Generating Two Independent Standard Normal Variables
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Image of problem
I have a problem with this explanation, the book says we can generate Y1 Y2 from X1 X2
but when we want to answer questions about the distribution of X1 X2 we take arctan which returns values in range (-PI/2,PI/2) even though we X1 X2 range is (0,1) I understand it makes no difference in the part when we calculate uniform pdf from inverse since it is 1 for each X1 X2 but I can't understand how it works in Jacobian. I think author made jump here and didn't want to go into details
statistics random-variables random
asked Dec 2 '18 at 1:46
Adam JochnaAdam Jochna
1
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add a comment |
$begingroup$
Image of problem
I have a problem with this explanation, the book says we can generate Y1 Y2 from X1 X2
but when we want to answer questions about the distribution of X1 X2 we take arctan which returns values in range (-PI/2,PI/2) even though we X1 X2 range is (0,1) I understand it makes no difference in the part when we calculate uniform pdf from inverse since it is 1 for each X1 X2 but I can't understand how it works in Jacobian. I think author made jump here and didn't want to go into details
statistics random-variables random
asked Dec 2 '18 at 1:46
Adam JochnaAdam Jochna
1
$endgroup$
1
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this process is Box-Muller transform
$endgroup$
– Adam Jochna
Dec 2 '18 at 1:47
add a comment |
$begingroup$
Image of problem
I have a problem with this explanation, the book says we can generate Y1 Y2 from X1 X2
but when we want to answer questions about the distribution of X1 X2 we take arctan which returns values in range (-PI/2,PI/2) even though we X1 X2 range is (0,1) I understand it makes no difference in the part when we calculate uniform pdf from inverse since it is 1 for each X1 X2 but I can't understand how it works in Jacobian. I think author made jump here and didn't want to go into details
statistics random-variables random
asked Dec 2 '18 at 1:46
Adam JochnaAdam Jochna
1
$endgroup$
Image of problem
I have a problem with this explanation, the book says we can generate Y1 Y2 from X1 X2
but when we want to answer questions about the distribution of X1 X2 we take arctan which returns values in range (-PI/2,PI/2) even though we X1 X2 range is (0,1) I understand it makes no difference in the part when we calculate uniform pdf from inverse since it is 1 for each X1 X2 but I can't understand how it works in Jacobian. I think author made jump here and didn't want to go into details
statistics random-variables random
statistics random-variables random
asked Dec 2 '18 at 1:46
Adam JochnaAdam Jochna
1
asked Dec 2 '18 at 1:46
Adam JochnaAdam Jochna
1
asked Dec 2 '18 at 1:46
Adam JochnaAdam Jochna
1
asked Dec 2 '18 at 1:46
Adam JochnaAdam Jochna
1
asked Dec 2 '18 at 1:46
Adam JochnaAdam Jochna
1
1
1
$begingroup$
this process is Box-Muller transform
$endgroup$
– Adam Jochna
Dec 2 '18 at 1:47
add a comment |
1
$begingroup$
this process is Box-Muller transform
$endgroup$
– Adam Jochna
Dec 2 '18 at 1:47
1
1
$begingroup$
this process is Box-Muller transform
$endgroup$
– Adam Jochna
Dec 2 '18 at 1:47
$begingroup$
this process is Box-Muller transform
$endgroup$
– Adam Jochna
Dec 2 '18 at 1:47
add a comment |
1 Answer
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$begingroup$
arctan which returns values in range (-PI/2,PI/2)
Actually the inverse trigonometric functions can be defined with several ranges. In your formula, it's clear that they are taking the range $[0,2pi]$
Regarding the Jacobian (and the proof in general) of this Box-Muller transform, there are many answers:
Box-Muller Transform Normality
proof of Box–Muller transform (polar form)
https://stats.stackexchange.com/questions/188183/some-details-about-the-box-muller-transform-method
answered Dec 2 '18 at 2:55
leonbloyleonbloy
41.1k645107
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add a comment |
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$begingroup$
arctan which returns values in range (-PI/2,PI/2)
Actually the inverse trigonometric functions can be defined with several ranges. In your formula, it's clear that they are taking the range $[0,2pi]$
Regarding the Jacobian (and the proof in general) of this Box-Muller transform, there are many answers:
Box-Muller Transform Normality
proof of Box–Muller transform (polar form)
https://stats.stackexchange.com/questions/188183/some-details-about-the-box-muller-transform-method
answered Dec 2 '18 at 2:55
leonbloyleonbloy
41.1k645107
$endgroup$
add a comment |
$begingroup$
arctan which returns values in range (-PI/2,PI/2)
Actually the inverse trigonometric functions can be defined with several ranges. In your formula, it's clear that they are taking the range $[0,2pi]$
Regarding the Jacobian (and the proof in general) of this Box-Muller transform, there are many answers:
Box-Muller Transform Normality
proof of Box–Muller transform (polar form)
https://stats.stackexchange.com/questions/188183/some-details-about-the-box-muller-transform-method
answered Dec 2 '18 at 2:55
leonbloyleonbloy
41.1k645107
$endgroup$
add a comment |
$begingroup$
arctan which returns values in range (-PI/2,PI/2)
Actually the inverse trigonometric functions can be defined with several ranges. In your formula, it's clear that they are taking the range $[0,2pi]$
Regarding the Jacobian (and the proof in general) of this Box-Muller transform, there are many answers:
Box-Muller Transform Normality
proof of Box–Muller transform (polar form)
https://stats.stackexchange.com/questions/188183/some-details-about-the-box-muller-transform-method
answered Dec 2 '18 at 2:55
leonbloyleonbloy
41.1k645107
$endgroup$
arctan which returns values in range (-PI/2,PI/2)
Actually the inverse trigonometric functions can be defined with several ranges. In your formula, it's clear that they are taking the range $[0,2pi]$
Regarding the Jacobian (and the proof in general) of this Box-Muller transform, there are many answers:
Box-Muller Transform Normality
proof of Box–Muller transform (polar form)
https://stats.stackexchange.com/questions/188183/some-details-about-the-box-muller-transform-method
answered Dec 2 '18 at 2:55
leonbloyleonbloy
41.1k645107
edited Dec 2 '18 at 21:30
answered Dec 2 '18 at 2:55
leonbloyleonbloy
41.1k645107
answered Dec 2 '18 at 2:55
leonbloyleonbloy
41.1k645107
answered Dec 2 '18 at 2:55
leonbloyleonbloy
41.1k645107
41.1k645107
add a comment |
add a comment |
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$begingroup$
this process is Box-Muller transform
$endgroup$
– Adam Jochna
Dec 2 '18 at 1:47