Suppose X,Y are independent standard normal rndom variable . Let $Z= \frac{|X|}{|Y|}$.find the pdf of Z

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Suppose X,Y are inddependent standard normal rndom variable . Let $Z= \frac{|X|}{|Y|}$

I am asked to find the pdf of Z

$Z= \frac{X}{Y} \quad X\ge 0 ; Y\ge 0 \\ =\frac{-X}{Y} \quad X\le 0 ; Y\ge 0 \\ = \frac{-X}{Y} \quad X\ge 0 ; Y\le 0\\ = \frac{X}{Y} \quad X\le 0 ; Y\le 0\\ $

Can I apply Jacobian method for this transformation? (I dont think jacobian can be applied since Z is not one-one function of x,y)

$Z= \frac{X}{Y} \quad ; \quad V=X\\ J= \begin{vmatrix} \frac{\partial X}{\partial Z} &\frac{\partial X}{\partial V}\\ \frac{\partial Y}{\partial Z} & \frac{\partial Y}{\partial V} \end{vmatrix}$

$|J|= \frac{1}{z^2}$

$f_{X,Y}(X,Y)= \frac{1}{2\pi} e^{\frac{x^2 + y^2}{2}}$

$\therefore f_{V,Z}(v,z)= \frac{1}{2\pi} e^{\frac{v^2}{2} + \frac{v^2}{2z^2}}$

$f_Z(z)= \int_V f_{V,Z}(v,z)dz$

this is the part i am stuck at, What are the limits of V?

I am not even sure if my method is correct , but this is the only way i can think of .

Any help ?? Thank You

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Hint: Easier to find $X/Y$ first, then take it's absolute value. Show that $X/Y$ is standard Cauchy, so that $|X/Y| $ is 'folded' Cauchy.

You are headed in the right direction but you forgot to multiply the |Jacobian| in the joint density of $(V,Z)$, which is actually $v/z^2$. Support of both $X$ and $Y$ is $\mathbb R$. So both $V$ and $Z$ also have support $\mathbb R$. The integration limits should be clear now. The transformation is one-to-one.

(For deriving the distribution of $X/Y$ , an easier choice of $V$ is $Y$.)