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
edited Nov 23 at 15:30
kjetil b halvorsen
4,72742638
asked Mar 9 '16 at 9:13
Koushik
788
add a comment |
up vote
3
down vote
favorite
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
edited Nov 23 at 15:30
kjetil b halvorsen
4,72742638
asked Mar 9 '16 at 9:13
Koushik
788
A more general version of this question: stats.stackexchange.com/questions/375989/…
– kjetil b halvorsen
Nov 16 at 16:32
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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
edited Nov 23 at 15:30
kjetil b halvorsen
4,72742638
asked Mar 9 '16 at 9:13
Koushik
788
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
edited Nov 23 at 15:30
kjetil b halvorsen
4,72742638
asked Mar 9 '16 at 9:13
Koushik
788
edited Nov 23 at 15:30
kjetil b halvorsen
4,72742638
asked Mar 9 '16 at 9:13
Koushik
788
edited Nov 23 at 15:30
kjetil b halvorsen
4,72742638
edited Nov 23 at 15:30
kjetil b halvorsen
4,72742638
edited Nov 23 at 15:30
kjetil b halvorsen
4,72742638
4,72742638
asked Mar 9 '16 at 9:13
Koushik
788
asked Mar 9 '16 at 9:13
Koushik
788
asked Mar 9 '16 at 9:13
Koushik
788
788
A more general version of this question: stats.stackexchange.com/questions/375989/…
– kjetil b halvorsen
Nov 16 at 16:32
add a comment |
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
add a comment |
2 Answers
2
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up vote
1
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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.
answered Nov 23 at 11:48
kjetil b halvorsen
4,72742638
add a comment |
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
$$
frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
$$
answered Mar 9 '16 at 9:33
joriki
170k10183342
This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
– kjetil b halvorsen
Nov 16 at 16:30
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 23 at 11:48
kjetil b halvorsen
4,72742638
add a comment |
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.
answered Nov 23 at 11:48
kjetil b halvorsen
4,72742638
add a comment |
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.
answered Nov 23 at 11:48
kjetil b halvorsen
4,72742638
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.
answered Nov 23 at 11:48
kjetil b halvorsen
4,72742638
answered Nov 23 at 11:48
kjetil b halvorsen
4,72742638
answered Nov 23 at 11:48
kjetil b halvorsen
4,72742638
answered Nov 23 at 11:48
kjetil b halvorsen
4,72742638
4,72742638
add a comment |
add a comment |
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
$$
frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
$$
answered Mar 9 '16 at 9:33
joriki
170k10183342
This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
– kjetil b halvorsen
Nov 16 at 16:30
add a comment |
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
$$
frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
$$
answered Mar 9 '16 at 9:33
joriki
170k10183342
This cannot be correct, take the limit $n toinfty$ the correlation becomes 1.
– kjetil b halvorsen
Nov 16 at 16:30
add a comment |
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
$$
frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
$$
answered Mar 9 '16 at 9:33
joriki
170k10183342
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
$$
frac{n-1}{sqrt{n^2-1}}=sqrt{frac{n-1}{n+1}}=sqrt{1-frac2{n+1}};.
$$
answered Mar 9 '16 at 9:33
joriki
170k10183342
answered Mar 9 '16 at 9:33
joriki
170k10183342
answered Mar 9 '16 at 9:33
joriki
170k10183342
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
add a comment |
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
add a comment |
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A more general version of this question: stats.stackexchange.com/questions/375989/…
– kjetil b halvorsen
Nov 16 at 16:32