1. ## Find PDF

Start at a point (0,1) in a rectangular coordinate system, choose an angle $\displaystyle \displaystyle\theta\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ and draw the line to cross the x-axis at X. The range for X is $\displaystyle (-\infty,\infty)$. Find the pdf of X.

I picked the angle $\displaystyle \displaystyle\frac{\pi}{4}$ and now have no idea what to do.

2. What an odd question. Anyway, you do not seem to have a useful idea. Let's see if I can make some sense of it...

I believe the intent is to choose ANY angle, not a specific one, but ALL of them in the given range. Perhaps examples would be useful.

P(X<0) = 1/2 and this corresponds to $\displaystyle \theta = \frac{\pi}{2}$

P(X<-1) = 1/4 and this corresponds to $\displaystyle \theta = \frac{\pi}{4}$

P(X<1) = 3/4 and this corresponds to $\displaystyle \theta = -\frac{\pi}{4}$

Essentially, what I am trying to define is a mapping from a uniform distribution on [-pi/2,pi/2] to the entire x-axis. Sounds like it might be related to a tangent, perhaps.

3. Originally Posted by dwsmith
Start at a point (0,1) in a rectangular coordinate system, choose an angle $\displaystyle \displaystyle\theta\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ and draw the line to cross the x-axis at X. The range for X is $\displaystyle (-\infty,\infty)$. Find the pdf of X.

I picked the angle $\displaystyle \displaystyle\frac{\pi}{4}$ and now have no idea what to do.

It reduces to the pdf of a standard Cauchy distribution. Angle OQP is $\displaystyle \theta$ in the figure and OP=x.

The pdf of $\displaystyle \theta$ can be written as: $\displaystyle f(\theta)=\frac{1}{\pi}$, when $\displaystyle -\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$.

Also, $\displaystyle -\infty<x<\infty$. Then the cdf of X can be given by:

$\displaystyle G(x)=P(X\leq x)$

$\displaystyle =P(\frac{OP}{OQ}\leq x)$

$\displaystyle =P(\tan\theta\leq x)$

$\displaystyle =P(\theta\leq \tan^{-1}x)$

$\displaystyle F(\tan^{-1}x)$ , $\displaystyle -\infty<x<\infty$, where $\displaystyle F$ is the cdf of $\displaystyle \theta$

So the pdf of X is given by:

$\displaystyle g(x)=\frac{d}{dx}G(x)$

$\displaystyle =f(\tan^{-1}x).\frac{d}{dx}\tan^{-1}x$, where $\displaystyle f$ is the pdf of $\displaystyle \theta$

$\displaystyle =f(\tan^{-1}x).\frac{1}{1+x^2}$

=$\displaystyle f(\theta)\frac{1}{1+x^2}$

=$\displaystyle \frac{1}{\pi}\frac{1}{1+x^2}$ which is a Standard Cauchy pdf.