Cauchy Distribution Homework: X/Y Derivation

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SUMMARY

The discussion focuses on deriving the distribution function of the quotient of two normally distributed random variables, X and Y. The correct approach involves using the Dirac delta function, integrating it with the probability density functions (pdfs) of X and Y. An alternative method is introduced by defining new variables W and V, where W = Z1/Z2 and V = Z2, and calculating the Jacobian determinant to facilitate the transformation of the joint distribution. The user ultimately seeks assistance in resolving an issue with evaluating the integral to obtain the probability density function of W.

PREREQUISITES
  • Understanding of probability density functions (pdfs)
  • Familiarity with the Dirac delta function
  • Knowledge of joint distributions and transformations
  • Experience with Jacobian determinants in multivariable calculus
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  • Study the properties and applications of the Dirac delta function in probability theory
  • Learn about joint distributions and their transformations in statistical analysis
  • Explore the derivation of the probability density function for the ratio of two random variables
  • Review examples of calculating Jacobian determinants in multivariable calculus
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Students and professionals in statistics, mathematicians working on probability theory, and anyone interested in advanced statistical methods for deriving distributions of random variables.

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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 \delta(Z-X/Y) 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)

<br /> W = \frac{Z_1}{Z_2}, \quad V = Z_2<br />

Then

<br /> Z_1 = V \cdot W, \quad Z_2 = V<br />

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

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