Distribution of angle of two dimensional normal vector

Question

The original subject is:

Suppose random variables $X$ and $Y$ are independent and both follow the Normal distribution $N(0,\sigma ^2)$.

1) Prove $U=X^2+Y^2$ and $V = \frac{X}{\sqrt{X^2+Y^2}}$ are independent.

2) suppose $\Theta=\arcsin(V)$, prove $\Theta$ follows the Uniform distribution on $(-\frac{\pi}{2},\frac{\pi}{2})$.

I am stuck in the second question. Allow me to demonstrate my answer to the first question:

$X$ and $Y$ are considered as coordinates, so we take the radius $R=X^2+Y^2$ and the angle $\Theta$ being random variables as well, with their ranges $0\le R < \infty$ and $0 \le \Theta < 2\pi$.

Take $0< r_0 < \infty$ and $0 < \theta_0 < 2\pi$. Notate the event $B=\{0\le R \le r_0, 0\le \Theta \le \theta_0\}$

Then, the joint distribution is: $$P(0\le R \le r_0, 0\le \Theta \le \theta_0) = \frac{1}{2\pi \sigma^2}\int_0^{r_0} \int_0^{\theta_0}\exp(-\frac{r^2}{2\sigma^2})rdrd\theta$$

and the joint density function is : $$f_{R\Theta}(r,\theta) = \frac{1}{2\pi \sigma^2}\exp(-\frac{r^2}{2\sigma^2})r \qquad for \space 0\le r < \infty \space and \space 0 \le \theta < 2\pi$$

and the two marginal density functions $$f_R(r)= \frac{1}{ \sigma^2}\exp(-\frac{r^2}{2\sigma^2})r \qquad for \space 0\le r < \infty $$ $$f_{\Theta}(\theta)= \frac{1}{2\pi} \qquad for \space 0 \le \theta < 2\pi $$

together combine the joint density function: $$f_{R\Theta}(r,\theta) = f_R(r)f_{\Theta}(\theta)$$

So, $R$ and $\Theta$ are independent.

And because $U=R$, and $V=\sin(\Theta)$ is an function only related to $\Theta$, we have $U$ and $V$ are also independent.

Till now, I finish the first question. Consequently, I get the idea that $\Theta$ are uniformly distributed on $[0, 2\pi]$, which is apparently at odds with the second question. Did I do something wrong? If not, how to prove the second problem with respect to my first answer?

Answer 1

There is nothing wrong with your answer and it does not constradicts the question.

Note when $\arcsin (V)$ in part 2, is not the same $\Theta$ YOU used in the integral.

Note $Y=y, X=x$ and $Y=-y, X=x$ gives the same value of $\arcsin(V)$, i.e. $\arcsin(x/\sqrt{x^2+y^2}) = \arcsin(x/\sqrt{x^2+(-y)^2})$ but $X=x$ and $Y=y$ and $Y=y, X=-x$ gives different value of $\Theta$ in your parametrisation.

e.g.

$y = 1, x = 1$ gives $\Theta = \pi/4$ but $y=-1, x=1$ gives $\Theta = \pi/4 $

Explicitly

$\arcsin (V) = \Theta$ for $\Theta \in [0,\pi/2]$

$\arcsin (V) = -\Theta + \pi$ for $\Theta \in [3\pi/2, 2\pi]$

$\arcsin (V) = \Theta - \pi$ for $\Theta \in (\pi/2, \pi)$

$\arcsin (V) = -\Theta +2\pi $ for $\Theta \in [3\pi/2, 2\pi)$

so $\arcsin (V)$ is a uniform distribution on $(-\pi/2, \pi/2)$

Answer 2

Maybe $V=\cos(\Theta)$ ?

$\arcsin (V) = \pi/2 - \Theta$ for $\Theta \in [0,\pi)$

$\arcsin (V) = \Theta - 3\pi/2$ for $\Theta \in [\pi, 2\pi)$

So $\arcsin (V)$ is a uniform distribution on $(-\pi/2, \pi/2)$, am I correct?