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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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