Find the distribution of $$ {{\mathrm X}\over{\vert{\mathrm Y}\vert}} $$ given that ${\mathrm X}$ and ${\mathrm Y}$ are independent standard normal variables.

${\mathbf {Note:}}$ I have had success with these problems and doing change of variable in the past. Though my struggle with this problem is doing the change of variable with an absolute value in the denominator. This is not for credit, but for my understanding. Help is appreciated.

  • $\begingroup$ The dash goes the other way, as in \frac. $\endgroup$ – Pedro Tamaroff Feb 25 '17 at 23:55
  • $\begingroup$ THANK YOU SO MUCH! It's always the easy things that trip me up, got it. $\endgroup$ – Joe Feb 25 '17 at 23:57
  • $\begingroup$ It's called the Cauchy distribution, wiki. $\endgroup$ – ThePuix Feb 26 '17 at 0:03
  • $\begingroup$ @ThePuix I do not see it in wiki, ratio of normals is Cauchy, but OP is about ratio of normal and half-normal. $\endgroup$ – Jan Feb 26 '17 at 0:11
  • $\begingroup$ Wont let me fix the above. I was trying to type this in my prior comment: I get something close to the Cauchy distribution. Though I am having an issue deriving it using change of variable method I have been studying. I have been setting ${\mathrm U} ={{\mathrm X}\over{\mathrm{|Y|}}} $ and ${\mathrm V} = | \mathrm Y |$ Though I was lost in getting the jacobian and the bounds for $\mathrm X$. $\endgroup$ – Joe Feb 26 '17 at 0:14

Let $Z=\frac{Y}{|X|}$. Then

$$ \mathbb{P}(Z\leq z)=\mathbb{P}(Y\leq |X|z)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{|x|z}e^{-\frac{x^2+y^2}{2}}\;dydx $$ Setting $y=|x|u$, this becomes $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^z|x|e^{-\frac{x^2(1+u^2)}{2}}\;dudx=\frac{1}{2\pi}\int_{-\infty}^{z}\int_{-\infty}^{\infty}|x|e^{-\frac{x^2(1+u^2)}{2}}\;dxdu $$

hence $Z$ has pdf $$ f_Z(z)=\frac{1}{2\pi}\int_{-\infty}^{\infty}|x|e^{-\frac{x^2(1+z^2)}{2}}\;dx=\frac{1}{\pi}\int_0^{\infty}xe^{-\frac{x^2(1+z^2)}{2}}\;dx$$ $$=\frac{1}{\pi}\frac{1}{1+z^2}\int_0^{\infty}e^{-u}\;du=\frac{1}{\pi}\frac{1}{1+z^2} $$ so $Z$ is Cauchy distributed.

  • $\begingroup$ This is actually a cool new method I have not seen before. Though if you wanted to use the jacobian with ${\mathrm Z}={{\mathrm Y}\over{\mathrm |X|}}$ and ${\mathrm U}={\mathrm Y}$? $\endgroup$ – Joe Feb 26 '17 at 0:24
  • $\begingroup$ Actually I just figured it out! Thank you, your answer helped me and now I know a new method as well!! I really appreciate it. $\endgroup$ – Joe Feb 26 '17 at 0:34
  • $\begingroup$ No problem, happy to help. $\endgroup$ – carmichael561 Feb 26 '17 at 0:36

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