Uniform random distribution on a unit disk

Question

a) A point is uniformly chosen in the unit disk $0 ≤ x^2 + y^2 ≤ 1$. Find the probability that its distance from the origin is less than $r$, for $0 ≤ r ≤ 1$.

b) Compute its expected distance from the origin.

c)Let the coordinates of the point be $(X, Y )$. Determine the marginal p.d.f. of $X$. Are $X$ and $Y$ independent?

I did the part a) using geometry that the area of the circle with a radius of $r$ is divided by the area of the unit circle, such that $$P(R\leq r)=\frac{\pi r^2}{\pi\cdot 1^2}=r^2$$

Part b) is attempted to solve by differentiating the cdf, such that $$f(r)=\frac{d}{dr}r^2=2r,\hspace{3mm} E(R)=\int_{0}^1r\times 2r\,dr=\frac{2}{3}.$$

But this result does not seem to be right... And I do not know about part c)

A simulation was done using python to visualize this distribution, and it gives image like this.

Why are the points concentrated near the central area?

from scipy.stats import uniform
import matplotlib.pyplot as plt
import math  
r = uniform.rvs(scale =1,size=5000)
pi = 3.14159265359
theta = uniform.rvs(scale =2*pi,size=5000)
x = []
y = []
for i in range (5000):
    x.append(r[i]*math.cos(theta[i]) )
    y.append(r[i]*math.sin(theta[i]) )
fig=plt.figure()
ax=fig.add_axes([0,0,2,3])
ax.scatter(x, y)```

Answer 1

Your results for parts a) and b) are correct.

For part c), note that $Y\le\sqrt{1-X^2}$. Thus, knowing $X$ reduces the possible range of $Y$, so the two variables can’t be independent.

To find the marginal distribution of $X$ you can use $P(x\le X\le X+\mathrm dx)=\frac1\pi\left(2\sqrt{1-x^2}\right)\mathrm dx$, since the density in the circle is uniformly $\frac1\pi$ and an infinitesimal strip of width $\mathrm dx$ at $x$ has area $\left(2\sqrt{1-x^2}\right)\mathrm dx$. Thus $f_X(x)=\frac2\pi\sqrt{1-x^2}$. (Incidentally, this tells you that $\int_{-1}^1\sqrt{1-x^2}=\frac\pi2$.)

Answer 2

1. The probability a point is closer to the origin than $r$ is the area of a disk of radius $r$ to that of the disk of radius $1$. That is, $\frac{\pi r^2}{ \pi 1^2} = r^2$.
2. The expected distance is $\int\limits_{s=0}^r s (2 \pi s)\ ds$ as can be easily seen:

3. $p(y\mid x) = {\cal U}[-\sqrt{1-x^2}, +\sqrt{1-x^2}]$

as can be easily seen here:

And clearly $X$ and $Y$ are not independent. If $X=.99$, then $Y$ must be small...