Joint distribution of $\max(X, Z)$ and $\max(Y, Z)$ where $X$, $Y$, $Z$ are independent exponential variables with mean $1$

Question

Assume there are three independent exponential random variables $X$, $Y$, and $Z$ with mean $1$. I am trying to look at the joint distribution of $[U = \max(X, Z), V = \max(Y, Z)]$.

I want to find $F_{\, U,V}(u, v)$ which is the cumulative distribution function of joint distribution $(U,V)$.

What I am doing is diving the possible relations of $X$, $Y$, $Z$ like $X < Y < Z$, $X < Z < Y$, etc. But I’m getting stuck at cases like $u < v$ and $Z < X < Y$. What is some other way(s) to find the cumulative distribution of this joint distribution?

Answer 1

$P(U \leq u, V \leq v) = P(\max(X,Z) \leq u, \max(Y,Z) \leq v) = P(X \leq u, Y \leq v, Z \leq \min(u,v)) = P(X \leq u) \times P(Y \leq v) \times P(Z \leq \min(u,v)) = (1-e^{-u}) \times (1-e^{-v}) \times (1-e^{-\min(u,v)})$

Answer 2

I like your solution. As you mentioned, $$P(U<V \textrm{ and } Z<X<Y)= P(Z<X<Y) = P(Z=\min(X,Y,Z))P(X<Y) = \frac{1}{3}\cdot \frac{1}{2}$$ because the event $Z<X<Y$ is contained in the event $U<V$.

Now, you can consider all 6 cases,

$$P(U<V \textrm{ and } Z<X<Y)$$

$$P(U<V \textrm{ and } X<Z<Y)$$

$$P(V<U \textrm{ and } Z<Y<X)$$

$$P(V<U \textrm{ and } Y<Z<X)$$

$$P(U<V \textrm{ and } X<Y<Z)$$

$$P(U<V \textrm{ and } Y<X<Z)$$

Does that make sense?

Here is another idea:

\begin{align} &P(U<u, V<v, z<Z<z+dz) \\& =P(U<u, V<v |z<Z<z+dz) P(z<Z<z+dz) =\\ &=P(U<u |z<Z<z+dz) P(V<v|z<Z<z+dz) P(z<Z<z+dz) \end{align}

Therefore,

\begin{align} &P(U<u, V<v, z<Z<z+dz) \\& =P(X<u, Z<u |z<Z<z+dz) P(Y<v, Z<v |z<Z<z+dz) \exp(-z)dz\end{align}

If $u>z$ and $v>z$

\begin{align} &P(U<u, V<v, z<Z<z+dz) \\& =(1-\exp(-u) )(1-\exp(-v) ) \exp(-z)dz\end{align}

Do the other cases follow similarly?