Proving the correlation $rho$ is always $|rho|leq 1$ and finding when $rho=-1$












0












$begingroup$


Let $xi$ and $eta$ be random variables. $mathbb{D}xi >0$ and $mathbb{D}eta>0$
Let $rho=rho(xi,eta)=frac{mathbb{E}(xi-mathbb{E}xi)(eta-mathbb{E}eta)} {sqrt{mathbb{D}xi mathbb{D}eta}}$.



I need to show that



1) $|rho|leq 1$



2) $rho=-1$ when $frac{eta -mathbb{E}eta}{sqrt{mathbb{D}eta}}=-frac{xi -mathbb{E}xi}{sqrt{mathbb{D}xi}}$



For 1) I think I should use $|mathbb{E}xieta|leq sqrt{mathbb{E}xi^2mathbb{E}eta^2}$ but actually I don't know how to do it correctly.



For 2) I have no Ideas. How should I start and what should I use?






probability-theory random-variables correlation







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$endgroup$








  • 1




    $begingroup$
    For (2), can't you "just plug it in"?
    $endgroup$
    – Rahul
    Dec 18 '18 at 13:03
















0












$begingroup$


Let $xi$ and $eta$ be random variables. $mathbb{D}xi >0$ and $mathbb{D}eta>0$
Let $rho=rho(xi,eta)=frac{mathbb{E}(xi-mathbb{E}xi)(eta-mathbb{E}eta)} {sqrt{mathbb{D}xi mathbb{D}eta}}$.



I need to show that



1) $|rho|leq 1$



2) $rho=-1$ when $frac{eta -mathbb{E}eta}{sqrt{mathbb{D}eta}}=-frac{xi -mathbb{E}xi}{sqrt{mathbb{D}xi}}$



For 1) I think I should use $|mathbb{E}xieta|leq sqrt{mathbb{E}xi^2mathbb{E}eta^2}$ but actually I don't know how to do it correctly.



For 2) I have no Ideas. How should I start and what should I use?






probability-theory random-variables correlation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For (2), can't you "just plug it in"?
    $endgroup$
    – Rahul
    Dec 18 '18 at 13:03














0












0








0





$begingroup$


Let $xi$ and $eta$ be random variables. $mathbb{D}xi >0$ and $mathbb{D}eta>0$
Let $rho=rho(xi,eta)=frac{mathbb{E}(xi-mathbb{E}xi)(eta-mathbb{E}eta)} {sqrt{mathbb{D}xi mathbb{D}eta}}$.



I need to show that



1) $|rho|leq 1$



2) $rho=-1$ when $frac{eta -mathbb{E}eta}{sqrt{mathbb{D}eta}}=-frac{xi -mathbb{E}xi}{sqrt{mathbb{D}xi}}$



For 1) I think I should use $|mathbb{E}xieta|leq sqrt{mathbb{E}xi^2mathbb{E}eta^2}$ but actually I don't know how to do it correctly.



For 2) I have no Ideas. How should I start and what should I use?






probability-theory random-variables correlation







share|cite|improve this question











$endgroup$




Let $xi$ and $eta$ be random variables. $mathbb{D}xi >0$ and $mathbb{D}eta>0$
Let $rho=rho(xi,eta)=frac{mathbb{E}(xi-mathbb{E}xi)(eta-mathbb{E}eta)} {sqrt{mathbb{D}xi mathbb{D}eta}}$.



I need to show that



1) $|rho|leq 1$



2) $rho=-1$ when $frac{eta -mathbb{E}eta}{sqrt{mathbb{D}eta}}=-frac{xi -mathbb{E}xi}{sqrt{mathbb{D}xi}}$



For 1) I think I should use $|mathbb{E}xieta|leq sqrt{mathbb{E}xi^2mathbb{E}eta^2}$ but actually I don't know how to do it correctly.



For 2) I have no Ideas. How should I start and what should I use?






probability-theory random-variables correlation




probability-theory random-variables correlation






share|cite|improve this question















share|cite|improve this question













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edited Dec 18 '18 at 15:12









Did

247k23223459




247k23223459










asked Dec 18 '18 at 12:47









AtstovasAtstovas

1109




1109








  • 1




    $begingroup$
    For (2), can't you "just plug it in"?
    $endgroup$
    – Rahul
    Dec 18 '18 at 13:03














  • 1




    $begingroup$
    For (2), can't you "just plug it in"?
    $endgroup$
    – Rahul
    Dec 18 '18 at 13:03








1




1




$begingroup$
For (2), can't you "just plug it in"?
$endgroup$
– Rahul
Dec 18 '18 at 13:03




$begingroup$
For (2), can't you "just plug it in"?
$endgroup$
– Rahul
Dec 18 '18 at 13:03










3 Answers
3






active

oldest

votes


















1












$begingroup$

For 1) you can apply the inequality you want to use on $xi-Exi$ and $eta-Eeta$ instead of $xi$ and $eta$.
To follow the statement from this, you will have to use that



$E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    1) Since zero-mean random variables comprise a vector space on which covariance is an inner product, $|rho|le 1$ is a restatement of that inner product space's Cauchy-Schwarz inequality. If this isn't intuitive, rewrite your favourite proof of the inequality, including its saturation condition (which we'll need in a moment), in the language of expectations, covariances etc.



    2) Since $rho=-1$ requires the unit vectors $frac{xi-Bbb Exi}{sqrt{Bbb Dxi}},,frac{eta-Bbb Eeta}{sqrt{Bbb Deta}}$ to be "antiparallel", their ratio is $-1$ as required.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      1) Let $z=(xi,eta)'$, then the variance matrix $V(z)=E[(z-Ez)(z-Ez)']$ is positive semidefinite and so $det(V(z)) geq 0$, which implies $|rho| leq 1$.



      2) If $a=-b$ then $a^2=-ab$ and $E[a^2]=-E[ab]$, and so $rho equiv E[ab] = -E[a^2] = -1$, by the definition of the variance of $a$.






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        For 1) you can apply the inequality you want to use on $xi-Exi$ and $eta-Eeta$ instead of $xi$ and $eta$.
        To follow the statement from this, you will have to use that



        $E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          For 1) you can apply the inequality you want to use on $xi-Exi$ and $eta-Eeta$ instead of $xi$ and $eta$.
          To follow the statement from this, you will have to use that



          $E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            For 1) you can apply the inequality you want to use on $xi-Exi$ and $eta-Eeta$ instead of $xi$ and $eta$.
            To follow the statement from this, you will have to use that



            $E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$






            share|cite|improve this answer











            $endgroup$



            For 1) you can apply the inequality you want to use on $xi-Exi$ and $eta-Eeta$ instead of $xi$ and $eta$.
            To follow the statement from this, you will have to use that



            $E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 18 '18 at 15:13

























            answered Dec 18 '18 at 15:06









            Student7Student7

            2089




            2089























                0












                $begingroup$

                1) Since zero-mean random variables comprise a vector space on which covariance is an inner product, $|rho|le 1$ is a restatement of that inner product space's Cauchy-Schwarz inequality. If this isn't intuitive, rewrite your favourite proof of the inequality, including its saturation condition (which we'll need in a moment), in the language of expectations, covariances etc.



                2) Since $rho=-1$ requires the unit vectors $frac{xi-Bbb Exi}{sqrt{Bbb Dxi}},,frac{eta-Bbb Eeta}{sqrt{Bbb Deta}}$ to be "antiparallel", their ratio is $-1$ as required.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  1) Since zero-mean random variables comprise a vector space on which covariance is an inner product, $|rho|le 1$ is a restatement of that inner product space's Cauchy-Schwarz inequality. If this isn't intuitive, rewrite your favourite proof of the inequality, including its saturation condition (which we'll need in a moment), in the language of expectations, covariances etc.



                  2) Since $rho=-1$ requires the unit vectors $frac{xi-Bbb Exi}{sqrt{Bbb Dxi}},,frac{eta-Bbb Eeta}{sqrt{Bbb Deta}}$ to be "antiparallel", their ratio is $-1$ as required.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    1) Since zero-mean random variables comprise a vector space on which covariance is an inner product, $|rho|le 1$ is a restatement of that inner product space's Cauchy-Schwarz inequality. If this isn't intuitive, rewrite your favourite proof of the inequality, including its saturation condition (which we'll need in a moment), in the language of expectations, covariances etc.



                    2) Since $rho=-1$ requires the unit vectors $frac{xi-Bbb Exi}{sqrt{Bbb Dxi}},,frac{eta-Bbb Eeta}{sqrt{Bbb Deta}}$ to be "antiparallel", their ratio is $-1$ as required.






                    share|cite|improve this answer









                    $endgroup$



                    1) Since zero-mean random variables comprise a vector space on which covariance is an inner product, $|rho|le 1$ is a restatement of that inner product space's Cauchy-Schwarz inequality. If this isn't intuitive, rewrite your favourite proof of the inequality, including its saturation condition (which we'll need in a moment), in the language of expectations, covariances etc.



                    2) Since $rho=-1$ requires the unit vectors $frac{xi-Bbb Exi}{sqrt{Bbb Dxi}},,frac{eta-Bbb Eeta}{sqrt{Bbb Deta}}$ to be "antiparallel", their ratio is $-1$ as required.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 18 '18 at 15:39









                    J.G.J.G.

                    24.9k22539




                    24.9k22539























                        0












                        $begingroup$

                        1) Let $z=(xi,eta)'$, then the variance matrix $V(z)=E[(z-Ez)(z-Ez)']$ is positive semidefinite and so $det(V(z)) geq 0$, which implies $|rho| leq 1$.



                        2) If $a=-b$ then $a^2=-ab$ and $E[a^2]=-E[ab]$, and so $rho equiv E[ab] = -E[a^2] = -1$, by the definition of the variance of $a$.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          1) Let $z=(xi,eta)'$, then the variance matrix $V(z)=E[(z-Ez)(z-Ez)']$ is positive semidefinite and so $det(V(z)) geq 0$, which implies $|rho| leq 1$.



                          2) If $a=-b$ then $a^2=-ab$ and $E[a^2]=-E[ab]$, and so $rho equiv E[ab] = -E[a^2] = -1$, by the definition of the variance of $a$.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            1) Let $z=(xi,eta)'$, then the variance matrix $V(z)=E[(z-Ez)(z-Ez)']$ is positive semidefinite and so $det(V(z)) geq 0$, which implies $|rho| leq 1$.



                            2) If $a=-b$ then $a^2=-ab$ and $E[a^2]=-E[ab]$, and so $rho equiv E[ab] = -E[a^2] = -1$, by the definition of the variance of $a$.






                            share|cite|improve this answer











                            $endgroup$



                            1) Let $z=(xi,eta)'$, then the variance matrix $V(z)=E[(z-Ez)(z-Ez)']$ is positive semidefinite and so $det(V(z)) geq 0$, which implies $|rho| leq 1$.



                            2) If $a=-b$ then $a^2=-ab$ and $E[a^2]=-E[ab]$, and so $rho equiv E[ab] = -E[a^2] = -1$, by the definition of the variance of $a$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 20 '18 at 20:18

























                            answered Dec 18 '18 at 19:37









                            BertrandBertrand

                            964




                            964






























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