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?
Best Answer
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 $(-\pi/2, \pi/2)$