Cauchy Distribution Homework: X/Y Derivation

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Homework Help Overview

The discussion revolves around deriving the distribution function of the quotient of two normally distributed random variables, X and Y. The original poster attempts to understand the correct approach to finding the probability density function (pdf) of the ratio X/Y.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods for deriving the distribution of the quotient, including the use of the Dirac delta function and transformations of random variables. Some participants question the validity of dividing the pdfs directly and explore the implications of using Jacobian determinants in their transformations.

Discussion Status

The discussion is ongoing, with participants providing different methods and insights. Some guidance has been offered regarding the use of transformations and integration techniques, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There are indications of confusion regarding the calculations involved in the transformation methods, particularly in relation to the Jacobian determinant and the evaluation of integrals. The original poster expresses uncertainty about the results obtained from their calculations.

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Homework Statement



http://img132.imageshack.us/img132/1/48572399ly5.png

The Attempt at a Solution


I tried dividing the two pdf's but that isn't right. If you have X normal and Y normal distributed how can you derive the distribution function of X/Y?
 
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Of course you don't divide the pdf's. To get the quotient pdf as a function of Z, you take the dirac delta function [itex]\delta(Z-X/Y)[/itex] and integrate it times the pdf's for X and Y, dXdY.
 
Another method (based techniques typically presented early in mathematical stats) is this:

Define these two variables (one you already have)

[tex] W = \frac{Z_1}{Z_2}, \quad V = Z_2[/tex]

Then

[tex] Z_1 = V \cdot W, \quad Z_2 = V[/tex]

By their definitions both new random variables range over [tex](-\infty, \infty)[/tex].

Use the basic ideas for transformation of a joint distribution to get the distribution of [tex]V[/tex] and [tex]W[/tex], then integrate out [tex]V[/tex].
 
Hi,

I'm actually going over some probability problems and I got a bit stuck in this one too.

If you let:

W=Z1/Z2 and V=Z2

Then truly Z1=V*W and Z2=V

And if you calculate the Jacobian determinant of such transformations you get:

Jacobian determinant = V (here we take the absolute value when sticking into formula below)

Therefore:

f(w,v) =[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP91219bfg00e7b70caif00005aa636h4egb540i1?MSPStoreType=image/gif&s=39&w=148&h=44

and so all you need to get the probability density function of W is to integrate the joint probability with respect to v as follows:

First note that: d/dv (e-v2(1+w2)/2) = -v(1+w2)*e-v2(1+w2)/2

=>[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP243119bff8ch835e1b7i00001639e2c5d96b97gg?MSPStoreType=image/gif&s=39&w=366&h=54

and here is where I seem to be overlooking something, in order to get f(w) you must evaluate the integral from minus infinity to plus infinity and so I believe you get:

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP237019bff8ch83i380ib0000641a5786ggg043f3?MSPStoreType=image/gif&s=39&w=124&h=43

Which is just plainly equal to zero, so I must've done something wrong, can anyone spot what was it? I would appreciate if someone did. Thanks.
 
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