what does it mean by "$\min\{X,Y\}$" where $X$ and $Y$ are random variable? I see this term often.
find $P(\min\{X,Y\}< z)  $      where $X$ and $Y$ are independent random variable .
From the online source, I see people saying this as $P( \{X < z\} \cup \{Y< z\} )$, but I don't see why.
Can you guys please define what it means? Also for maximum?
 A: $P(\min{X,Y}<z)$ is the probability that a realization of $X$ and $Y$ from their distributions will be such that the minimum of those two numbers is less than $z$.
For instance, for two uniformly distributed random variables from $[0,1]$, how likely is it that their minimum is less than $1/2$?  (Equivalently, how likely is it that both are $\geq 1/2$, which is $1/2 \times 1/2 = 1/4$.)
If the minimum of two numbers is less than $z$ then either $X < z$ or $Y < z$.  (If neither is less than $z$, then the minimum is not less than $z$.)  So the universes with minimum less than $z$ is the union of universes where one is less with the universes where the other is less.  (This includes universes where both are less than $z$, but their minimum is still less than $z$.  Via union, we only count such universes once, instead of twice.)
A: This is more a statement about sets than probabilities.
$\{\omega | \min(X(\omega), Y(\omega)) < z \} = \{\omega | X(\omega) < z \text{ or } Y(\omega) < z \} = \{\omega | X(\omega) < z \} \cup \{\omega | Y(\omega) < z \}$.
You can show this in the following way if you prefer, but it is essentially the same as above: Suppose $\alpha \in \{\omega | \min(X(\omega), Y(\omega)) < z \}$, then  $\min(X(\alpha), Y(\alpha)) < z$ and so we must have either $X(\alpha) <z$ or $Y(\alpha) < z$ in which case we have $\alpha \in \{\omega | X(\omega) < z \} $ or $\alpha \in \{\omega | Y(\omega) < z \} $, or in other words $\alpha \in \{\omega | X(\omega) < z \} \cup \{\omega | Y(\omega) < z \}$.
Since $\min(X(\alpha), Y(\alpha)) \le X(\alpha)$, we see that if $\alpha \in \{\omega | X(\omega) < z \} $ then we must have $\alpha \in \{\omega | \min(X(\omega), Y(\omega)) < z \}$. The same applies to $Y$.
For $\max$, you can use $\max(a,b) = - \min (-a,-b)$, or replicate the reasoning above with $\text{or}$ replaced by $\text{and}$:
$\{\omega | \max(X(\omega), Y(\omega)) < z \} = \{\omega | X(\omega) < z \text{ and } Y(\omega) < z \} = \{\omega | X(\omega) < z \} \cap \{\omega | Y(\omega) < z \}$.
A: Well, if the minimum of $X$ and $Y$ is less than $z$ then the union you state must be true, correct? 
If the union statement is true, then the minimum of $X$ and $Y$ must be less than $z,$ right? 
So if each implies the other, the two statements must be equivalent. 
