# Thread:

1. ## Joint probability distribution of function of two r.vs

If X and Y are independent standard random variables, determine the joint density function of
$U=X, V=\frac{X}{Y}$
Then use your result to show that $\frac{X}{Y}$ is a Cauchy distribution.

I get:

$f(x,y)=f_X(x)f_Y(y)$
$\\\\\\= \frac{1}{2\pi}e^{-(x^2+y^2)/2}$

now $u=x \ and \ v=\frac{x}{y} \rightarrow y=\frac{u}{v} \ and \ x=u$

$J(x,y)= \begin{vmatrix} 1 & 0 \\ \frac{1}{y} & \frac{-x}{y^2} \\ \end{vmatrix} =\frac{-x}{y^2}$

so $J(x,y)^-^1 = \frac{y^2}{x}$

now $f_U_,_V(u,v)=f_X_,_Y(x,y)|J(x,y)|^-^1$
$= f_X_,_Y(u,\frac{u}{v^2})$
$f_U_,_V(u,v)=\frac{1}{2\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{u}{v^2}$

$f_V(v)= \int^\infty_{-\infty} \frac{1}{2\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{u}{v^2}\ du$

However when you evaluate the integral the answer seems to be meaningless

2. Hello,

The mistake is in there : |J(x,y)|=|x|/y²
The absolute value will change the boundaries.

I trust you for the formula of the change of variables. I've never seen it this way, so I can't really check.

3. It works

$f_V(v)= \int^\infty_{-\infty} \frac{1}{2\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{|u|}{v^2}\ du$

$f_V(v)=2 \int^\infty_0 \frac{1}{2\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{u}{v^2}\ du$

$f_V(v)= \int^\infty_0 \frac{1}{\pi}e^{-u^2/2}^{(1+\frac{1}{v^2})}.\frac{u}{v^2}\ du$

Now integrate...

$w=(-u^2/2)(1+\frac{1}{v^2})$
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