A point chosen at random from a disc

Question

I've been working on this question and have managed to complete parts (i) through (iii) but am struggling with the last two parts.
For (iv) I end up getting this when trying to find the distribution:

$\sin(\theta _1)\leq$$\frac{Y}{\sqrt{X^2+Y^2}}$$\leq \sin(\theta _2)$

But I am not sure how to turn this into an integral in terms of the area of the circle? Any help would be appreciated.

Question 6: A point chosen at random from the disc

Suppose that a point with co-ordinates $(X,Y)$ is chosen uniformly at random from the disc $\{(x,y) \in \mathbb{R}^2 \mid x^2 + y^2 \leq 1\}$. Let $R = \sqrt{X^2 + Y^2}$ and $\Theta = \arcsin(Y/R)$.

i) What is the joint p.d.f. of $X$ and $Y$?

ii) What are the marginal p.d.f.s of $X$ and $Y$? Are $X$ and $Y$ independent?

iii) By considering the probability $\textrm{P}(R \leq r)$ for suitable $r$ show the distribution of $R$ is continuous and find its density.

iv) By considering the probability $\textrm{P}(\theta_1 \leq \Theta \leq \theta_2)$ for suitable $\theta_1 \leq \theta_2$ show the distribution of $\Theta$ is continuous and find its density.

v) Show the joint distribution of $R$ and $\Theta$ is continuous and find its density. Are $R$ and $\Theta$ independent?

Answer 1

André's comment addresses (iv). This addresses (v).

Some notation first: Let $\overline{B}$ be the closed unit disk. Let $f_1(x) = \sqrt{x_1^2+x_2^2}$, $f_2(x) = \arcsin {x_2 \over\sqrt{x_1^2+x_2^2} } $. We have $f_1(\mathbb{R}^2) = [0,\infty)$, $f_2(\mathbb{R}^2) = [-{\pi \over 2} , {\pi \over 2} ]$. The probability measure that is uniform on $\overline{B}$ can be written as $P A = {1 \over \pi} \int_\overline{B} 1_A$.

We wish to compute $F(\alpha,\beta) = P \{ x | f_1(x) \in (-\infty, \alpha], f_2(x) \in (-\infty, \beta] \}$. The picture shows the area when $\alpha \in (0,1), \beta \in (-{\pi \over 2} , {\pi \over 2})$, and from geometric considerations we get $F(\alpha,\beta) = {2 \over \pi} \alpha^2 \beta$.

If $\alpha \le 0$ or $\beta \le -{\pi \over 2} $ we have $F(\alpha,\beta) = 0$, if $\alpha \ge 1$ and $\beta \ge {\pi \over 2} $ we have $F(\alpha,\beta) = 1$ and for other values the computation follows the same lines. Putting these together, we can write $F(\alpha,\beta) = \int 1_{(-\infty,\alpha] \times (-\infty, \beta]}((r,\theta)) \left( 2 r 1_{(0,1)}(r) \right) \left( {1 \over \pi} 1_{(-{\pi \over 2} , {\pi \over 2})}(\theta) \right) dr d \theta $, from which we can see that the density is $f(r,\theta) = 2 r 1_{(0,1)}(r) {1 \over \pi} 1_{(-{\pi \over 2} , {\pi \over 2})}(\theta) $. Since we can write $f(r,\theta) = g_1(r) g_2(\theta)$ some some $g_1,g_2$ we see (from Fubini) that $f_1,f_2$ are independent.

Here is an alternative masochistic analytic approach using the change of variables theorem. Analysis is my security blanket.

Let $B_+ = \{ x | \|x\|<1, x_1 > 0 \}$ and similarly for $B_-$. Note that $P A = P (A \cap B_+) + P (A \cap B_-)$ for any measurable set $A$.

Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ given by $f(x) = (f_1(x),f_2(x))^T$, and note that the restrictions $f:B_+ \to (0,1) \times (-{\pi \over 2} , {\pi \over 2})$ and $f:B_- \to (0,1) \times (-{\pi \over 2} , {\pi \over 2})$ are injective and $C^1$. A tedious computation shows that $J_f(x) = \det Df(x) = {1 \over f_1(x)}$.

Let $R = (-\infty,\alpha] \times (-\infty, \beta]$, and $A=\{ x | f(x) \in R \}$. We want to compute $F(\alpha, \beta) = P A = P (A \cap B_+) + P (A \cap B_-) $. Note that the map $T:A \cap B_+ \to A \cap B_-$ given by $T(x) = \begin{bmatrix} -1 & 0 \\ 0 & 1\end{bmatrix} x$ is a bijection, we see that $P (A \cap B_+) = P (A \cap B_-) $, and so $P A = 2 P (A \cap B_+)$.

We have $P A = {2 \over \pi} \int_\overline{B} 1_{A \cap B_+} = {2 \over \pi} \int_{B_+} 1_A = {2 \over \pi} \int_{B_+} 1_R \circ f$. This is close to the form we need to apply the change of variables theorem (see, for example, Rudin's "Real & Complex Analysis", 3rd Ed., Theorem 7.26), except that it is missing a $|J_f|$ term. We can write $(1_R \circ f) (x) = 1_R(f(x)) = 1_R(f(x)) f_1(x) {1 \over f_1(x)} = (g \circ f)(x) |J_f(x)|$, where $g(x) = x_1 1_R(x)$.

The change of variables theorem gives $\int_{B_+} 1_R \circ f = \int_{B_+} (g \circ f) |J_f| = \int_{f(B_+)} g = \int_{ (0,1) \times (-{\pi \over 2} , {\pi \over 2}) } g$, and so we have \begin{eqnarray} F(\alpha, \beta) &=& P A \\ &=& {2 \over \pi} \int_{ (0,1) \times (-{\pi \over 2} , {\pi \over 2}) } g \\ &=& {2 \over \pi} \int_{ (0,1) \times (-{\pi \over 2} , {\pi \over 2}) } r 1_R((r, \theta)) d r d \theta \\ &=& \int 1_R((r,\theta)) \left( 2 r 1_{(0,1)}(r) \right) \left( {1 \over \pi} 1_{(-{\pi \over 2} , {\pi \over 2})}(\theta) \right) dr d \theta \end{eqnarray} which is the same as the answer above, of course.