If $(x, y)$ is a uniformly randomly chosen point on a unit circle with radius $1$, then the distance of said point to origin is $D = \sqrt{x^2+y^2}$
How would I find the $CDF$ of $D$?
I think the PDF is of $D$ is $2D$ (using the approach outlined here: Distribution of distance from origin for uniformly randomly chosen point in circle) but am not sure how to get the $CDF$ from this. If I try taking the CDF by taking the double integral of the $PDF$, I get a really complicated integral evaluation. Am I doing this right or is there a more simple and intuitive way to get the $CDF$ of $D$?
Thanks in advance for your help.
Best Answer
(I'm assuming that you meant the point was chosen uniformly in the unit disc)
The area of a circle of radius $r$ is $\pi r^2$, and the uniform measure on the circle is a constant density of $1/\pi$. So the probability of being inside a circle of radius $r$ should be $r^2$.
Thus the cdf $P(D <= r) = r^2$ for $r\in[0,1]$ (and it is $0$ for $r<0$ and $1$ for $r> 1$). From this you can get the pdf is $2r 1_{[0,1]}(r)$.