Proving Sum of 2 Indep. Cauchy RVs is Cauchy

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SUMMARY

The discussion focuses on proving that the sum of two independent Cauchy random variables, X and Y, is also a Cauchy random variable, denoted as Z = X + Y. The density function for X and Y is given by f(x) = 1/(π(1+x²)). The convolution integral for independent variables is utilized, specifically ∫f(x)f(y-x)dx, along with a hint involving a specific form of the convolution. The hint provided simplifies the convolution integral, leading to the conclusion that Z retains the Cauchy distribution properties.

PREREQUISITES
  • Understanding of Cauchy random variables and their properties
  • Familiarity with convolution integrals in probability theory
  • Knowledge of characteristic functions and Fourier transforms
  • Basic skills in calculus, particularly integration techniques
NEXT STEPS
  • Study the properties of Cauchy distributions in detail
  • Learn about convolution integrals and their applications in probability
  • Explore characteristic functions and their role in proving distribution properties
  • Investigate partial fraction decomposition techniques in integration
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This discussion is beneficial for statisticians, mathematicians, and students studying probability theory, particularly those interested in the properties of Cauchy distributions and convolution methods.

glacier302
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Given the fact that X and Y are independent Cauchy random variables, I want to show that Z = X+Y is also a Cauchy random variable.

I am given that X and Y are independent and identically distributed (both Cauchy), with density function
f(x) = 1/(∏(1+x2)) . I also use the fact the convolution integral for X and Y is ∫f(x)f(y-x)dx .

My book says to use the following hint:

f(x)f(y-x) = (f(x)+f(y-x))/(∏(4+y2)) + 2/(∏y(4+y2))(xf(x)+(y-x)f(y-x)) .

Using this hint, I'm able to solve the rest of the problem, but I can't figure out how to prove that this hint is true.

Any help would be much appreciated : )
 
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I can't see what to do with the hint. There is an alternate approach which might be easier if you have been exposed to characteristic functions (Fourier transform of densities).
If you have, then get the characteristic function of Cauchy, square it and then get the inverse transform. This is the density that you want.
 


I notice than in several statistics texts, this result is asserted or assigned as an exercise and the indicated method is expand the integrand in partial fractions. Perhaps the identity would be the result of that.
 

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