Question

Stuck on this joint continuous distribution question

Answer 1

Let X and Y have the following density \[f(x,y) = \lambda^2e^{-\lambda*y}, 0 \le x \le y; 0 \space otherwise\]
Find joint distribution and density of (Y, X+Y)

Answer 2

wow complex

Answer 3

I found the distribution of X+Y as below:\[F_Z(z)=P(Z\le z)=P(Y \le z-X) = ...\]\[=\lambda e^{\frac{-\lambda}{2}*z}-\lambda e^{-\lambda z}\]

Answer 4

For joint distribution I started out as follows:\[F(a,b) = P(Y \le a, X+Y \le b)=P(Y \le a, Y \le X-B)=\]\[=\frac{P(Y \le a \cap Y \le b-X)}{P(Y \le b-X)}\]

Answer 5

And I'm totally stumped at this point, anyone have any idea what to do?
@ganeshie8 @Zarkon

Answer 6

i think its way to complex

Answer 7

Wait, the below was density \[=\lambda e^{\frac{-\lambda}{2}*z}-\lambda e^{-\lambda z}\]The distribution that I found was:\[1 + e^{-\lambda * z}-2e^{\frac{-\lambda}{2}z}\]

Answer 8

@Kainui @robtobey