If X and Y has joint pdf, f(x,y) =1 0 < x < 1, 0 < y < 1, and we want to find the pdf of Z = X +Y what is an easy way to do this? The hard part about this problem is determining the limits. When I employ the Jacobian method, I create another variable Q = Y and find the pdf of Z and Q. But when I find the marginal of Z, the bounds for Q is between 0 and z where 0 < z < 1 since Q = Y. But what happens if 1 < z < 2? I am stuck in finding the pdf.
How to easily find limits in transformation problems
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probability
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1Draw a square with opposite vertices $(0,0)$ and $(1,1)$. Draw a line $x+y=z$ where $1 < z < 2$. This line will cross the square, dividing it into a triangle and a pentagon. Then, $P\{X+Y > z\} = 1 - F_{X+Y}(z)$ is the area of the triangle (in this simple case) which you should be able to express as a function of $z$, and you can find that for $1 < z < 2$, $$f_{X+Y}(z) = -\frac{d}{dz}P\{X+Y > z\} = 2-z.$$ If you draw the line $x+y=z$ for $0 < z < 1$, you can similarly get $P\{X+Y
as the area of a triangle, and $f_{X+Y}(z) = z$ for $0 < z < 1$. No muss, no fuss, no $Q=Y$, no Jacobians. – 2012-02-29 -
0Why couldn't you do this using the Jacobian method? Will the Jacobian method not work here? – 2012-02-29
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1You tried and failed in using the Jacobian method. I suggested to you a simpler method, **which I had also suggested to you in response to a [question of yours](http://math.stackexchange.com/q/74085/15941) about five months ago in an answer that you accepted**. Yes, the Jacobian method will work here. **You** can use the Jacobian method if you like; **I** will refuse to use the Jacobian method in solving this problem, and I will not help you with doing it that way either. Wait for someone else to post the complete answer via the Jacobian method. – 2012-02-29