1. The problem statement, all variables and given/known data Let f(x,y) = e^(-x-y), 0<x< infinity, 0<y<infinity, zero elsewhere, be the pdf of X and Y. Then if Z = X + Y, compute P(Z<=0), P(Z,<=6), and, more generally P(Z<=z), for 0<z<infinity. What is the pdf of Z? 2. Relevant equations P(x,y) = ∫∫(f(x,y) dxdy 3. The attempt at a solution so, P(Z<= 0 ) is pretty obviously 0 P(Z<=6) = P(X+Y <= 6) =P(X<=6-Y) into equation ∫from (0 to infinty) ∫ from (0 to 6-Y ) e^(-x-y) dx dy -e^(-x-y) eval from 0 to 6-Y = (-e^-6 ) + e^-y ∫from (0 to infinity) (-e^-6 ) + e^-y dy = -ye^(-6) -e^-y eval from (0 to infinity) = -infinity -0 - ( 0 - 1) = - infinity so, im definitly going wrong somewhere because a probability of negative infinity makes no sense... I know this could be done using a different method of tranformations, but i think im supposed to do something along these lines because thats what is taught in the preceeding chapter.