Prove $P({Xge t}cup{Yge t})le t^{-2}(1+sqrt{1-r^2})$ where $r$ is the correlation of $X$ and $Y$ centered...
$begingroup$
Let $xi$ and $eta$ be random variables with variances $mathbb{D}xi$ and $mathbb{D}eta$, and correlation coefficient $rho$. Show that $$mathbb{P}({xi - mathbb{E}xi geq varepsilon sqrt{mathbb{D}xi} } cup { eta - mathbb{E}eta geq varepsilon sqrt{mathbb{D}eta}) leq frac{1}{varepsilon^2}(1+sqrt{1-rho^2}).$$
So, if I am right, I see a connection with $$mathbb{E}max{xi^2,eta^2}leq 1 + sqrt{1-rho^2}.$$ But actually don't know how to use it here.
probability-theory random-variables correlation
edited Dec 18 '18 at 20:14
Did
249k23226466
asked Dec 18 '18 at 19:36
AtstovasAtstovas
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$endgroup$
add a comment |
$begingroup$
Let $xi$ and $eta$ be random variables with variances $mathbb{D}xi$ and $mathbb{D}eta$, and correlation coefficient $rho$. Show that $$mathbb{P}({xi - mathbb{E}xi geq varepsilon sqrt{mathbb{D}xi} } cup { eta - mathbb{E}eta geq varepsilon sqrt{mathbb{D}eta}) leq frac{1}{varepsilon^2}(1+sqrt{1-rho^2}).$$
So, if I am right, I see a connection with $$mathbb{E}max{xi^2,eta^2}leq 1 + sqrt{1-rho^2}.$$ But actually don't know how to use it here.
probability-theory random-variables correlation
edited Dec 18 '18 at 20:14
Did
249k23226466
asked Dec 18 '18 at 19:36
AtstovasAtstovas
1139
$endgroup$
add a comment |
$begingroup$
Let $xi$ and $eta$ be random variables with variances $mathbb{D}xi$ and $mathbb{D}eta$, and correlation coefficient $rho$. Show that $$mathbb{P}({xi - mathbb{E}xi geq varepsilon sqrt{mathbb{D}xi} } cup { eta - mathbb{E}eta geq varepsilon sqrt{mathbb{D}eta}) leq frac{1}{varepsilon^2}(1+sqrt{1-rho^2}).$$
So, if I am right, I see a connection with $$mathbb{E}max{xi^2,eta^2}leq 1 + sqrt{1-rho^2}.$$ But actually don't know how to use it here.
probability-theory random-variables correlation
edited Dec 18 '18 at 20:14
Did
249k23226466
asked Dec 18 '18 at 19:36
AtstovasAtstovas
1139
$endgroup$
Let $xi$ and $eta$ be random variables with variances $mathbb{D}xi$ and $mathbb{D}eta$, and correlation coefficient $rho$. Show that $$mathbb{P}({xi - mathbb{E}xi geq varepsilon sqrt{mathbb{D}xi} } cup { eta - mathbb{E}eta geq varepsilon sqrt{mathbb{D}eta}) leq frac{1}{varepsilon^2}(1+sqrt{1-rho^2}).$$
So, if I am right, I see a connection with $$mathbb{E}max{xi^2,eta^2}leq 1 + sqrt{1-rho^2}.$$ But actually don't know how to use it here.
probability-theory random-variables correlation
probability-theory random-variables correlation
edited Dec 18 '18 at 20:14
Did
249k23226466
asked Dec 18 '18 at 19:36
AtstovasAtstovas
1139
edited Dec 18 '18 at 20:14
Did
249k23226466
asked Dec 18 '18 at 19:36
AtstovasAtstovas
1139
edited Dec 18 '18 at 20:14
Did
249k23226466
edited Dec 18 '18 at 20:14
Did
249k23226466
edited Dec 18 '18 at 20:14
Did
249k23226466
249k23226466
asked Dec 18 '18 at 19:36
AtstovasAtstovas
1139
asked Dec 18 '18 at 19:36
AtstovasAtstovas
1139
asked Dec 18 '18 at 19:36
AtstovasAtstovas
1139
1139
add a comment |
add a comment |
1 Answer
1
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$begingroup$
WLOG after renormalizing, both variables have mean zero and dispersions equal to one, so we can write this as
$$
defe{varepsilon}
P(xige ecup etage e)=P(max(xi,eta)gee)le P(max(xi^2,eta^2)ge e^2).
$$
Now, use Markov's inequality to relate this to $E[max(xi^2,eta^2)]$.
answered Dec 18 '18 at 20:20
Mike EarnestMike Earnest
26.1k22151
$endgroup$
$begingroup$
Thank you. How to prove right inequality's side I know, but do I need to find $E(max{ (xi,eta) }$? If yes, then I have problems with it... I did everything the same, but don't know how much is $mathbb{E}(sqrt{xi}sqrt{eta})$. Is it equal to $sqrt{rho}$?
$endgroup$
– Atstovas
Dec 18 '18 at 20:51
$begingroup$
No, you only need to deal with $E[max(xi^2,eta^2)]$.
$endgroup$
– Mike Earnest
Dec 18 '18 at 20:52
$begingroup$
Oh, ok. Then I done with it. Thank you for your help
$endgroup$
– Atstovas
Dec 18 '18 at 20:54
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
WLOG after renormalizing, both variables have mean zero and dispersions equal to one, so we can write this as
$$
defe{varepsilon}
P(xige ecup etage e)=P(max(xi,eta)gee)le P(max(xi^2,eta^2)ge e^2).
$$
Now, use Markov's inequality to relate this to $E[max(xi^2,eta^2)]$.
answered Dec 18 '18 at 20:20
Mike EarnestMike Earnest
26.1k22151
$endgroup$
$begingroup$
Thank you. How to prove right inequality's side I know, but do I need to find $E(max{ (xi,eta) }$? If yes, then I have problems with it... I did everything the same, but don't know how much is $mathbb{E}(sqrt{xi}sqrt{eta})$. Is it equal to $sqrt{rho}$?
$endgroup$
– Atstovas
Dec 18 '18 at 20:51
$begingroup$
No, you only need to deal with $E[max(xi^2,eta^2)]$.
$endgroup$
– Mike Earnest
Dec 18 '18 at 20:52
$begingroup$
Oh, ok. Then I done with it. Thank you for your help
$endgroup$
– Atstovas
Dec 18 '18 at 20:54
add a comment |
$begingroup$
WLOG after renormalizing, both variables have mean zero and dispersions equal to one, so we can write this as
$$
defe{varepsilon}
P(xige ecup etage e)=P(max(xi,eta)gee)le P(max(xi^2,eta^2)ge e^2).
$$
Now, use Markov's inequality to relate this to $E[max(xi^2,eta^2)]$.
answered Dec 18 '18 at 20:20
Mike EarnestMike Earnest
26.1k22151
$endgroup$
$begingroup$
Thank you. How to prove right inequality's side I know, but do I need to find $E(max{ (xi,eta) }$? If yes, then I have problems with it... I did everything the same, but don't know how much is $mathbb{E}(sqrt{xi}sqrt{eta})$. Is it equal to $sqrt{rho}$?
$endgroup$
– Atstovas
Dec 18 '18 at 20:51
$begingroup$
No, you only need to deal with $E[max(xi^2,eta^2)]$.
$endgroup$
– Mike Earnest
Dec 18 '18 at 20:52
$begingroup$
Oh, ok. Then I done with it. Thank you for your help
$endgroup$
– Atstovas
Dec 18 '18 at 20:54
add a comment |
$begingroup$
WLOG after renormalizing, both variables have mean zero and dispersions equal to one, so we can write this as
$$
defe{varepsilon}
P(xige ecup etage e)=P(max(xi,eta)gee)le P(max(xi^2,eta^2)ge e^2).
$$
Now, use Markov's inequality to relate this to $E[max(xi^2,eta^2)]$.
answered Dec 18 '18 at 20:20
Mike EarnestMike Earnest
26.1k22151
$endgroup$
WLOG after renormalizing, both variables have mean zero and dispersions equal to one, so we can write this as
$$
defe{varepsilon}
P(xige ecup etage e)=P(max(xi,eta)gee)le P(max(xi^2,eta^2)ge e^2).
$$
Now, use Markov's inequality to relate this to $E[max(xi^2,eta^2)]$.
answered Dec 18 '18 at 20:20
Mike EarnestMike Earnest
26.1k22151
answered Dec 18 '18 at 20:20
Mike EarnestMike Earnest
26.1k22151
answered Dec 18 '18 at 20:20
Mike EarnestMike Earnest
26.1k22151
answered Dec 18 '18 at 20:20
Mike EarnestMike Earnest
26.1k22151
26.1k22151
$begingroup$
Thank you. How to prove right inequality's side I know, but do I need to find $E(max{ (xi,eta) }$? If yes, then I have problems with it... I did everything the same, but don't know how much is $mathbb{E}(sqrt{xi}sqrt{eta})$. Is it equal to $sqrt{rho}$?
$endgroup$
– Atstovas
Dec 18 '18 at 20:51
$begingroup$
No, you only need to deal with $E[max(xi^2,eta^2)]$.
$endgroup$
– Mike Earnest
Dec 18 '18 at 20:52
$begingroup$
Oh, ok. Then I done with it. Thank you for your help
$endgroup$
– Atstovas
Dec 18 '18 at 20:54
add a comment |
$begingroup$
Thank you. How to prove right inequality's side I know, but do I need to find $E(max{ (xi,eta) }$? If yes, then I have problems with it... I did everything the same, but don't know how much is $mathbb{E}(sqrt{xi}sqrt{eta})$. Is it equal to $sqrt{rho}$?
$endgroup$
– Atstovas
Dec 18 '18 at 20:51
$begingroup$
No, you only need to deal with $E[max(xi^2,eta^2)]$.
$endgroup$
– Mike Earnest
Dec 18 '18 at 20:52
$begingroup$
Oh, ok. Then I done with it. Thank you for your help
$endgroup$
– Atstovas
Dec 18 '18 at 20:54
$begingroup$
Thank you. How to prove right inequality's side I know, but do I need to find $E(max{ (xi,eta) }$? If yes, then I have problems with it... I did everything the same, but don't know how much is $mathbb{E}(sqrt{xi}sqrt{eta})$. Is it equal to $sqrt{rho}$?
$endgroup$
– Atstovas
Dec 18 '18 at 20:51
$begingroup$
Thank you. How to prove right inequality's side I know, but do I need to find $E(max{ (xi,eta) }$? If yes, then I have problems with it... I did everything the same, but don't know how much is $mathbb{E}(sqrt{xi}sqrt{eta})$. Is it equal to $sqrt{rho}$?
$endgroup$
– Atstovas
Dec 18 '18 at 20:51
$begingroup$
No, you only need to deal with $E[max(xi^2,eta^2)]$.
$endgroup$
– Mike Earnest
Dec 18 '18 at 20:52
$begingroup$
No, you only need to deal with $E[max(xi^2,eta^2)]$.
$endgroup$
– Mike Earnest
Dec 18 '18 at 20:52
$begingroup$
Oh, ok. Then I done with it. Thank you for your help
$endgroup$
– Atstovas
Dec 18 '18 at 20:54
$begingroup$
Oh, ok. Then I done with it. Thank you for your help
$endgroup$
– Atstovas
Dec 18 '18 at 20:54
add a comment |
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