If $X$ and $Y$ are independent exponential then $\{\min(X,Y)>z\}$ and $\{X
Given that $X$ and $Y$ are exponential random variables, possibly with different rates. Also, $Z = \min(X,Y)$. Show that the event $\{Z>z\}$ is independent of $\{X<Y\}$.
So, I've found the probabilities for both $P(Z>z)$ and $P(X<Y)$. But I don't know how to show that they are independent of each other. I know that $P(A \cap B) = P(A)P(B)$ means that $A$ and $B$ are independent of each other, but I don't know how I can apply that to $X,Y,Z$.
P.S.: I rarely use this website, so I apologise if I'm doing some things wrong. Please tell me how I can fix my ways to do better in the future. Thank you.
 A: I assume that $X$ and $Y$ are independent. Let $\lambda_X$ and $\lambda_Y$ be the rates of $X$ and $Y$, respectively.
We want to find the following probability:
$$P(Z > z \wedge X < Y) = P(\min(X,Y) > z \wedge X <Y).$$
The set $\mathcal{S}_z = \{(X,Y) : \min(X,Y) > z \wedge X <Y\}$ is defined by inequalities $X < Y$ and $z < X$ since $\min(X,Y) = X$ when $X<Y$. Then:
$$\begin{align}P((X,Y) \in \mathcal{S}_z) ~=~& \int_{X=z}^{+\infty}\left[ \int_{Y=X}^{+\infty}\lambda_Ye^{-\lambda_Y Y} dY \right]\lambda_Xe^{-\lambda_X X}dX \\[1ex] =~& \int_{X=z}^{+\infty}e^{-\lambda_Y X}\lambda_Xe^{-\lambda_X X}dX \\[1ex] =~& \frac{\lambda_X}{\lambda_X+\lambda_Y}e^{-(\lambda_X+\lambda_Y)z}\end{align}$$
Now, we need to find also the probabilities $P(X < Y)$ and $P(Z > z)$:
$$\begin{align}P(X < Y) ~=~& \int_{X=0}^{+\infty}\left[ \int_{Y=X}^{+\infty}\lambda_Ye^{-\lambda_Y Y} dY \right]\lambda_Xe^{-\lambda_X X}dX \\[1ex] =~& \frac{\lambda_X}{\lambda_X+\lambda_Y}
\\[2ex] P(\min(X,Y) > z) ~=~& \int_{X=z}^{+\infty}\left[ \int_{Y=z}^{+\infty}\lambda_Ye^{-\lambda_Y Y} dY \right]\lambda_Xe^{-\lambda_X X}dX \\[1ex]
=~& \int_{X=z}^{+\infty}e^{-\lambda_Y z}\lambda_Xe^{-\lambda_X X}dX \\[1ex] =~& e^{-(\lambda_X+\lambda_Y)z}\end{align}$$
Finally:
$$P(Z > z \wedge X <Y ) =\frac{\lambda_X}{\lambda_X+\lambda_Y}e^{-(\lambda_X+\lambda_Y)z} = P(Z > z) \cdot P(X <Y)$$
This means that the two events are independent.
