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Let X and Y be normal random variables with mean 0, variance $\sigma^2$ and correlation coefficient $\rho \in (-1,1)$, so that the density is given by

$$f(x,y) = \cfrac{1}{2\pi \sigma^2\sqrt{a-\rho^2}}\,\,\exp{\{-\cfrac{1}{2\sigma^2(1-\rho^2)}[x^2 - 2\rho xy+y^2]\}}$$

How to determine the distribution of $Z=\cfrac{X}{Y}$ ?

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closed as off-topic by heropup, colormegone, Leucippus, Shailesh, choco_addicted Apr 10 '16 at 1:53

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  • $\begingroup$ Do you know how you in general get the pdf of a random variable $Z=X/Y$ when you have the joint pdf $f_{XY}(x,y)$? $\endgroup$ – JKnecht Apr 9 '16 at 19:49
  • $\begingroup$ Yes, there is the specific formula $f_Z(z)=\int_{-\infty}^{+\infty}{|y| f_X(z y)f_Y(y)dy}$ $\endgroup$ – mic Apr 9 '16 at 19:55
  • $\begingroup$ But how do I find marginals of $f_{X,Y}$? I don't know how to integrate analytically a normal pdf. $\endgroup$ – mic Apr 9 '16 at 20:01
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Let $Z=X/Y$ and $W=Y$.

The transformation $z=x/y,\: w=y$ has the inverse transformation

$x=zw, \: y=w$, and

$$ \bar{J}(z,w) =\begin{vmatrix} \frac{\partial{x}}{\partial{z}} & \frac{\partial{x}}{\partial{w}} \\ \frac{\partial{y}}{\partial{z}} & \frac{\partial{y}}{\partial{w}} \\ \end{vmatrix} $$

$$ \bar{J}(z,w)=\begin{vmatrix} w & z \\ 0 & 1 \\ \end{vmatrix} = w $$

Now

$$f_{ZW}(z,w) = |w|f_{XY}(zw,w)$$

and the marginal pdf of $Z$ is

$$f_{Z}(z) = \int_{-\infty}^{\infty}|w|f_{XY}(zw,w)dw$$

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  • $\begingroup$ Perfect, I got $f_Z(z) = \cfrac{1}{z^2 - 2\rho z +1}$. $\endgroup$ – mic Apr 9 '16 at 20:36

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