Probability of normal distribution larger than half normal distribution
$begingroup$
I am pretty new to normal distribution and stuck on this question:
X ~ N(0, 1)
Y ~ N(0, 1)
X independent of Y
What is the probability of P(X>|2Y|)?
I know linear combination of normal distribution is normal distribution but what about linear combination of normal distribution and absolute value of normal distribution?
probability normal-distribution
asked Dec 29 '18 at 2:10
Ziyue JinZiyue Jin
31
$endgroup$
add a comment |
$begingroup$
I am pretty new to normal distribution and stuck on this question:
X ~ N(0, 1)
Y ~ N(0, 1)
X independent of Y
What is the probability of P(X>|2Y|)?
I know linear combination of normal distribution is normal distribution but what about linear combination of normal distribution and absolute value of normal distribution?
probability normal-distribution
asked Dec 29 '18 at 2:10
Ziyue JinZiyue Jin
31
$endgroup$
add a comment |
$begingroup$
I am pretty new to normal distribution and stuck on this question:
X ~ N(0, 1)
Y ~ N(0, 1)
X independent of Y
What is the probability of P(X>|2Y|)?
I know linear combination of normal distribution is normal distribution but what about linear combination of normal distribution and absolute value of normal distribution?
probability normal-distribution
asked Dec 29 '18 at 2:10
Ziyue JinZiyue Jin
31
$endgroup$
I am pretty new to normal distribution and stuck on this question:
X ~ N(0, 1)
Y ~ N(0, 1)
X independent of Y
What is the probability of P(X>|2Y|)?
I know linear combination of normal distribution is normal distribution but what about linear combination of normal distribution and absolute value of normal distribution?
probability normal-distribution
probability normal-distribution
asked Dec 29 '18 at 2:10
Ziyue JinZiyue Jin
31
asked Dec 29 '18 at 2:10
Ziyue JinZiyue Jin
31
asked Dec 29 '18 at 2:10
Ziyue JinZiyue Jin
31
asked Dec 29 '18 at 2:10
Ziyue JinZiyue Jin
31
asked Dec 29 '18 at 2:10
Ziyue JinZiyue Jin
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can view $(X,Y)$ as a random point on the plane $mathbb{R}^2$. The joint distribution is the standard bivariate normal distribution and is radially symmetric about the origin.
Plot the region consisting of pairs $(x,y)$ satisfying $x > |2y|$; you will find this is an infinite cone (an infinite slice of pizza or pie, if you will) with vertex at the origin. By the symmetry mentioned above, the probability of the random vector $(X,Y)$ being in this slice is equal to the relative angle of the slice (relative to a full turn, i.e. $2pi$ radians or $360$ degrees).
answered Dec 29 '18 at 2:31
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
Sorry, I don't understand why it is an infinite cone... Is there any mathematical way to calculate this probability?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 5:47
$begingroup$
@ZiyueJin "I don't understand why it is an infinite cone..." Did you plot the region of interest? It might help you understand why it is a cone...
$endgroup$
– Did
Dec 29 '18 at 6:33
$begingroup$
Should any region fall within 1? Because the variance is 1?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:35
$begingroup$
This explanation really helps me understand what happened but I still want a solution related to integral over PDF.
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:36
$begingroup$
"I still want a solution related to integral over PDF" Why? Sorry but this is absurd.
$endgroup$
– Did
Dec 29 '18 at 11:24
|
show 2 more comments
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
You can view $(X,Y)$ as a random point on the plane $mathbb{R}^2$. The joint distribution is the standard bivariate normal distribution and is radially symmetric about the origin.
Plot the region consisting of pairs $(x,y)$ satisfying $x > |2y|$; you will find this is an infinite cone (an infinite slice of pizza or pie, if you will) with vertex at the origin. By the symmetry mentioned above, the probability of the random vector $(X,Y)$ being in this slice is equal to the relative angle of the slice (relative to a full turn, i.e. $2pi$ radians or $360$ degrees).
answered Dec 29 '18 at 2:31
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
Sorry, I don't understand why it is an infinite cone... Is there any mathematical way to calculate this probability?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 5:47
$begingroup$
@ZiyueJin "I don't understand why it is an infinite cone..." Did you plot the region of interest? It might help you understand why it is a cone...
$endgroup$
– Did
Dec 29 '18 at 6:33
$begingroup$
Should any region fall within 1? Because the variance is 1?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:35
$begingroup$
This explanation really helps me understand what happened but I still want a solution related to integral over PDF.
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:36
$begingroup$
"I still want a solution related to integral over PDF" Why? Sorry but this is absurd.
$endgroup$
– Did
Dec 29 '18 at 11:24
|
show 2 more comments
$begingroup$
You can view $(X,Y)$ as a random point on the plane $mathbb{R}^2$. The joint distribution is the standard bivariate normal distribution and is radially symmetric about the origin.
Plot the region consisting of pairs $(x,y)$ satisfying $x > |2y|$; you will find this is an infinite cone (an infinite slice of pizza or pie, if you will) with vertex at the origin. By the symmetry mentioned above, the probability of the random vector $(X,Y)$ being in this slice is equal to the relative angle of the slice (relative to a full turn, i.e. $2pi$ radians or $360$ degrees).
answered Dec 29 '18 at 2:31
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
Sorry, I don't understand why it is an infinite cone... Is there any mathematical way to calculate this probability?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 5:47
$begingroup$
@ZiyueJin "I don't understand why it is an infinite cone..." Did you plot the region of interest? It might help you understand why it is a cone...
$endgroup$
– Did
Dec 29 '18 at 6:33
$begingroup$
Should any region fall within 1? Because the variance is 1?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:35
$begingroup$
This explanation really helps me understand what happened but I still want a solution related to integral over PDF.
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:36
$begingroup$
"I still want a solution related to integral over PDF" Why? Sorry but this is absurd.
$endgroup$
– Did
Dec 29 '18 at 11:24
|
show 2 more comments
$begingroup$
You can view $(X,Y)$ as a random point on the plane $mathbb{R}^2$. The joint distribution is the standard bivariate normal distribution and is radially symmetric about the origin.
Plot the region consisting of pairs $(x,y)$ satisfying $x > |2y|$; you will find this is an infinite cone (an infinite slice of pizza or pie, if you will) with vertex at the origin. By the symmetry mentioned above, the probability of the random vector $(X,Y)$ being in this slice is equal to the relative angle of the slice (relative to a full turn, i.e. $2pi$ radians or $360$ degrees).
answered Dec 29 '18 at 2:31
angryavianangryavian
42.2k23481
$endgroup$
You can view $(X,Y)$ as a random point on the plane $mathbb{R}^2$. The joint distribution is the standard bivariate normal distribution and is radially symmetric about the origin.
Plot the region consisting of pairs $(x,y)$ satisfying $x > |2y|$; you will find this is an infinite cone (an infinite slice of pizza or pie, if you will) with vertex at the origin. By the symmetry mentioned above, the probability of the random vector $(X,Y)$ being in this slice is equal to the relative angle of the slice (relative to a full turn, i.e. $2pi$ radians or $360$ degrees).
answered Dec 29 '18 at 2:31
angryavianangryavian
42.2k23481
answered Dec 29 '18 at 2:31
angryavianangryavian
42.2k23481
answered Dec 29 '18 at 2:31
angryavianangryavian
42.2k23481
answered Dec 29 '18 at 2:31
angryavianangryavian
42.2k23481
42.2k23481
$begingroup$
Sorry, I don't understand why it is an infinite cone... Is there any mathematical way to calculate this probability?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 5:47
$begingroup$
@ZiyueJin "I don't understand why it is an infinite cone..." Did you plot the region of interest? It might help you understand why it is a cone...
$endgroup$
– Did
Dec 29 '18 at 6:33
$begingroup$
Should any region fall within 1? Because the variance is 1?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:35
$begingroup$
This explanation really helps me understand what happened but I still want a solution related to integral over PDF.
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:36
$begingroup$
"I still want a solution related to integral over PDF" Why? Sorry but this is absurd.
$endgroup$
– Did
Dec 29 '18 at 11:24
|
show 2 more comments
$begingroup$
Sorry, I don't understand why it is an infinite cone... Is there any mathematical way to calculate this probability?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 5:47
$begingroup$
@ZiyueJin "I don't understand why it is an infinite cone..." Did you plot the region of interest? It might help you understand why it is a cone...
$endgroup$
– Did
Dec 29 '18 at 6:33
$begingroup$
Should any region fall within 1? Because the variance is 1?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:35
$begingroup$
This explanation really helps me understand what happened but I still want a solution related to integral over PDF.
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:36
$begingroup$
"I still want a solution related to integral over PDF" Why? Sorry but this is absurd.
$endgroup$
– Did
Dec 29 '18 at 11:24
$begingroup$
Sorry, I don't understand why it is an infinite cone... Is there any mathematical way to calculate this probability?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 5:47
$begingroup$
Sorry, I don't understand why it is an infinite cone... Is there any mathematical way to calculate this probability?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 5:47
$begingroup$
@ZiyueJin "I don't understand why it is an infinite cone..." Did you plot the region of interest? It might help you understand why it is a cone...
$endgroup$
– Did
Dec 29 '18 at 6:33
$begingroup$
@ZiyueJin "I don't understand why it is an infinite cone..." Did you plot the region of interest? It might help you understand why it is a cone...
$endgroup$
– Did
Dec 29 '18 at 6:33
$begingroup$
Should any region fall within 1? Because the variance is 1?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:35
$begingroup$
Should any region fall within 1? Because the variance is 1?
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:35
$begingroup$
This explanation really helps me understand what happened but I still want a solution related to integral over PDF.
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:36
$begingroup$
This explanation really helps me understand what happened but I still want a solution related to integral over PDF.
$endgroup$
– Ziyue Jin
Dec 29 '18 at 6:36
$begingroup$
"I still want a solution related to integral over PDF" Why? Sorry but this is absurd.
$endgroup$
– Did
Dec 29 '18 at 11:24
$begingroup$
"I still want a solution related to integral over PDF" Why? Sorry but this is absurd.
$endgroup$
– Did
Dec 29 '18 at 11:24
|
show 2 more comments
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