# Find PDF

• April 9th 2011, 04:11 PM
dwsmith
Find PDF
Start at a point (0,1) in a rectangular coordinate system, choose an angle $\displaystyle\theta\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ and draw the line to cross the x-axis at X. The range for X is $(-\infty,\infty)$. Find the pdf of X.

I picked the angle $\displaystyle\frac{\pi}{4}$ and now have no idea what to do.
• April 9th 2011, 07:40 PM
TKHunny
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 $\theta = \frac{\pi}{2}$

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

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

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.
• April 9th 2011, 08:15 PM
Sambit
Quote:

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

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

http://img163.imageshack.us/img163/2677/unledvf.png

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

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

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

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

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

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

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

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

So the pdf of X is given by:

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

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

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

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

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