- #1
Gooolati
- 22
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Hello all,
I am currently working on studying for my P actuary exam and had some questions regarding using convolution for the continuous case of the sum of two independent random variables. I have no problem with the actual integration, but what is troubling me is finding the bounds.
Consider two independent random variables X and Y. Let fx(x) = 1 - (x/2) for 0<=x<=2 and 0 otherwise. Let fy(y) = 2-2y for 0<=y<=1 and 0 otherwise. Find the probability density function of X+Y.
for a = x+y
(fx*fy)(a) (the convolution) is the integral from negative infinity to infinity of fx(a-y)fy(y)dy or fy(a-x)fxdx
For 0<=a<=1 I integrate using the above formula and get 2a-(3/2)a^2 + (1/6)a^3
I would have thought the bounds to be from 0 to 1 but they are actually from 0 to a. And for the other cases of a (where 1<=a<=2 and 2<=a<=3) I am having trouble finding out where the bounds are. Some insight would greatly be appreciated! Thanks!
I am currently working on studying for my P actuary exam and had some questions regarding using convolution for the continuous case of the sum of two independent random variables. I have no problem with the actual integration, but what is troubling me is finding the bounds.
Homework Statement
Consider two independent random variables X and Y. Let fx(x) = 1 - (x/2) for 0<=x<=2 and 0 otherwise. Let fy(y) = 2-2y for 0<=y<=1 and 0 otherwise. Find the probability density function of X+Y.
Homework Equations
for a = x+y
(fx*fy)(a) (the convolution) is the integral from negative infinity to infinity of fx(a-y)fy(y)dy or fy(a-x)fxdx
The Attempt at a Solution
For 0<=a<=1 I integrate using the above formula and get 2a-(3/2)a^2 + (1/6)a^3
I would have thought the bounds to be from 0 to 1 but they are actually from 0 to a. And for the other cases of a (where 1<=a<=2 and 2<=a<=3) I am having trouble finding out where the bounds are. Some insight would greatly be appreciated! Thanks!