For two random variables $X$ and $Y$, considering $W=\max (X,Y)$ or $W=\min (X,Y)$, I have learnt to rewrite $W$ in following way $$W=\max(X,Y)=\begin{cases} X & \textrm{ if } X\geq Y \\ Y & \textrm{ if } X<Y \end{cases}.\bigg\}\to (1)\\ W=\min(X,Y)=\begin{cases} Y & \textrm{ if } X\geq Y \\ X & \textrm{ if } X<Y \end{cases}.\bigg\} \to (2)$$.
I can find out the distribution function $F_W(w)$ for $W=\max(X,Y)$ or $W=\min(X,Y)$.
Now, specifically, I want to learn how to rewrite $\max (X,Y,Z), \min (X,Y,Z), \max(X, \min(Y,Z)), \min(Z, \max(X,Y))$. So that I can plot the graph of each cases (such as $X\geq Y\geq Z$, $X<Y<Z$.. etc) to understand the problems visually and to find out the distribution function $F_W(w)=P(W\leq w)$ at each case, considering $$(i).\ W=\max (X,Y,Z),\\(ii) \ W=\min(X,Y,Z), \\(iii) \ W= \max(X, \min(Y,Z))\\ (iv) \ W= \min(Z, \max(X,Y))$$
. But how to rewrite $W$ for three random variables $X$, $Y$, $Z$, where it is given that
$W=\max(X,Y,Z)$.
Here is my first approach (don't know if it is correct or not) to rewrite $(i)$ ..
$$W=\max(X,Y,Z)=\begin{cases} X & \textrm{ if } X\geq Y\geq Z\\X & \textrm{ if } X\geq Z\geq Y \\ \max(X,Z)=Z & \textrm{ if } X\geq Y\leq Z\\\max(X,Y)=Y & \textrm{ if } X\geq Z\leq Y \\\min(X,Z)=X & \textrm{ if } X< Y> Z \\\min(X,Y)=X & \textrm{ if } X< Z> Y \\Y & \textrm{ if } Y\geq Z\geq X \\Y & \textrm{ if } Y\geq X\geq Z \\\max(Y,Z)=Z & \textrm{ if } Y\geq X\leq Z\\\max(X,Y)=X & \textrm{ if } Y\geq Z\leq X \\\min(Y,Z)=Y & \textrm{ if } Y< X> Z\\\min(X,Y)=Y & \textrm{ if } Y< Z> X \\ Z & \textrm{ if } Z\geq X\geq Y \\ Z & \textrm{ if } Z\geq Y\geq X\\ \max(X,Z)=X & \textrm{ if } Z\geq Y\leq X \\ \max(Y,Z)=Y & \textrm{ if } Z\geq X\leq Y \\ \min(Y,Z)=Z & \textrm{ if } Z< X> Y \\ \min(X,Z)=Z & \textrm{ if } Z< Y> X \end{cases}.$$
And a second approach to rewrite $W=\max(X,Y,Z)$.
Define events $$(X\geq Y) \textrm{ as } \mathbb {A}\ \textrm{&} \space (X<Y) \textrm{ as } \mathbb {A'}\\ (Y\geq Z) \textrm{ as } \mathbb {B}\ \textrm{&} \space (Y<Z) \textrm{ as } \mathbb {B'}\\ (Z\geq X) \textrm{ as } \mathbb {C}\ \textrm{&} \space (Z<X) \textrm{ as } \mathbb {C'}$$
$\therefore$ $$W=\max(X,Y,Z)=\begin{cases} X & \textrm{ if } \mathbb {A} \cup \mathbb{B}\\Y & \textrm{ if } \mathbb {B} \cup \mathbb{C}\\Z & \textrm{ if } \mathbb {C} \cup \mathbb{A}\\ Z & \textrm{ if } \mathbb {A} \cup \mathbb{B'}\\ X & \textrm{ if } \mathbb {A} \cup \mathbb{C'}\\ Y & \textrm{ if } \mathbb {B} \cup \mathbb{A'}\\X & \textrm{ if } \mathbb {B} \cup \mathbb{C'}\\Y & \textrm{ if } \mathbb {C} \cup \mathbb{A'}\\Z & \textrm{ if } \mathbb {C} \cup \mathbb{B'}\\Z & \textrm{ if } \mathbb {A'} \cup \mathbb{B'} \\X & \textrm{ if } \mathbb {B'} \cup \mathbb{C'}\\Y & \textrm{ if } \mathbb {C'} \cup \mathbb{A'} \end{cases}.$$
Due to little knowledge I am confused and stuck to rewrite $(iii)$ and $(iv)$. If rewriting $(i)$ is correct, then $(ii)$ can be defined as the reverse of $(i)$.
Any help or explanation is valuable for learning and highly appreciated.
There are only six cases you need to consider, corresponding to total orderings of $(X,Y,Z)$; if there can be ties, then these cases are not mutually exclusive, but this doesn't matter for calculating maxes and mins. Here is the table of outcomes: \begin{array}{ c | c } \text{Case} & \max(X,Y,Z) & \min(X,Y,Z) & \max(X,\min(Y,Z)) & \min(Z,\max(X,Y))\\ \hline X \le Y \le Z & Z & X & Y & Y \\ \hline X \le Z \le Y & Y & X & Z & Z \\ \hline Y \le X \le Z & Z & Y & X & X \\ \hline Y \le Z \le X & X & Y & X & Z \\ \hline Z \le X \le Y & Y & Z & X & Z \\ \hline Z \le Y \le X & X & Z & X & Z \end{array} Note that there are $3^6=729$ possible "columns" for this table. It's an interesting exercise to think about whether you can write an expression corresponding to each. Thus far you have $3$ for single variables, $6$ for the min or max of a pair, $2$ for the min or max of all three, and $6$ more for the min (max) of one variable with the max (min) of the other two... only $17$!