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Transformation of variable (what happens with the area?)

  • Thread starter Zenga
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Homework Statement



Dear friends,

let U and V independent variables that are both defined on [-∏, ∏] and are uniformly distributed.
If x = cos(U + V) and y = sin(U-V), what is the area where the variables X and Y are defined?

Homework Equations


U + V = arccos(x)
U - V = arcsin(y)
For a test if we have chosen the right area, we can deploy probability density function f_(x,y) = f(u(x,y),v(x,y)*J(x,y), where J(x,y) is a Jacobian determinant.

The Attempt at a Solution


My attempt: x and y have minimum and maximum at -1 and 1, respectively.

I would be really grateful if anyone could help me, since this exercise is beyond my level of knowledge. If I have to remove the exercise to any other forum, feel free to remove the issue. Wish you all a pleasant day!:)
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Dear friends,

let U and V independent variables that are both defined on [-∏, ∏] and are uniformly distributed.
If x = cos(U + V) and y = sin(U-V), what is the area where the variables X and Y are defined?

Homework Equations


U + V = arccos(x)
U - V = arcsin(y)
For a test if we have chosen the right area, we can deploy probability density function f_(x,y) = f(u(x,y),v(x,y)*J(x,y), where J(x,y) is a Jacobian determinant.

The Attempt at a Solution


My attempt: x and y have minimum and maximum at -1 and 1, respectively.

I would be really grateful if anyone could help me, since this exercise is beyond my level of knowledge. If I have to remove the exercise to any other forum, feel free to remove the issue. Wish you all a pleasant day!:)
The equation ##f_{X,Y}(x,y) = f(u(x,y),v(x,y)) J(x,y)## is not correct in this case, because the transformation is not "monotone" (always increasing or always decreasing). The functions ##\sin## and ##\cos## have graphs that oscillate, sometimes going up and sometimes down, and that means that a single point (x,y) can have several "preimages" (u,v).

However, if all you want to know is the range of (X,Y) then you are correct: it is a square made from the two intervals ##X \in [-1,1], \: Y \in [-1,1]##.
 
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  • #3
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I am really thankful for your response!
Now it is getting exciting!!! :) I am interested how would we calculate ##f_{X,Y}(x,y)##?
 
  • #4
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Dear all, I have another question. :) How many pre-images has each (x,y)?
 
  • #5
Ray Vickson
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Dear all, I have another question. :) How many pre-images has each (x,y)?
For ##0 < x,y < 1## there are four solutions of ##\cos(s) = x, \; s \in [-2 \pi,2 \pi]## and four solutions of ##\sin(t) = y, \; t \in [-2 \pi,2 \pi]##. These are ##s_1 = a \equiv \arccos(x), s_2 = -a, s_3 = 2\pi - a, s_4 = a - 2 \pi## and ##t_1 = b \equiv \arcsin(y), t_2 = \pi-b, t_3 = -\pi-b, t_4 = b- 2\pi##. Therefore, there are 16 points ##(u,v)## of the form ##(s_i,t_j)##. However, for any choice of ##x,y \in (0,1)## only 8 of these points will also satisfy the individual bounds ##u,v \in [-\pi,\pi]##. In other words, for each ##(x,y)## there are 8 feasible ##(u,v)##, and all have the same absolute value of the Jacobian determinant. This implies that
[tex] f_{X,Y}(x,y) = \frac{8 f_{U,V}(u_i,v_i)}{J\left(\frac{x,y}{u,v}\right)}[/tex]
Since
[tex] f_{U,V}(u,v) = \frac{1}{4 \pi^2}[/tex]
and
[tex] J\left(\frac{x,y}{u,v}\right) = 2|\sin(u+v)||\cos(v-u)| = 2 \sqrt{1-x^2}\sqrt{1-y^2}[/tex]
we finally obtain
[tex] f_{X,Y}(x,y) = \frac{1}{\pi^2 \sqrt{1-x^2} \sqrt{1-y^2}}. [/tex]

Note: showing that 8 of the 16 points (u,v) are feasible is a tedious case-by-case verification exercise.

All the above was for ##0 < x,y < 1##. The other cases can be shown by the same type of argument all over again, or more simply by transforming them to the first case (by using ##-U## or ##-V## instead of ##U## and ##V##, etc.

Altogether, we see that the formula above for ##f_{X,Y}(x,y)## holds for all ##x,y \in (-1,1)##.
 
  • #6
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I am really thankful for your response!
 

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