I'm trying to solve an exercise, which I think I have almost managed to solve but not quite. Any help would be appreciated!
So, what we have is a vector which we obtain by norming the vector $\lambda=(x, 1)$, where $x \in (-\infty, \infty)$, into "half unit circle". I assume that this means that the vector we get is:
$$\lambda^*=\left(\frac{x}{\sqrt{x^2+1}}, \frac{1}{\sqrt{x^2+1}}\right)$$
Expressed in terms of the polar coordinates, the vector is
$$ \lambda^*=(\cos y, \sin y) $$ where $y\in[0,\pi)$. If $y$ is assigned the uniform prior, what is the implied prior for $x$? A hint is that $y=\text{arccot}x$.
My way of reasoning is this. First, the cdf of $y$ is $$ F_Y(y)=\frac{y}{\pi}, \quad 0\leq y <\pi. $$
Furthermore, we have the following relationship: $$ \sin y=\frac{1}{\sqrt{x^2+1}} $$ which means
$$ \sqrt{x^2+1}=\frac{1}{\sin y}\\ x=\left(\frac{1}{\sin y}\right)^2-1 $$
Using the transformation theorem, we can obtain
$$ F_X(x)=Pr(X\leq x)=Pr\left(\frac{1}{(\sin Y)^2}-1\leq x\right)=Pr\left(\frac{1}{x+1}\leq(\sin Y)^2\right) $$ This is where I think I go wrong. To continue and get $Y$ by itself, I need to take the square root and then the inverse of sin. But then I'll have $\sqrt{x+1}$, which will yield non-real numbers for $x<-1$ (remember that $x\in(-\infty, \infty)$.
Any ideas how to attack this?
One considers $Y$ uniform on the interval $(0,\pi)$ and $X=\cot Y$. There are several ways to compute the distribution of $X$, one is to start from the fact that $[X\leqslant x]=[Y\geqslant\cot^{-1} x]$ for every real number $x$, hence $$ F_X(x)=1-\frac1\pi\cot^{-1}x, $$ and, by differentiation, $$ f_X(x)=\frac1\pi\frac1{1+x^2}. $$