Random variables with arbitrary positive correlation but without arbitrary negative correlation. Then...












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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












  • $begingroup$
    @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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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












  • $begingroup$
    @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














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













    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: "69"
    };
    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: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    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
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035378%2frandom-variables-with-arbitrary-positive-correlation-but-without-arbitrary-negat%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









    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 17:25









        Marcus MMarcus M

        8,7931947




        8,7931947






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • 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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035378%2frandom-variables-with-arbitrary-positive-correlation-but-without-arbitrary-negat%23new-answer', 'question_page');
            }
            );

            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







            -

            Popular posts from this blog

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more