- #1
glacier302
- 35
- 0
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 : )
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 : )