Are all symmetric matrices with diagonal elements 1 and other values between -1 and 1 correlation matrices?
A question for the statisticians and other math lovers: Are all symmetric matrices with diagonal elements 1 and other values between $-1$ and 1 correlation matrices?
correlation mathematical-statistics multivariate-analysis covariance covariance-matrix
edited Jan 6 at 15:30
kjetil b halvorsen
29k980209
asked Jan 6 at 14:57
Math123Math123
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add a comment |
A question for the statisticians and other math lovers: Are all symmetric matrices with diagonal elements 1 and other values between $-1$ and 1 correlation matrices?
correlation mathematical-statistics multivariate-analysis covariance covariance-matrix
edited Jan 6 at 15:30
kjetil b halvorsen
29k980209
asked Jan 6 at 14:57
Math123Math123
311
New contributor
Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
No, it must also be positive definite.
– hard2fathom
Jan 6 at 15:01
@hard2fathom, thank you for your answer! What is this?
– Math123
Jan 6 at 15:06
Related: stats.stackexchange.com/questions/72790/…
– Julius
Jan 6 at 17:47
add a comment |
A question for the statisticians and other math lovers: Are all symmetric matrices with diagonal elements 1 and other values between $-1$ and 1 correlation matrices?
correlation mathematical-statistics multivariate-analysis covariance covariance-matrix
edited Jan 6 at 15:30
kjetil b halvorsen
29k980209
asked Jan 6 at 14:57
Math123Math123
311
New contributor
Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A question for the statisticians and other math lovers: Are all symmetric matrices with diagonal elements 1 and other values between $-1$ and 1 correlation matrices?
correlation mathematical-statistics multivariate-analysis covariance covariance-matrix
correlation mathematical-statistics multivariate-analysis covariance covariance-matrix
edited Jan 6 at 15:30
kjetil b halvorsen
29k980209
asked Jan 6 at 14:57
Math123Math123
311
New contributor
Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 6 at 15:30
kjetil b halvorsen
29k980209
asked Jan 6 at 14:57
Math123Math123
311
New contributor
Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 6 at 15:30
kjetil b halvorsen
29k980209
edited Jan 6 at 15:30
kjetil b halvorsen
29k980209
edited Jan 6 at 15:30
kjetil b halvorsen
29k980209
29k980209
asked Jan 6 at 14:57
Math123Math123
311
New contributor
Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 6 at 14:57
Math123Math123
311
asked Jan 6 at 14:57
Math123Math123
311
311
New contributor
Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
No, it must also be positive definite.
– hard2fathom
Jan 6 at 15:01
@hard2fathom, thank you for your answer! What is this?
– Math123
Jan 6 at 15:06
Related: stats.stackexchange.com/questions/72790/…
– Julius
Jan 6 at 17:47
add a comment |
5
No, it must also be positive definite.
– hard2fathom
Jan 6 at 15:01
@hard2fathom, thank you for your answer! What is this?
– Math123
Jan 6 at 15:06
Related: stats.stackexchange.com/questions/72790/…
– Julius
Jan 6 at 17:47
5
5
No, it must also be positive definite.
– hard2fathom
Jan 6 at 15:01
No, it must also be positive definite.
– hard2fathom
Jan 6 at 15:01
@hard2fathom, thank you for your answer! What is this?
– Math123
Jan 6 at 15:06
@hard2fathom, thank you for your answer! What is this?
– Math123
Jan 6 at 15:06
Related: stats.stackexchange.com/questions/72790/…
– Julius
Jan 6 at 17:47
Related: stats.stackexchange.com/questions/72790/…
– Julius
Jan 6 at 17:47
add a comment |
1 Answer
1
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votes
I thought this must be asked & answered before, but cannot find it, so here it goes ... Let $S$ be a covariance matrix (for the algebra that follows it does not matter if it is theoretical or empirical). Let $D$ be a diagonal matrix with the diagonal of $S$. Then the correlation matrix $R$ is given by
$$
R= D^{-1/2} S D^{-1/2}
$$ (how you can see this directly is explained here.)
To see that $S$ must be positive (semi)-definite (abbreviated posdef), let $X$ be a random variable with covariance amtrix $S$, and $c$ a vector. Then
$$ DeclareMathOperator{var}{mathbb{V}ar}
var(c^T X)= c^T S c ge 0
$$ since variance is always nonnegative. Then this transfers to the correlation matrix:
$$
c^T R c = c^T D^{-1/2} S D^{-1/2} c = (D^{-1/2} c)^T S (D^{-1/2} c) ge 0
$$
Armed with this it is easy to make an counterexample, the following is not a correlation matrix:
$$
begin{pmatrix} 1 & -0.9 & -0.9 \
-0.9& 1 & -0.9 \
-0.9 & -0.9 & 1 end{pmatrix}
$$
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29k980209
add a comment |
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1 Answer
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I thought this must be asked & answered before, but cannot find it, so here it goes ... Let $S$ be a covariance matrix (for the algebra that follows it does not matter if it is theoretical or empirical). Let $D$ be a diagonal matrix with the diagonal of $S$. Then the correlation matrix $R$ is given by
$$
R= D^{-1/2} S D^{-1/2}
$$ (how you can see this directly is explained here.)
To see that $S$ must be positive (semi)-definite (abbreviated posdef), let $X$ be a random variable with covariance amtrix $S$, and $c$ a vector. Then
$$ DeclareMathOperator{var}{mathbb{V}ar}
var(c^T X)= c^T S c ge 0
$$ since variance is always nonnegative. Then this transfers to the correlation matrix:
$$
c^T R c = c^T D^{-1/2} S D^{-1/2} c = (D^{-1/2} c)^T S (D^{-1/2} c) ge 0
$$
Armed with this it is easy to make an counterexample, the following is not a correlation matrix:
$$
begin{pmatrix} 1 & -0.9 & -0.9 \
-0.9& 1 & -0.9 \
-0.9 & -0.9 & 1 end{pmatrix}
$$
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29k980209
add a comment |
I thought this must be asked & answered before, but cannot find it, so here it goes ... Let $S$ be a covariance matrix (for the algebra that follows it does not matter if it is theoretical or empirical). Let $D$ be a diagonal matrix with the diagonal of $S$. Then the correlation matrix $R$ is given by
$$
R= D^{-1/2} S D^{-1/2}
$$ (how you can see this directly is explained here.)
To see that $S$ must be positive (semi)-definite (abbreviated posdef), let $X$ be a random variable with covariance amtrix $S$, and $c$ a vector. Then
$$ DeclareMathOperator{var}{mathbb{V}ar}
var(c^T X)= c^T S c ge 0
$$ since variance is always nonnegative. Then this transfers to the correlation matrix:
$$
c^T R c = c^T D^{-1/2} S D^{-1/2} c = (D^{-1/2} c)^T S (D^{-1/2} c) ge 0
$$
Armed with this it is easy to make an counterexample, the following is not a correlation matrix:
$$
begin{pmatrix} 1 & -0.9 & -0.9 \
-0.9& 1 & -0.9 \
-0.9 & -0.9 & 1 end{pmatrix}
$$
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29k980209
add a comment |
I thought this must be asked & answered before, but cannot find it, so here it goes ... Let $S$ be a covariance matrix (for the algebra that follows it does not matter if it is theoretical or empirical). Let $D$ be a diagonal matrix with the diagonal of $S$. Then the correlation matrix $R$ is given by
$$
R= D^{-1/2} S D^{-1/2}
$$ (how you can see this directly is explained here.)
To see that $S$ must be positive (semi)-definite (abbreviated posdef), let $X$ be a random variable with covariance amtrix $S$, and $c$ a vector. Then
$$ DeclareMathOperator{var}{mathbb{V}ar}
var(c^T X)= c^T S c ge 0
$$ since variance is always nonnegative. Then this transfers to the correlation matrix:
$$
c^T R c = c^T D^{-1/2} S D^{-1/2} c = (D^{-1/2} c)^T S (D^{-1/2} c) ge 0
$$
Armed with this it is easy to make an counterexample, the following is not a correlation matrix:
$$
begin{pmatrix} 1 & -0.9 & -0.9 \
-0.9& 1 & -0.9 \
-0.9 & -0.9 & 1 end{pmatrix}
$$
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29k980209
I thought this must be asked & answered before, but cannot find it, so here it goes ... Let $S$ be a covariance matrix (for the algebra that follows it does not matter if it is theoretical or empirical). Let $D$ be a diagonal matrix with the diagonal of $S$. Then the correlation matrix $R$ is given by
$$
R= D^{-1/2} S D^{-1/2}
$$ (how you can see this directly is explained here.)
To see that $S$ must be positive (semi)-definite (abbreviated posdef), let $X$ be a random variable with covariance amtrix $S$, and $c$ a vector. Then
$$ DeclareMathOperator{var}{mathbb{V}ar}
var(c^T X)= c^T S c ge 0
$$ since variance is always nonnegative. Then this transfers to the correlation matrix:
$$
c^T R c = c^T D^{-1/2} S D^{-1/2} c = (D^{-1/2} c)^T S (D^{-1/2} c) ge 0
$$
Armed with this it is easy to make an counterexample, the following is not a correlation matrix:
$$
begin{pmatrix} 1 & -0.9 & -0.9 \
-0.9& 1 & -0.9 \
-0.9 & -0.9 & 1 end{pmatrix}
$$
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29k980209
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29k980209
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29k980209
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29k980209
29k980209
add a comment |
add a comment |
Math123 is a new contributor. Be nice, and check out our Code of Conduct.
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5
No, it must also be positive definite.
– hard2fathom
Jan 6 at 15:01
@hard2fathom, thank you for your answer! What is this?
– Math123
Jan 6 at 15:06
Related: stats.stackexchange.com/questions/72790/…
– Julius
Jan 6 at 17:47