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\begin{document}
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\heading{E050: Changing random variables $x=\cos(\theta)$}
\auname{Elkana Porat}
{\bf The problem: }
\Dn
Assume that the random phase ${\theta}$ has a uniform distribution over the range $[0,2\pi]$.
Define a new random variable ${x = \cos \left(\theta\right)}$. What
is the probability distribution of ${x}$ ?
\Dn
\Dn\Dn
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{\bf The solution:}
\Dn
We use the formula:
$$\rho\left(x\right) = \rho\left({\theta}\right)\left|\frac{d{\theta}}{dx}\right|$$
For $\theta$ we have:
$$\theta = \cos^{-1}\left(x\right)$$
$$\frac{d{\theta}}{dx} = -\frac{1}{\sqrt{1-x^2}}$$
and the probability distribution:
$$\rho\left(\theta\right) = \left\{
{\begin{array}{ccc}
\frac{1}{2\pi}& ;& 0 < \theta < 2\pi\\
0 & ;& o.w.\\
\end{array}}\right.$$
So for the probability distribution of $x$ we have:
$$\begin{array}{ccl}
\rho\left(x\right)& = &\frac{1}{\sqrt{1-x^2}}\cdot\left\{
\begin{array}{ccc}
\frac{1}{2\pi}& ;& 0 < \cos^{-1}\left(x\right) < 2\pi\\
0 & ;& o.w.\\
\end{array}\right.\\
\\
& = & \left\{
\begin{array}{ccc}
\frac{1}{\pi\sqrt{1-x^2}}& ;& -1 < x < 1\\
0 & ;& o.w.\\
\end{array}\right.
\end{array}$$
When in the last step we add a factor of 2 because of the degeneracy of the solution of $\theta = \cos^{-1}\left(x\right)$ in the range $\theta \in \left[0,2\pi\right]$.
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\end{document}