Uniform distribution setup

  • A
  • Thread starter mathman
  • Start date
  • #1
mathman
Science Advisor
7,984
513
Summary:
Random variable uniform on circle. Joint distribution of coordinates?
A random variable is distributed uniformly over a circle of radius R. What does the cdf ##F(x,y)## look like as a function of the Cartesian coordinates? The pdf can be expressed as ##f(x,y)=\frac{\delta(\sqrt{x^2+y^2}-R)}{2\pi R}##, where ##\delta## is Dirac delta function. Integration is confusing.
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,679
673
Can you describe in words how F(x,y) relates to the geometry of the circle?
 
  • #3
Stephen Tashi
Science Advisor
7,709
1,516
The pdf can be expressed as ##f(x,y)=\frac{\delta(\sqrt{x^2+y^2}-R)}{2\pi R}##, where ##\delta## is Dirac delta function.

Can it really? We have to consider the general definition of a pdf.

As I make it out, the pdf is a measure defined relative to some other measure on a probability space and it is defined as a generalized derivative (i.e. a Radon-Nikodym derivative) of that other measure. So a pdf exists or fails to exist relative to the other measure. If the other measure is taken to be volume in 2-D then it is possible to have a sequence of smaller and smaller volumes that converge to an arc on the circle. The limit of the measures of the volumes is zero, but we don't want the probability of an arc of finite length on the circle to be zero So the measure we seek from a pdf is not absolutely continuous with respect to volume measure. So the Radon-Nikodym theorem does not apply. If a pdf is defined as type of Radon-Nikodym derivative, we have to ask if a Radon-Nikodym derivative can exist (i.e. be defined) in situations where the theorem does not apply.

Of course is practice, we would compute the cumulative distribution ##F(a,b)## by ##\frac{1}{2 \pi R}## times the arc length of the part of the circle contained in ##\{(x,y): x \le a, y \le b\}## However, aren't the partial derivatives of that function always zero? [Edit: I should say "zero or nonexistent"]
 
Last edited:
  • #4
mathman
Science Advisor
7,984
513
To avoid the delta, how about a thin ring and points uniformly distributed over the ring. The pdf in polar coordinates is ##\frac{1}{\pi (r_2^2-r_1^2)}##. Question: get pdf and cdf in Cartesian coordinates.
 
  • #5
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,679
673
The cdf really is pretty simple, and I'll reiterate my question. What does ##F(x,y)## mean in terms of the geometry of the circle? What is ##F(0,0)## and what is ##F(1,0)##? Your answer should say something about where those points are relative to the circle.
 
  • #6
mathman
Science Advisor
7,984
513
F(0,0)=1/4. F(1,0)=F(0,1)=1/2. (assuming a radius=1).
 
  • #7
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,679
673
Ok, and the more important general question is what does F(x,y) have to do with the position of the point relative to the circle? Just in words describe it.
 
  • #8
mathman
Science Advisor
7,984
513
F(x,y)=Prob((X< x) and(Y< y))

Should be \le, but Latex didn't work.
 
  • #9
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,679
673
The answer I was looking for was something like
"It's the fraction of the circle that is below and to the left of the point".

Now if you have a point (x,y), you need to find the two points of the circle that are below and to the left of it. Just as an example of one case, if ##-\sqrt{1-y^2} < x < \sqrt{1-y^2}## and ##-\sqrt{1-x^2} < y < \sqrt{1-x^2}##, then the point is inside the circle, and you can find there is one point on the circle to the left of the point, and one below it. These are ##(-\sqrt{1-y^2},y)## and ##(x,-\sqrt{1-x^2}))##. Then you need to compute what fraction of the circle is between those points.

There are also some cases where there are either two points below or two points to the left of (x,y), can you describe them!
 
Last edited:
Top