Proving Sum of 2 Indep. Cauchy RVs is Cauchy

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To prove that the sum of two independent Cauchy random variables, Z = X + Y, is also Cauchy, one can utilize the convolution integral of their density functions. The provided hint simplifies the convolution integral, leading to the necessary density function for Z. While the hint is useful for solving the problem, its verification remains unclear to some participants. An alternative method involves using characteristic functions, where the characteristic function of a Cauchy random variable is squared to derive the density of the sum. This approach is often referenced in statistics texts as a standard exercise.
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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