Uniform Distribution on unit Circle

  • #1
190
0

Main Question or Discussion Point

I keep reading that a random vector (X, Y) uniformly distributed over the unit circle has probability density [itex]\frac{1}{\pi}[/itex]. The only proof I've seen is that
[tex]f_{X,Y}(x,y) = \begin{cases} c, &\text{if }x^2 + y^2 \leq 1 \\ 0 &\text{otherwise}\end{cases} [/tex]

And then you solve for [itex]c[/itex] by integrating to 1. This does not seem self-evident to me. Can anyone please offer a more detailed proof? Thanks!
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
132
Hey IniquiTrance.

With a uniform distribution, each realization or outcome has exactly the same chance as every other outcome.

Now in this problem, we assign the same probability outcome for every single point and equal sized interval, which means we assign a probability of c for every point inside the region of the circle (including it's boundary).

Now since this is a probability, we need to make sure that integrating the PDF over the entire region gives us 1. Since the PDF is a constant (i.e. c) and does not depend on x or y, we can take it out of the integral.

Now the integral over the region is simply the area of the region. The area of a circle is given by pi*r2 but r = 1 so we just get pi. This means 1 = c*pi since the integral over the whole space must equal 1.

This means by re-arranging, we get c = 1/pi.
 
  • #3
190
0
Thank you.
 

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