# Calculating Cauchy's PDF, using domain definition drawings of F_X/Y(z)

#### vitalipom

##### New Member
Hi,

I am trying to draw F (CDF of Cauchy) using X/Y relation for X,Y~N(0,1) iids and at some point very shortly I get lost. Especially because I don't know if I'm on the right direction.

Could you direct me to the right way of solving this using the way I chose?

So.

Problem: Given X,Y~N(0,1) iids find PDF of X/Y.
(And I know it is Cauchy's disterbution.)

What I try do:

I want to find F_Z whilest Z=X/Y and derive it, whilest I want to find it using explicit drawing of definition areas on X,Y axis. Like that:

For X,Y,z>0 we get

F_Z(z) = P(z<Z) = P(z<X/Y) = P (Y<X/z)

This is the area below Y=X/z for X,Y,z>0 on X,Y axis graph. (May I not destroy the text with ugly picture?)
Picture is below.

I think we can use Proposition formula to find how much is it exactly. (?)
We can use double integrals here.

Now, when z>0 it's all summerizes beautifully and if z would be only >0. Meaning by that, if F_Z(z) would be defined only on the right side of Y axis, we would finish this and I would even do the calculations.

BUT

z (small z) might be <0.

So for example, for X,Y>0 and z<0, we get

F_Z(z) = P(z<Z) = P(z<X/Y) = 1.

My question is - How to apply the complete disterbution function while taking in mind small z? P(X>0)=P(Y>0)=P(X<0)=P(Y<0)=1/2

Or how do I actually calculate it when for F_Z(z), z<0 and F_Z(z), z>0 the formulas are different?

Last edited:

#### vitalipom

##### New Member
Ok.

(Some breakthrough found. All I want is to solve this problem in a way I -- simple student can understand. So your help IS NEEDED.)

I complained I don't know P_Z(z<0) & P_Z(z>0). Well I know them, since P_Z=X/Y(z<X/Y<0)=P{X<0,Y>0 U X>0, Y<0}=1/2, I can assume I am able to do something with complete probability function, though running not on X or Y but on Z=X/Y. Remember that I wanted to calculate the integeral over X/z using X/z's disterbution which I think can be given by the proposition(?) ((superposition???)) function.

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