1. ## Independence of events

Let X and Y be random variables that both have an exponential distribution with parameters lambda and mu respectively. X and Y are independent.

Define Z = min{X, Y}

(1) Prove that the event {X < Y} is independent of the event {Z > z}

(2) Find E[ e^(-z) | X ].

2. You can calculate the three probabilities.
That's probably not a clever way, but it must work.

$\displaystyle P(X<Y,\min\{X,Y\}>z)=P(X<Y,X>z)$

$\displaystyle ={1\over \mu\lambda}\int_z^{\infty}\int_z^y e^{-x/\lambda}e^{-y/\mu}dxdy$

Then get

$\displaystyle P(X<Y)={1\over \mu\lambda}\int_0^{\infty}\int_0^y e^{-x/\lambda}e^{-y/\mu}dxdy$

and

$\displaystyle P(\min\{X,Y\}>z)= P(X>z,Y>z)= P(X>z)P(Y>z)$

$\displaystyle ={1\over \mu\lambda}\int_z^{\infty}e^{-x/\lambda}dx\int_z^y e^{-y/\mu}dy=e^{-z/\lambda}e^{-z/\mu}$

Then compare to see if P(AB)=P(A)P(B).