Mixing
One-to-One Transformation in Bi-variate Probability Density Function
Clash Royale CLAN TAG#URR8PPP
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I am trying to solve this problem:
"A point is generated at random in the plane according to the following polar
scheme. A radius R is chosen, where the distribution of R2
is Ç2 with 2 degrees of
freedom. Independently, an angle θ is chosen, where θ ~ uniform(0, 2À). Find the
joint distribution of X = Rcosθ and Y = Rsinθ."
I am off the final answer by a factor. I thought the transformation from (X,Y) to (R,θ) is not a one-to-one transformation due to the nature of inverse tangent function θ = tan-1(Y/X). Therefore I need to separate the support and calculate probability density function. Apparently, this is not true. A little bit thought makes me realize that I can identify the angle using X and Y. However, this is not obvious from θ = tan-1(Y/X) which I used to calculate the transformed probability density function. Any argument to reconciles this or clarify it?
Edit:
I don't think my question is really about deriving Box-Muller but rather to understand the concept of one-to-one transformation. Take a simple example:
$Y = sqrt X $ is one-to-one transformation. However, $Y=X^2$ is not a one-to-one transformation.
Take another example:
$$Y_1 = X_1^2 + X_2^2 quadtextand quad Y_2 = fracX_1sqrt Y_1$$
is not one-to-to transformation so that when calculate its PDF we need to account the non one-to-one transformation. In solving for PDF of X and Y in the problem above. It is normal to think that:
$$f_R^2,theta = frac14pie^-R^2/2, quad 0<t<inf, quad 0<theta<2pi$$
the transformation is :
$$X = Rcostheta,quad Y=Rsintheta.quad $$
The inverse transformation is:
$$R^2=X^2+Y^2,quad theta = tan^-1(Y/X)$$
Here comes the question that the inverse transformation given above is not one-to-one in inverse tangent. Do I need to add another contribution to the final probability as in the situation of finding the PDF of $Y$ if $Y=X^2$?
calculus probability probability-theory statistics derivatives
asked Jul 27 at 3:21
Owls
183
add a comment |Â
up vote
1
down vote
favorite
I am trying to solve this problem:
"A point is generated at random in the plane according to the following polar
scheme. A radius R is chosen, where the distribution of R2
is Ç2 with 2 degrees of
freedom. Independently, an angle θ is chosen, where θ ~ uniform(0, 2À). Find the
joint distribution of X = Rcosθ and Y = Rsinθ."
I am off the final answer by a factor. I thought the transformation from (X,Y) to (R,θ) is not a one-to-one transformation due to the nature of inverse tangent function θ = tan-1(Y/X). Therefore I need to separate the support and calculate probability density function. Apparently, this is not true. A little bit thought makes me realize that I can identify the angle using X and Y. However, this is not obvious from θ = tan-1(Y/X) which I used to calculate the transformed probability density function. Any argument to reconciles this or clarify it?
Edit:
I don't think my question is really about deriving Box-Muller but rather to understand the concept of one-to-one transformation. Take a simple example:
$Y = sqrt X $ is one-to-one transformation. However, $Y=X^2$ is not a one-to-one transformation.
Take another example:
$$Y_1 = X_1^2 + X_2^2 quadtextand quad Y_2 = fracX_1sqrt Y_1$$
is not one-to-to transformation so that when calculate its PDF we need to account the non one-to-one transformation. In solving for PDF of X and Y in the problem above. It is normal to think that:
$$f_R^2,theta = frac14pie^-R^2/2, quad 0<t<inf, quad 0<theta<2pi$$
the transformation is :
$$X = Rcostheta,quad Y=Rsintheta.quad $$
The inverse transformation is:
$$R^2=X^2+Y^2,quad theta = tan^-1(Y/X)$$
Here comes the question that the inverse transformation given above is not one-to-one in inverse tangent. Do I need to add another contribution to the final probability as in the situation of finding the PDF of $Y$ if $Y=X^2$?
calculus probability probability-theory statistics derivatives
asked Jul 27 at 3:21
Owls
183
This is essentially the Box-Muller transformation used to generate normal random variables from uniform random variables. The angle $Theta$ is uniform and the squared distance $R^2$ from the origin is $mathsfChisq(2) equiv mathsfExp(rate=1/2),$ which can be generated in terms of a uniform random variable on $(0,1).$ The demonstration is a straightforward change from polar to rectangular coordinates.
– BruceET
Jul 27 at 4:40
Also perhaps look at this Q @ A on our site.
– BruceET
Jul 27 at 5:02
Also see this post:math.stackexchange.com/questions/2845710/….
– StubbornAtom
Jul 30 at 16:11
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to solve this problem:
"A point is generated at random in the plane according to the following polar
scheme. A radius R is chosen, where the distribution of R2
is Ç2 with 2 degrees of
freedom. Independently, an angle θ is chosen, where θ ~ uniform(0, 2À). Find the
joint distribution of X = Rcosθ and Y = Rsinθ."
I am off the final answer by a factor. I thought the transformation from (X,Y) to (R,θ) is not a one-to-one transformation due to the nature of inverse tangent function θ = tan-1(Y/X). Therefore I need to separate the support and calculate probability density function. Apparently, this is not true. A little bit thought makes me realize that I can identify the angle using X and Y. However, this is not obvious from θ = tan-1(Y/X) which I used to calculate the transformed probability density function. Any argument to reconciles this or clarify it?
Edit:
I don't think my question is really about deriving Box-Muller but rather to understand the concept of one-to-one transformation. Take a simple example:
$Y = sqrt X $ is one-to-one transformation. However, $Y=X^2$ is not a one-to-one transformation.
Take another example:
$$Y_1 = X_1^2 + X_2^2 quadtextand quad Y_2 = fracX_1sqrt Y_1$$
is not one-to-to transformation so that when calculate its PDF we need to account the non one-to-one transformation. In solving for PDF of X and Y in the problem above. It is normal to think that:
$$f_R^2,theta = frac14pie^-R^2/2, quad 0<t<inf, quad 0<theta<2pi$$
the transformation is :
$$X = Rcostheta,quad Y=Rsintheta.quad $$
The inverse transformation is:
$$R^2=X^2+Y^2,quad theta = tan^-1(Y/X)$$
Here comes the question that the inverse transformation given above is not one-to-one in inverse tangent. Do I need to add another contribution to the final probability as in the situation of finding the PDF of $Y$ if $Y=X^2$?
calculus probability probability-theory statistics derivatives
asked Jul 27 at 3:21
Owls
183
I am trying to solve this problem:
"A point is generated at random in the plane according to the following polar
scheme. A radius R is chosen, where the distribution of R2
is Ç2 with 2 degrees of
freedom. Independently, an angle θ is chosen, where θ ~ uniform(0, 2À). Find the
joint distribution of X = Rcosθ and Y = Rsinθ."
I am off the final answer by a factor. I thought the transformation from (X,Y) to (R,θ) is not a one-to-one transformation due to the nature of inverse tangent function θ = tan-1(Y/X). Therefore I need to separate the support and calculate probability density function. Apparently, this is not true. A little bit thought makes me realize that I can identify the angle using X and Y. However, this is not obvious from θ = tan-1(Y/X) which I used to calculate the transformed probability density function. Any argument to reconciles this or clarify it?
Edit:
I don't think my question is really about deriving Box-Muller but rather to understand the concept of one-to-one transformation. Take a simple example:
$Y = sqrt X $ is one-to-one transformation. However, $Y=X^2$ is not a one-to-one transformation.
Take another example:
$$Y_1 = X_1^2 + X_2^2 quadtextand quad Y_2 = fracX_1sqrt Y_1$$
is not one-to-to transformation so that when calculate its PDF we need to account the non one-to-one transformation. In solving for PDF of X and Y in the problem above. It is normal to think that:
$$f_R^2,theta = frac14pie^-R^2/2, quad 0<t<inf, quad 0<theta<2pi$$
the transformation is :
$$X = Rcostheta,quad Y=Rsintheta.quad $$
The inverse transformation is:
$$R^2=X^2+Y^2,quad theta = tan^-1(Y/X)$$
Here comes the question that the inverse transformation given above is not one-to-one in inverse tangent. Do I need to add another contribution to the final probability as in the situation of finding the PDF of $Y$ if $Y=X^2$?
calculus probability probability-theory statistics derivatives
asked Jul 27 at 3:21
Owls
183
edited Jul 27 at 14:54
asked Jul 27 at 3:21
Owls
183
asked Jul 27 at 3:21
Owls
183
asked Jul 27 at 3:21
Owls
183
183
This is essentially the Box-Muller transformation used to generate normal random variables from uniform random variables. The angle $Theta$ is uniform and the squared distance $R^2$ from the origin is $mathsfChisq(2) equiv mathsfExp(rate=1/2),$ which can be generated in terms of a uniform random variable on $(0,1).$ The demonstration is a straightforward change from polar to rectangular coordinates.
– BruceET
Jul 27 at 4:40
Also perhaps look at this Q @ A on our site.
– BruceET
Jul 27 at 5:02
Also see this post:math.stackexchange.com/questions/2845710/….
– StubbornAtom
Jul 30 at 16:11
add a comment |Â
This is essentially the Box-Muller transformation used to generate normal random variables from uniform random variables. The angle $Theta$ is uniform and the squared distance $R^2$ from the origin is $mathsfChisq(2) equiv mathsfExp(rate=1/2),$ which can be generated in terms of a uniform random variable on $(0,1).$ The demonstration is a straightforward change from polar to rectangular coordinates.
– BruceET
Jul 27 at 4:40
Also perhaps look at this Q @ A on our site.
– BruceET
Jul 27 at 5:02
Also see this post:math.stackexchange.com/questions/2845710/….
– StubbornAtom
Jul 30 at 16:11
This is essentially the Box-Muller transformation used to generate normal random variables from uniform random variables. The angle $Theta$ is uniform and the squared distance $R^2$ from the origin is $mathsfChisq(2) equiv mathsfExp(rate=1/2),$ which can be generated in terms of a uniform random variable on $(0,1).$ The demonstration is a straightforward change from polar to rectangular coordinates.
– BruceET
Jul 27 at 4:40
This is essentially the Box-Muller transformation used to generate normal random variables from uniform random variables. The angle $Theta$ is uniform and the squared distance $R^2$ from the origin is $mathsfChisq(2) equiv mathsfExp(rate=1/2),$ which can be generated in terms of a uniform random variable on $(0,1).$ The demonstration is a straightforward change from polar to rectangular coordinates.
– BruceET
Jul 27 at 4:40
Also perhaps look at this Q @ A on our site.
– BruceET
Jul 27 at 5:02
Also perhaps look at this Q @ A on our site.
– BruceET
Jul 27 at 5:02
Also see this post:math.stackexchange.com/questions/2845710/….
– StubbornAtom
Jul 30 at 16:11
Also see this post:math.stackexchange.com/questions/2845710/….
– StubbornAtom
Jul 30 at 16:11
add a comment |Â
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This is essentially the Box-Muller transformation used to generate normal random variables from uniform random variables. The angle $Theta$ is uniform and the squared distance $R^2$ from the origin is $mathsfChisq(2) equiv mathsfExp(rate=1/2),$ which can be generated in terms of a uniform random variable on $(0,1).$ The demonstration is a straightforward change from polar to rectangular coordinates.
– BruceET
Jul 27 at 4:40
Also perhaps look at this Q @ A on our site.
– BruceET
Jul 27 at 5:02
Also see this post:math.stackexchange.com/questions/2845710/….
– StubbornAtom
Jul 30 at 16:11