91Ó°ÊÓ

X and Y are both normal random variables.

Let,\[U = \frac{X}{Y}\] and V = Y.

Since P(Y = 0) = 0 then, the possibility of Y = 0 can be excluded.

Then the transformation from X and Y to U and V is one-to-one.

The inverse transformation is

\[\begin{array}{l}X = UV\\Y = V\end{array}\]

The jacobian is

\[\begin{array}{c}J = \det \left[ {\begin{array}{*{20}{c}}{\frac{{dx}}{{du}}}&{\frac{{dx}}{{dv}}}\\{\frac{{dy}}{{du}}}&{\frac{{dy}}{{dv}}}\end{array}} \right]\\ = \det \left[ {\begin{array}{*{20}{c}}v&u\\0&1\end{array}} \right]\\ = v\end{array}\]

Since X and Y both are independent and both follow the standard normal distribution, the joint p.d.f is

\[f\left( {x,y} \right) = \frac{1}{{2\pi }}{e^{ - \frac{1}{2}\left( {{x^2} + {y^2}} \right)}}; - \infty < x < \infty , - \infty < y < \infty \]

Then the joint p.d.f of U and V becomes,

\[g\left( {u,v} \right) = \frac{{\left| v \right|}}{{2\pi }}{e^{ - \frac{1}{2}\left( {{u^2} + 1} \right){v^2}}}; - \infty < u < \infty , - \infty < v < \infty \]

Now,

\[\begin{array}{c}{g_1}\left( u \right) = \int\limits_{ - \infty }^\infty {\frac{{\left| v \right|}}{{2\pi }}{e^{ - \frac{1}{2}\left( {{u^2} + 1} \right){v^2}}}dv} \\ = \int\limits_0^\infty {\frac{1}{{2\pi }}{e^{ - \frac{1}{2}\left( {{u^2} + 1} \right){v^2}}}vdv} \\ = \frac{1}{{\pi \left( {{u^2} + 1} \right)}}; - \infty < u < \infty \end{array}\]

Hence, the quotient\[\frac{X}{Y}\]has the Cauchy distribution.

Hence, proved.