Mixing
Are all symmetric matrices with diagonal elements 1 and other values between -1 and 1 correlation matrices?
$begingroup$
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
29.5k980216
asked Jan 6 at 14:57
Math123Math123
311
$endgroup$
add a comment |
$begingroup$
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
29.5k980216
asked Jan 6 at 14:57
Math123Math123
311
$endgroup$
6
$begingroup$
No, it must also be positive definite.
$endgroup$
– hard2fathom
Jan 6 at 15:01
$begingroup$
@hard2fathom, thank you for your answer! What is this?
$endgroup$
– Math123
Jan 6 at 15:06
$begingroup$
Related: stats.stackexchange.com/questions/72790/…
$endgroup$
– Julius
Jan 6 at 17:47
add a comment |
$begingroup$
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
29.5k980216
asked Jan 6 at 14:57
Math123Math123
311
$endgroup$
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
29.5k980216
asked Jan 6 at 14:57
Math123Math123
311
edited Jan 6 at 15:30
kjetil b halvorsen
29.5k980216
asked Jan 6 at 14:57
Math123Math123
311
edited Jan 6 at 15:30
kjetil b halvorsen
29.5k980216
edited Jan 6 at 15:30
kjetil b halvorsen
29.5k980216
edited Jan 6 at 15:30
kjetil b halvorsen
29.5k980216
29.5k980216
asked Jan 6 at 14:57
Math123Math123
311
asked Jan 6 at 14:57
Math123Math123
311
asked Jan 6 at 14:57
Math123Math123
311
311
6
$begingroup$
No, it must also be positive definite.
$endgroup$
– hard2fathom
Jan 6 at 15:01
$begingroup$
@hard2fathom, thank you for your answer! What is this?
$endgroup$
– Math123
Jan 6 at 15:06
$begingroup$
Related: stats.stackexchange.com/questions/72790/…
$endgroup$
– Julius
Jan 6 at 17:47
add a comment |
6
$begingroup$
No, it must also be positive definite.
$endgroup$
– hard2fathom
Jan 6 at 15:01
$begingroup$
@hard2fathom, thank you for your answer! What is this?
$endgroup$
– Math123
Jan 6 at 15:06
$begingroup$
Related: stats.stackexchange.com/questions/72790/…
$endgroup$
– Julius
Jan 6 at 17:47
6
6
$begingroup$
No, it must also be positive definite.
$endgroup$
– hard2fathom
Jan 6 at 15:01
$begingroup$
No, it must also be positive definite.
$endgroup$
– hard2fathom
Jan 6 at 15:01
$begingroup$
@hard2fathom, thank you for your answer! What is this?
$endgroup$
– Math123
Jan 6 at 15:06
$begingroup$
@hard2fathom, thank you for your answer! What is this?
$endgroup$
– Math123
Jan 6 at 15:06
$begingroup$
Related: stats.stackexchange.com/questions/72790/…
$endgroup$
– Julius
Jan 6 at 17:47
$begingroup$
Related: stats.stackexchange.com/questions/72790/…
$endgroup$
– Julius
Jan 6 at 17:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
29.5k980216
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f385852%2fare-all-symmetric-matrices-with-diagonal-elements-1-and-other-values-between-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
29.5k980216
$endgroup$
add a comment |
$begingroup$
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
29.5k980216
$endgroup$
add a comment |
$begingroup$
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
29.5k980216
$endgroup$
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
29.5k980216
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29.5k980216
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29.5k980216
answered Jan 6 at 16:05
kjetil b halvorsenkjetil b halvorsen
29.5k980216
29.5k980216
add a comment |
add a comment |
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f385852%2fare-all-symmetric-matrices-with-diagonal-elements-1-and-other-values-between-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
6
$begingroup$
No, it must also be positive definite.
$endgroup$
– hard2fathom
Jan 6 at 15:01
$begingroup$
@hard2fathom, thank you for your answer! What is this?
$endgroup$
– Math123
Jan 6 at 15:06
$begingroup$
Related: stats.stackexchange.com/questions/72790/…
$endgroup$
– Julius
Jan 6 at 17:47