Proving the correlation $rho$ is always $|rho|leq 1$ and finding when $rho=-1$
$begingroup$
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
edited Dec 18 '18 at 15:12
Did
249k23226466
asked Dec 18 '18 at 12:47
AtstovasAtstovas
1139
$endgroup$
add a comment |
$begingroup$
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
edited Dec 18 '18 at 15:12
Did
249k23226466
asked Dec 18 '18 at 12:47
AtstovasAtstovas
1139
$endgroup$
1
$begingroup$
For (2), can't you "just plug it in"?
$endgroup$
– Rahul
Dec 18 '18 at 13:03
add a comment |
$begingroup$
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
edited Dec 18 '18 at 15:12
Did
249k23226466
asked Dec 18 '18 at 12:47
AtstovasAtstovas
1139
$endgroup$
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
edited Dec 18 '18 at 15:12
Did
249k23226466
asked Dec 18 '18 at 12:47
AtstovasAtstovas
1139
edited Dec 18 '18 at 15:12
Did
249k23226466
asked Dec 18 '18 at 12:47
AtstovasAtstovas
1139
edited Dec 18 '18 at 15:12
Did
249k23226466
edited Dec 18 '18 at 15:12
Did
249k23226466
edited Dec 18 '18 at 15:12
Did
249k23226466
249k23226466
asked Dec 18 '18 at 12:47
AtstovasAtstovas
1139
asked Dec 18 '18 at 12:47
AtstovasAtstovas
1139
asked Dec 18 '18 at 12:47
AtstovasAtstovas
1139
1139
1
$begingroup$
For (2), can't you "just plug it in"?
$endgroup$
– Rahul
Dec 18 '18 at 13:03
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
$E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$
answered Dec 18 '18 at 15:06
Student7Student7
1839
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 18 '18 at 15:39
J.G.J.G.
32.1k23250
$endgroup$
add a comment |
$begingroup$
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 $a$.
answered Dec 18 '18 at 19:37
BertrandBertrand
45815
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$
answered Dec 18 '18 at 15:06
Student7Student7
1839
$endgroup$
add a comment |
$begingroup$
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
$E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$
answered Dec 18 '18 at 15:06
Student7Student7
1839
$endgroup$
add a comment |
$begingroup$
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
$E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$
answered Dec 18 '18 at 15:06
Student7Student7
1839
$endgroup$
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
$E((xi-Exi)^2)=E(xi^2-2xi E(xi)+E(xi)^2)=E(xi^2)-2E(xi)E(xi)+E(xi)^2=E(xi^2)-E(xi)^2=Dxi.$
answered Dec 18 '18 at 15:06
Student7Student7
1839
edited Dec 18 '18 at 15:13
answered Dec 18 '18 at 15:06
Student7Student7
1839
answered Dec 18 '18 at 15:06
Student7Student7
1839
answered Dec 18 '18 at 15:06
Student7Student7
1839
1839
add a comment |
add a comment |
$begingroup$
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.
answered Dec 18 '18 at 15:39
J.G.J.G.
32.1k23250
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 18 '18 at 15:39
J.G.J.G.
32.1k23250
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 18 '18 at 15:39
J.G.J.G.
32.1k23250
$endgroup$
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.
answered Dec 18 '18 at 15:39
J.G.J.G.
32.1k23250
answered Dec 18 '18 at 15:39
J.G.J.G.
32.1k23250
answered Dec 18 '18 at 15:39
J.G.J.G.
32.1k23250
answered Dec 18 '18 at 15:39
J.G.J.G.
32.1k23250
32.1k23250
add a comment |
add a comment |
$begingroup$
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 $a$.
answered Dec 18 '18 at 19:37
BertrandBertrand
45815
$endgroup$
add a comment |
$begingroup$
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 $a$.
answered Dec 18 '18 at 19:37
BertrandBertrand
45815
$endgroup$
add a comment |
$begingroup$
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 $a$.
answered Dec 18 '18 at 19:37
BertrandBertrand
45815
$endgroup$
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 $a$.
answered Dec 18 '18 at 19:37
BertrandBertrand
45815
edited Dec 20 '18 at 20:18
answered Dec 18 '18 at 19:37
BertrandBertrand
45815
answered Dec 18 '18 at 19:37
BertrandBertrand
45815
answered Dec 18 '18 at 19:37
BertrandBertrand
45815
45815
add a comment |
add a comment |
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1
$begingroup$
For (2), can't you "just plug it in"?
$endgroup$
– Rahul
Dec 18 '18 at 13:03