)/2$. By exchangeability we have that $DeclareMathOperator{E}{mathbb{E}} E X_1 R(X_1) =E X_j R(X_j)$ for $j=1, dotsc,n$. Then we find
      $$
      E X_1 R(X_1) =frac1{n}sum_{j=1}^n E X_j R(X_j) = \
      frac1{n}sum_j X_{(j)} R(X_{(j)}) =frac1{n}sum_j E X_{(j)} j =\
      frac1{n} E sum_j j X_j=frac1{n} mu sum_j j =\
      frac{n(n+1)mu}{2 n}=frac{mu (n+1)}{2}
      $$
      and then it follows fast that the correlation is zero, under much greater generality than in the question.






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        A generalization of the problem statement is that $X_1, dotsc,X_n$ is exchangeable. Then we will also get $R_1, dotsc, R_n$ is exchangeable, and even the bivariate vectors
        $$
        left( (begin{smallmatrix} X_1\R_1end{smallmatrix}), dotsc, (begin{smallmatrix} X_n\R_nend{smallmatrix}) right)
        $$         is exchangeable. Let $mu$ be the common mean of the $X_j$. The common mean of the $R_j$ is $(n+1)/2$. By exchangeability we have that $DeclareMathOperator{E}{mathbb{E}} E X_1 R(X_1) =E X_j R(X_j)$ for $j=1, dotsc,n$. Then we find
        $$
        E X_1 R(X_1) =frac1{n}sum_{j=1}^n E X_j R(X_j) = \
        frac1{n}sum_j X_{(j)} R(X_{(j)}) =frac1{n}sum_j E X_{(j)} j =\
        frac1{n} E sum_j j X_j=frac1{n} mu sum_j j =\
        frac{n(n+1)mu}{2 n}=frac{mu (n+1)}{2}
        $$
        and then it follows fast that the correlation is zero, under much greater generality than in the question.






        share|cite|improve this answer












        A generalization of the problem statement is that $X_1, dotsc,X_n$ is exchangeable. Then we will also get $R_1, dotsc, R_n$ is exchangeable, and even the bivariate vectors
        $$
        left( (begin{smallmatrix} X_1\R_1end{smallmatrix}), dotsc, (begin{smallmatrix} X_n\R_nend{smallmatrix}) right)
        $$         is exchangeable. Let $mu$ be the common mean of the $X_j$. The common mean of the $R_j$ is $(n+1)/2$. By exchangeability we have that $DeclareMathOperator{E}{mathbb{E}} E X_1 R(X_1) =E X_j R(X_j)$ for $j=1, dotsc,n$. Then we find
        $$
        E X_1 R(X_1) =frac1{n}sum_{j=1}^n E X_j R(X_j) = \
        frac1{n}sum_j X_{(j)} R(X_{(j)}) =frac1{n}sum_j E X_{(j)} j =\
        frac1{n} E sum_j j X_j=frac1{n} mu sum_j j =\
        frac{n(n+1)mu}{2 n}=frac{mu (n+1)}{2}
        $$
        and then it follows fast that the correlation is zero, under much greater generality than in the question.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 at 11:48









        kjetil b halvorsen

        4,72742638




        4,72742638






















            up vote
            -2
            down vote













            begin{align}
            E[X_1R(X_1)]=frac1nsum_{r=1}^nrfrac{int_0^1xx^{r-1}(1-x)^{n-r}mathrm dx}{int_0^1x^{r-1}(1-x)^{n-r}mathrm dx}=frac1nsum_{r=1}^nfrac{r^2}{n+1}=frac{2n+1}6;,
            end{align}



            so with $E[X_1]=frac12$ and $E[R(X_1)]=frac{n+1}2$ we have



            $$
            operatorname{Cov}(X_1,R(X_1))=frac{2n+1}6-frac12cdotfrac{n+1}2=frac{n-1}{12};.
            $$



            With $operatorname{Var}(X_1)=frac1{12}$ and $operatorname{Var}(R(X_1))=frac{n^2-1}{12}$ the correlation coefficient is



            $$
            frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
            $$






            share|cite|improve this answer





















            • This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
              – kjetil b halvorsen
              Nov 16 at 16:30















            up vote
            -2
            down vote













            begin{align}
            E[X_1R(X_1)]=frac1nsum_{r=1}^nrfrac{int_0^1xx^{r-1}(1-x)^{n-r}mathrm dx}{int_0^1x^{r-1}(1-x)^{n-r}mathrm dx}=frac1nsum_{r=1}^nfrac{r^2}{n+1}=frac{2n+1}6;,
            end{align}



            so with $E[X_1]=frac12$ and $E[R(X_1)]=frac{n+1}2$ we have



            $$
            operatorname{Cov}(X_1,R(X_1))=frac{2n+1}6-frac12cdotfrac{n+1}2=frac{n-1}{12};.
            $$



            With $operatorname{Var}(X_1)=frac1{12}$ and $operatorname{Var}(R(X_1))=frac{n^2-1}{12}$ the correlation coefficient is



            $$
            frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
            $$






            share|cite|improve this answer





















            • This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
              – kjetil b halvorsen
              Nov 16 at 16:30













            up vote
            -2
            down vote










            up vote
            -2
            down vote









            begin{align}
            E[X_1R(X_1)]=frac1nsum_{r=1}^nrfrac{int_0^1xx^{r-1}(1-x)^{n-r}mathrm dx}{int_0^1x^{r-1}(1-x)^{n-r}mathrm dx}=frac1nsum_{r=1}^nfrac{r^2}{n+1}=frac{2n+1}6;,
            end{align}



            so with $E[X_1]=frac12$ and $E[R(X_1)]=frac{n+1}2$ we have



            $$
            operatorname{Cov}(X_1,R(X_1))=frac{2n+1}6-frac12cdotfrac{n+1}2=frac{n-1}{12};.
            $$



            With $operatorname{Var}(X_1)=frac1{12}$ and $operatorname{Var}(R(X_1))=frac{n^2-1}{12}$ the correlation coefficient is



            $$
            frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
            $$






            share|cite|improve this answer












            begin{align}
            E[X_1R(X_1)]=frac1nsum_{r=1}^nrfrac{int_0^1xx^{r-1}(1-x)^{n-r}mathrm dx}{int_0^1x^{r-1}(1-x)^{n-r}mathrm dx}=frac1nsum_{r=1}^nfrac{r^2}{n+1}=frac{2n+1}6;,
            end{align}



            so with $E[X_1]=frac12$ and $E[R(X_1)]=frac{n+1}2$ we have



            $$
            operatorname{Cov}(X_1,R(X_1))=frac{2n+1}6-frac12cdotfrac{n+1}2=frac{n-1}{12};.
            $$



            With $operatorname{Var}(X_1)=frac1{12}$ and $operatorname{Var}(R(X_1))=frac{n^2-1}{12}$ the correlation coefficient is



            $$
            frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 9 '16 at 9:33









            joriki

            170k10183342




            170k10183342












            • This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
              – kjetil b halvorsen
              Nov 16 at 16:30


















            • This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
              – kjetil b halvorsen
              Nov 16 at 16:30
















            This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
            – kjetil b halvorsen
            Nov 16 at 16:30




            This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
            – kjetil b halvorsen
            Nov 16 at 16:30


















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