Proving Cauchy Density Function for Z = X+Y

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SUMMARY

The discussion focuses on proving that the sum of two independent random variables, X and Y, each with a Cauchy density function defined as f(x) = 1/(π(1+x²)), results in another random variable Z = X + Y that also has a Cauchy density function. The convolution integral, ∫f(x)f(y-x)dx, is central to this proof. A hint provided in the discussion simplifies the process by suggesting to expand and simplify the right-hand side of the equation f(x)f(y-x) = (f(x)+f(y-x))/(π(4+y²)) + 2/(πy(4+y²))(xf(x)+(y-x)f(y-x)).

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Homework Statement



Let X and Y be independent random variables each having the Cauchy density function f(x)=1/(∏(1+x2)), and let Z = X+Y. Show that Z also has a Cauchy density function.

Homework Equations



Density function for X and Y is f(x)=1/(∏(1+x2)) .
Convolution integral = ∫f(x)f(y-x)dx .

The Attempt at a Solution



My book gives the following hint, saying to "check it":

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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Just expand out the righthand side and simplify.
 

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