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

Question

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?

Answer 1

Your reasoning is incorrect. First of all, by the definition of CDF, $$ F_R(r)=P(R\leq r)=P(X^2+Y^2\leq r^2)\neq 0 $$ when $0<r<1$.

Second, for a probability density $f_R(r)$, one must have $$\int_{\mathbb{R}} f_R(r)\ dr=1$$ which contradicts your calculation.

Answer 2

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 $r\le1$,

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

Answer 3

One of the reasons I love probability is that there are usual a million ways to do a problem.

I provide a heuristic alternative.

We recognize that we are interested in the event $R\in dr$. In words, this means we want the radius to fall in an infinitesimal annulus, with infinitesimal width $dr$, and area $2\pi rdr$. Since we were told that the points are uniformly distributed, then density is flat over the region of interest, call this $h$. We know the entire volume must be $1$, hence $$h\cdot \pi 1^2 = 1\implies h = \frac{1}{\pi}.$$ This tells us that $$P(R\in dr) = \frac{1}{\pi}\cdot 2\pi r\,dr = 2r\,dr.$$ This implies that the density is $f_R(r) = 2r$ over the region of interest.

Thus, for $0\leq r\leq 1$, $$F_R(r) = \int_0^r f(t)\,dt = r^2.$$