Solving Joint Distribution of Exponential RVs: X, Y, Z & U

[lolypop]

Hi ,
I got this question in my midterm today but up till now I don't know how to solve it ,

The Question is as follow :
If X and Y are two exponential Rv with different lambda . and there's a new Rvs Z and U are defined such that :

Z= 0 : X<Y and 1 : X=> Y

U= min(X,Y)

and the Question asked to proof that Z and U are independent .

So I started my solution by deriving the pdf of U since I know how to then tried to derive the pdf of Z but didn't know where to start and got stuck , since I didn't how to get the pdf function when the boundary of the Rv is other RV.

Can anyone tell me of a way to derive the pdf of Z . or is there another way to solve the problem ?

lolypop

 

[kurt.physics]

13To solve this problem, you need to use the concept of conditional probability. First, we need to find the probability of Z given U, which can be expressed as P(Z=z|U=u). The value of P(Z=z|U=u) is equal to 1 if X>Y, since in this case Z will always have the value 1. If X<Y, then the probability of Z=z is equal to 0, since Z will always have the value 0 in this case. Now, we need to find the probability of U given Z, which can be expressed as P(U=u|Z=z). In this case, the probability of U=u is equal to the probability of X<Y or X>Y, depending on the value of z. If z=0, then the probability of U=u is the same as the probability of X<Y. Similarly, if z=1, then the probability of U=u is the same as the probability of X>Y. Now, the independence of Z and U can be easily demonstrated by the fact that P(Z=z|U=u) = P(U=u|Z=z), which implies that Z and U are independent RVs.