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{\bf Question}

Suppose that $X$ is uniform in the interval $(0,2\pi)$ and $Y$,
independent of $X$, is exponential with parameter 1.

\begin{description}
\item[(a)]
Find the joint pdf of $U$ and $V$ defined by

$$U=\sqrt{2Y}\cos(X),\ V=\sqrt{2Y}\sin(X).$$

\item[(b)]
Show that $U$ and $V$ are independent, each having a standard
normal distribution.

\item[(c)]
State the distribution of $\tan(X)$.
\end{description}




\vspace{.25in}

{\bf Answer}

$f(x,y)=\ds\frac{1}{2\pi}e^{-y},\ \ x \in (0,2\pi),\ y>0$

Therefore $\left.\begin{array} {ccc} x & = & \tan^{-1}
\left(\frac{v}{u})\right)\\ y & = & \frac{u^2 + v^2}{2}
\end{array} \right|$ Also $\begin{array} {ccccc} -\infty & < & u &
< \infty\\ -\infty & < & v & < \infty \end{array}$

$J=\left|\begin{array}{cc} \frac{\partial x}{\partial u} &
\frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} &
\frac{\partial y}{\partial
v}\end{array}\right|=\left|\begin{array}{cc}
\frac{1}{1+\frac{v^2}{u^2}}\frac{-v}{u^2} &
\frac{1}{1+\frac{v^2}{u^2}}\frac{1}{u}\\ u & v
\end{array}\right| = -1$

Therefore \begin{eqnarray*} f(u,v) & = &
\frac{1}{2\pi}e^{-\frac{1}{2}(u^2+v^2)} \cdot |-1|,\
-\infty<u<\infty,\ -\infty<v<\infty\\ & = &
\frac{1}{2\pi}e^{-\frac{1}{2}(u^2+v^2)}\\ & = &
\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}u^2} \cdot
\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}v^2} \end{eqnarray*}

Therefore $U$ and $V$ are independent standard normals.

$\tan(X)=\ds\frac{V}{U}$ where $U$ and $V$ are $indep\ N(0,1).$

Therefore $\tan(X) \sim$ Cauchy distribution.



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