Random variables with arbitrary positive correlation but without arbitrary negative correlation. Then...
$begingroup$
Let $X_i in L_2$ be a sequence of pairwise correlated random variables. The random variables can have arbitrary positive correlation but can't have arbitrary negative correlation.
How can I show that for $(X_1,ldots,X_n)$ and $forall i,j in {1,ldots,n}, ineq j$
$$mathrm{Cor}(X_i,X_j)<frac{-1}{n-1}$$
is not possible
probability statistics measure-theory
asked Dec 11 '18 at 15:04
conradconrad
757
$endgroup$
add a comment |
$begingroup$
Let $X_i in L_2$ be a sequence of pairwise correlated random variables. The random variables can have arbitrary positive correlation but can't have arbitrary negative correlation.
How can I show that for $(X_1,ldots,X_n)$ and $forall i,j in {1,ldots,n}, ineq j$
$$mathrm{Cor}(X_i,X_j)<frac{-1}{n-1}$$
is not possible
probability statistics measure-theory
asked Dec 11 '18 at 15:04
conradconrad
757
$endgroup$
$begingroup$
Are you sure that your claim is correct ? What if $n = 2$, $X_1 = -X_2$ with variance equal to $1$ ?
$endgroup$
– dallonsi
Dec 11 '18 at 15:27
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@dallonsi, it is correct; note the strict inequality in the question, whereas your example has an equality.
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– Marcus M
Dec 11 '18 at 17:21
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yes Marcus, thanks. My mistake
$endgroup$
– dallonsi
Dec 13 '18 at 13:11
add a comment |
$begingroup$
Let $X_i in L_2$ be a sequence of pairwise correlated random variables. The random variables can have arbitrary positive correlation but can't have arbitrary negative correlation.
How can I show that for $(X_1,ldots,X_n)$ and $forall i,j in {1,ldots,n}, ineq j$
$$mathrm{Cor}(X_i,X_j)<frac{-1}{n-1}$$
is not possible
probability statistics measure-theory
asked Dec 11 '18 at 15:04
conradconrad
757
$endgroup$
Let $X_i in L_2$ be a sequence of pairwise correlated random variables. The random variables can have arbitrary positive correlation but can't have arbitrary negative correlation.
How can I show that for $(X_1,ldots,X_n)$ and $forall i,j in {1,ldots,n}, ineq j$
$$mathrm{Cor}(X_i,X_j)<frac{-1}{n-1}$$
is not possible
probability statistics measure-theory
probability statistics measure-theory
asked Dec 11 '18 at 15:04
conradconrad
757
asked Dec 11 '18 at 15:04
conradconrad
757
asked Dec 11 '18 at 15:04
conradconrad
757
asked Dec 11 '18 at 15:04
conradconrad
757
asked Dec 11 '18 at 15:04
conradconrad
757
757
$begingroup$
Are you sure that your claim is correct ? What if $n = 2$, $X_1 = -X_2$ with variance equal to $1$ ?
$endgroup$
– dallonsi
Dec 11 '18 at 15:27
$begingroup$
@dallonsi, it is correct; note the strict inequality in the question, whereas your example has an equality.
$endgroup$
– Marcus M
Dec 11 '18 at 17:21
$begingroup$
yes Marcus, thanks. My mistake
$endgroup$
– dallonsi
Dec 13 '18 at 13:11
add a comment |
$begingroup$
Are you sure that your claim is correct ? What if $n = 2$, $X_1 = -X_2$ with variance equal to $1$ ?
$endgroup$
– dallonsi
Dec 11 '18 at 15:27
$begingroup$
@dallonsi, it is correct; note the strict inequality in the question, whereas your example has an equality.
$endgroup$
– Marcus M
Dec 11 '18 at 17:21
$begingroup$
yes Marcus, thanks. My mistake
$endgroup$
– dallonsi
Dec 13 '18 at 13:11
$begingroup$
Are you sure that your claim is correct ? What if $n = 2$, $X_1 = -X_2$ with variance equal to $1$ ?
$endgroup$
– dallonsi
Dec 11 '18 at 15:27
$begingroup$
Are you sure that your claim is correct ? What if $n = 2$, $X_1 = -X_2$ with variance equal to $1$ ?
$endgroup$
– dallonsi
Dec 11 '18 at 15:27
$begingroup$
@dallonsi, it is correct; note the strict inequality in the question, whereas your example has an equality.
$endgroup$
– Marcus M
Dec 11 '18 at 17:21
$begingroup$
@dallonsi, it is correct; note the strict inequality in the question, whereas your example has an equality.
$endgroup$
– Marcus M
Dec 11 '18 at 17:21
$begingroup$
yes Marcus, thanks. My mistake
$endgroup$
– dallonsi
Dec 13 '18 at 13:11
$begingroup$
yes Marcus, thanks. My mistake
$endgroup$
– dallonsi
Dec 13 '18 at 13:11
add a comment |
1 Answer
1
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oldest
votes
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The idea is to realize that covariance matrices are positive semi-definite. Define variables $Y_j = X_j/ sqrt{operatorname{Var}(X_j)}.$ Seeking a contradiction, suppose that the condition you mention holds, i.e. $$operatorname{Cor}(X_i,X_j) = operatorname{Cov}(Y_i,Y_j) < -frac{1}{n-1}.$$
Let $Sigma$ be the covariance matrix of $(Y_1,ldots,Y_n)$. Then $Sigma$ has $1$'s on the diagonal, and each off-diagonal entry is less than $-frac{1}{n-1}$. If we define $x = (1,1,ldots,1)^T$, then we see that $x^T Sigma x < 0$, which contradicts the fact that covariance matrices are positive semi-definite.
answered Dec 11 '18 at 17:25
Marcus MMarcus M
8,7931947
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add a comment |
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1 Answer
1
active
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1 Answer
1
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oldest
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$begingroup$
The idea is to realize that covariance matrices are positive semi-definite. Define variables $Y_j = X_j/ sqrt{operatorname{Var}(X_j)}.$ Seeking a contradiction, suppose that the condition you mention holds, i.e. $$operatorname{Cor}(X_i,X_j) = operatorname{Cov}(Y_i,Y_j) < -frac{1}{n-1}.$$
Let $Sigma$ be the covariance matrix of $(Y_1,ldots,Y_n)$. Then $Sigma$ has $1$'s on the diagonal, and each off-diagonal entry is less than $-frac{1}{n-1}$. If we define $x = (1,1,ldots,1)^T$, then we see that $x^T Sigma x < 0$, which contradicts the fact that covariance matrices are positive semi-definite.
answered Dec 11 '18 at 17:25
Marcus MMarcus M
8,7931947
$endgroup$
add a comment |
$begingroup$
The idea is to realize that covariance matrices are positive semi-definite. Define variables $Y_j = X_j/ sqrt{operatorname{Var}(X_j)}.$ Seeking a contradiction, suppose that the condition you mention holds, i.e. $$operatorname{Cor}(X_i,X_j) = operatorname{Cov}(Y_i,Y_j) < -frac{1}{n-1}.$$
Let $Sigma$ be the covariance matrix of $(Y_1,ldots,Y_n)$. Then $Sigma$ has $1$'s on the diagonal, and each off-diagonal entry is less than $-frac{1}{n-1}$. If we define $x = (1,1,ldots,1)^T$, then we see that $x^T Sigma x < 0$, which contradicts the fact that covariance matrices are positive semi-definite.
answered Dec 11 '18 at 17:25
Marcus MMarcus M
8,7931947
$endgroup$
add a comment |
$begingroup$
The idea is to realize that covariance matrices are positive semi-definite. Define variables $Y_j = X_j/ sqrt{operatorname{Var}(X_j)}.$ Seeking a contradiction, suppose that the condition you mention holds, i.e. $$operatorname{Cor}(X_i,X_j) = operatorname{Cov}(Y_i,Y_j) < -frac{1}{n-1}.$$
Let $Sigma$ be the covariance matrix of $(Y_1,ldots,Y_n)$. Then $Sigma$ has $1$'s on the diagonal, and each off-diagonal entry is less than $-frac{1}{n-1}$. If we define $x = (1,1,ldots,1)^T$, then we see that $x^T Sigma x < 0$, which contradicts the fact that covariance matrices are positive semi-definite.
answered Dec 11 '18 at 17:25
Marcus MMarcus M
8,7931947
$endgroup$
The idea is to realize that covariance matrices are positive semi-definite. Define variables $Y_j = X_j/ sqrt{operatorname{Var}(X_j)}.$ Seeking a contradiction, suppose that the condition you mention holds, i.e. $$operatorname{Cor}(X_i,X_j) = operatorname{Cov}(Y_i,Y_j) < -frac{1}{n-1}.$$
Let $Sigma$ be the covariance matrix of $(Y_1,ldots,Y_n)$. Then $Sigma$ has $1$'s on the diagonal, and each off-diagonal entry is less than $-frac{1}{n-1}$. If we define $x = (1,1,ldots,1)^T$, then we see that $x^T Sigma x < 0$, which contradicts the fact that covariance matrices are positive semi-definite.
answered Dec 11 '18 at 17:25
Marcus MMarcus M
8,7931947
answered Dec 11 '18 at 17:25
Marcus MMarcus M
8,7931947
answered Dec 11 '18 at 17:25
Marcus MMarcus M
8,7931947
answered Dec 11 '18 at 17:25
Marcus MMarcus M
8,7931947
8,7931947
add a comment |
add a comment |
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$begingroup$
Are you sure that your claim is correct ? What if $n = 2$, $X_1 = -X_2$ with variance equal to $1$ ?
$endgroup$
– dallonsi
Dec 11 '18 at 15:27
$begingroup$
@dallonsi, it is correct; note the strict inequality in the question, whereas your example has an equality.
$endgroup$
– Marcus M
Dec 11 '18 at 17:21
$begingroup$
yes Marcus, thanks. My mistake
$endgroup$
– dallonsi
Dec 13 '18 at 13:11