PDF of Jakes' Model $\cos(\theta)$

Question

If we have an angle $\theta$ follows a uniformly distribution $[-\pi,\pi]$, the PDF of $\theta$ will be $p(\theta)=\frac{1}{\pi-(-\pi)}$ in $[-\pi,\pi]$, if we consider the Jakes' model $f=f_{\max}\cos(\theta)$, where $f_{\max}$ is just a constant. what will be the PDF $p(f)$ of this Jakes' model? I guess we might need the Bessel function.

Answer 1

Is there a mistake here? I didn't use the Bessel function.

So $p(\theta)=1/(2\pi)$, constant. For the CDF $F_f$ of $f = f_{\rm max}\cos(\theta)$, let's assume $f_{\rm max}$ is positive. Note $\cos$ is increasing on $[-\pi,0]$ and decreasing on $[0,\pi]$.

The values of $f$ are in $[-f_{\rm max},f_{\rm max}]$. If $t<-f_{\rm max}$, then $F_f(t)=0$. If $t>f_{\rm max}$, then $F_f(t) = 1$.

For $-f_{\rm max}\le t \le f_{\rm max}$, \begin{align} F_f(t) &= \mathbb P[f \le t] = \mathbb P[f_{\rm max}\cos\theta \le t] =\mathbb P\big[\cos\theta \le \frac{t}{f_{\rm max}}\big] \\ &=\mathbb P\left[\theta \ge \arccos \frac{t}{f_{\rm max}}\right] +\left[\theta \le -\arccos \frac{t}{f_{\rm max}}\right] =2\mathbb P\left[\theta \ge \arccos \frac{t}{f_{\rm max}}\right] \\ &=\frac{2}{2\pi}\left(\pi - \arccos \frac{t}{f_{\rm max}}\right) = 1 - \frac{1}{\pi}\arccos \frac{t}{f_{\rm max}} = \frac{1}{2} +\frac{1}{\pi}\arcsin\frac{t}{f_{\rm max}} \end{align} So this CDF is merely a rearrangement of the arccsin graph. See arcsine distribution

Then the PDF is the derivative of this, $$ p_f(t) = \frac{1}{\pi\sqrt{f_{\rm max}^2-t^2}} $$