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Let X and Y be independent standard normal random variables.

What is the distribution of $\large \frac{X}{|Y|}$?

Attempt:

Let $\large U = \frac{X}{|Y|}$ and $ V = |Y|$.

This transformation is not one-to-one. We can make it one-to-one by restricting consideration to either positive or negative values of y. Let

$A_1 = $ {$(x,y): y > 0$} $, A_2 = ${$(x,y): y < 0$}, $A_0 = $ {$(x,y): y = 0$}.

$B = ${$ (u,v): v > 0$} is the image of both $A_1$ and $A_2$. The inverse transformations are given by:

$y = v, x = uv$ for B to $A_1$, and $ y = -v, x = -uv$ for B to $A_2$.

Both Jacobians $J_1$ and $J_2$ are equal to v.

Then the joint distribution of U,V is given by:

$f_{U,V}(u,v) = \Large \frac{1}{2\pi}e^{-uv^2/2}e^{-v^2/2}*v+\frac{1}{2\pi}e^{-uv^2/2}e^{-(-v^2/2)}*v $

$= \Large \frac{v}{\pi}e^{-(u^2+1)v^2/2}$, $-\infty < u < \infty, 0 < v < \infty$.

To find the marginal pdf of U, integrate out V:

$ f_{U}(U) = \Large \int_0^\infty\frac{v}{\pi}e^{-(u^2+1)v^2/2}dv$

$ = \Large \frac{1}{2\pi}\int_0^\infty e^{-(u^2+1)z/2}dz$, where $z = v^2$

$ = \Large \frac{1}{2\pi} \frac{2}{(u^2+1)}$ (integrand is kernel of exponential ($\beta = 2/(u^2+1)))$

$= \Large \frac{1}{\pi (u^2+1)},$ $-\infty < u < \infty$, which is a Cauchy random variable.

StubbornAtom
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statsguyz
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    Because $X$ and $Y$ are each symmetric about $0$, this is the same as looking at the distribution of $\frac{X}{Y}$, which for standard normal distributions does indeed give a Cauchy distribution. – Henry Sep 27 '14 at 20:58
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    If one insists on using a change of variables, why not use U=X/|Y|, V=Y? – Did Sep 28 '14 at 08:59
  • https://math.stackexchange.com/questions/79583/the-probability-density-function-of-the-ratio-of-two-normal-r-v-s/ – StubbornAtom Dec 18 '17 at 15:42
  • @Did is there an alternate way to do this without a change of variables?? – Governor Oct 28 '21 at 14:12

1 Answers1

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Notice, $|Y|=\sqrt{\frac{Y^2}{1}}$. So you have $\frac{Z_1}{\sqrt{\frac{{Z_2}^2}{1}}}$. So it is $t_1$, which is a Cauchy distribution.

ztyh
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