Intuition of joint density of min(X,Y) and max(X,Y) The problem is to find the joint density of $U = min(X,Y)$ and $V=max(X,Y)$ when both are exponential random variables.
The solution to it is:



I can finish it after the first step but I don't understand the intuition behind
$$P(U \le u, V \le v) = P(V\le v)-P(U>u, V\le v) $$
I understand that it's because $P(A\cap B) = P(B) -  P(A^c\cap B)$ but what would make you think that it's a good idea to start off by rewriting it like this?
 A: The problem with the joint distribution of $U$ and $Y$ is that you do not know which of $X$ and $Y$ is $U$ and which is $V$.  Similarly the difficulty with $P(U \le u, V \le v) $, i.e. at least one of $X$ and $Y$ is below $u$ and both below $v$, is handling the question of which one is below $u$ (taking into account that they both may be).  
The trick in the book solution is spotting that $$P(U \le u, V \le v) = P(V\le v)-P(u \lt U \le v, V\le v)$$ gives you two terms which each involve $X$ and $Y$ being in the same interval.
It is in fact easy to go the naive route:  with $0 \le u \le v$, you could say $$P(U \le u, V \le v)  = P(X \le u, u \lt Y \le v)+ P( u \lt X \le v, Y \le u) + P(X \le u, Y \le u) $$ $$= P(X \le u)P( u \lt Y \le v)+ P( u \lt X \le v)P(Y \le u) + P(X \le u)P( Y \le u)$$ by independence and derive the answer that way. You will get the same answer since this is obviously equal to $$P(X \le v)P(Y \le v) - P(u \lt X \le v)P( u \lt Y \le v).$$
A: Rewriting it like this guarantees that both random variables $X$ and $Y$ are wedged between two numbers, and the probability that this happens can be calculated from their distribution.
In particular, $\max (X,Y)\le v$ guarantees that both $X$ and $Y$ are not larger than $v$, while $u<\min(X,Y)\le\max(X,Y)\le v$ guarantees that $X$ and $Y$ lie both between $u$ and $v$. But the probability that this is the case factorizes because $X$ and $Y$ are independent by assumption.
Therefore, splitting the probabilities like that allows to employ the independence of $X$ and $Y$, while $U$ and $V$ are not independent.
A: Precision 1st and then intuition.
Precision:

*

*$$(U \le u) \cup (U > u) = \Omega \tag{A}$$


*Intersect both sides with $(V \le v)$. By distributive property of intersections and unions we get
$$(U \le u, V \le v) \cup (U > u, V \le v) = (V \le v) \tag{C}$$


*Apply probability to both sides. Since the LHS of $(C)$ is a disjoint union (actually the LHS of $(A)$ is a disjoint union too), we simply replace $\cup$ with $+$:

$$P(U \le u, V \le v) + P(U > u, V \le v) = P(V \le v) \tag{B}$$
Intuition:
$(B)$ is like $(A)$ but $(V \le v)$ is like the new $\Omega$. Hopefully this answers 'what would make you think that it's a good idea to start off by rewriting it like this'
