Proving Cauchy Density Function for Z = X+Y

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To prove that Z = X + Y has a Cauchy density function when X and Y are independent random variables with the Cauchy density function f(x) = 1/(π(1+x²)), one must utilize the convolution integral. The hint provided suggests verifying the expression f(x)f(y-x) and simplifying it. By expanding the right-hand side of the hint and simplifying, one can demonstrate that the resulting function retains the form of a Cauchy density function. This approach confirms that the sum of two independent Cauchy-distributed variables is also Cauchy-distributed.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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