[Math] Uniform random distribution on a unit disk

geometric-probabilityindependenceprobabilityprobability distributions

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. Distribution plot

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)```

Best Answer

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 $\int_{-1}^1\sqrt{1-x^2}=\frac\pi2$.)

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