What is the distribution of the impact point of a Random Ray

Question

In the $\displaystyle (O,x,y)$ plane, a random ray emerges from a light source at the point $\displaystyle (-1,0)$, towards the $\displaystyle (O,y)$ axis. The angle with the $\displaystyle (O,x)$ axis is uniform on $\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. What is the distribution of the impact point with the $\displaystyle (O,y)$ axis?

Answer 1

First we have our uniform: X~Uni$\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.Next let us set $Y=\tan X$. This will be the distribution which we are looking for. Thus we will have:

$$F(y)=P(Y \le y)=P(\tan X \le y)=P(X \le \tan^{-1}y).$$

Because X is uniform on $(-\pi/2,\pi/2)$ we can sub its CDF in here:

$$F(y)=\frac{\tan^{-1}y+\pi/2}{\pi},$$

Thus this is the CDF, the PDF is below

$$f(y)=\frac{1}{\pi}\frac{1}{1+y^2}.$$