[Math] Probability Question for Random Variable $R = \sqrt{X^2 + Y^2}$

density functionplane-geometryprobabilityprobability distributionsprobability theory

Problem: Let $(X, Y)$ be uniformly distributed on the unit disk $\{ (x,y) : x^2 + y^2 \le 1\}$. Let $R = \sqrt{X^2 + Y^2}$. Find the CDF and PDF of $R$.

Attempted Solution: First note that $r \in R = \sqrt{X^2 + Y^2}$ represents a point on $\mathbb{R}^2$ with radius $r$ about the origin. Since only points with radius $1$ had probability greater than $0$ on $(X, Y)$ we have that

$$
F_R(r) = \begin{cases} 0 & r < 1 \\ 1 & r \ge 1 \end{cases}
$$

so that

$$
f_R(r) = F_R'(r) = \begin{cases} 0 & r < 1 \\ \text{undefined} & r = 0 \\ 0 & r > 1 \end{cases}
$$

since

  1. $F'_R$ is discontinuous at $r = 0$.
  2. $F_R$ is constant everywhere else (so that the derivative of a constant is $0$, and hence $F_R'$ is $0$).

Question: Is my reasoning correct here?

Best Answer

As the random variable is uniformly distributed, the probability of $R$ not exceeding a given $r$ is proportional to the enclosed area.

$$P(R\le r)\propto r^2.$$

As the probability is exactly $1$ for the radius $r=1$, the constant of proportionality is $1$.

For $r\le1$,

$$F_R(r)=r^2,\\f_R(r)=2r.$$

Related Question