At first, we check that f(x,y) is a valid joint probability density. We have to prove that ∫0+∞∫0+∞e−(x+y)dxdy=1. Compute the integral at the left side: ∫0+∞∫0+∞e−(x+y)dxdy=∫0+∞−e−(x+y)∣0+ ∞dy=∫0+∞e−ydy=−e−y∣0+∞=1.
Thus, it is the valid distribution. In order to find the joint distribution, consider the probability: P{X<z,X+Y<w}. In case z<0 or w<0, P{X<z,X+Y<w}=0. In case z>0,w>0, we get: P{Z<z,X+Y<w}=∬x<z,x+y<wf(x,y)dxdy=∫0min(z,w)(∫0w−xe−(x+y)dy)dx=∫0min(z,w)−e−(x+y)∣0w−xdx=∫0min(z,w)(e−x−e−w)dx=−e−x∣0min(z,w)−min(z,w)e−w=1−e−min(z,w)−min(z,w)e−w.
Thus, in case z<w, we get: P{Z<z,X+Y<w}=1−e−z−ze−w. For z>w we have: P{Z<z,X+Y<w}=1−e−w−we−w. Denote F(z,w)=P{Z<z,X+Y<w}
The joint probability density is: p(z,w)=∂z∂w∂2F(z,w)=e−w, z<w, z>0,w>0.
Answer: The joint probability density function is: p(z,w)=e−w, z<w, z>0,w>0.
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