# Maximum/minimum of two random variables is a random variable

Suppose $X,Y$ are random variables. I'm trying to understand why $\max\{X,Y\}$ and $\min\{X,Y\}$ are also random variables. The proof in the book that I'm using states that for each $t$,

$\{ \max\{X,Y\} \leq t \} = \{X \leq t \} \cap \{Y \leq t \}$,

and

$\{ \min\{X,Y\} \leq t \} = \{X \leq t \} \cup \{Y \leq t \}$.

I can't make the mental leap. Could someone explain this more explicitly?

Edit:

We use the following definition for a random variable: A random variable is a function on $X: \Omega \rightarrow \mathbb{R}$ such that

$X^{-1}((-\infty,t]) := \{\omega \in \Omega : X(\omega) \leq t\} \in \mathcal{F}$

for all $t\in\mathbb{R}$, where $\mathcal{F}$ is the set of all events.

The obvious first step that I made was

$\{ \max\{X,Y\} \leq t \} = \{ \omega : \max\{X(\omega),Y(\omega)\} \leq t \}$.

I just don't see why the next equality follows.

Edit2: Fixed typo.

• What are the tools you know, to prove that something is a random variable? – Did Mar 30 '13 at 14:54
• If you have several random variables, then practically any combination of them will be a random variable. :) – Caran-d'Ache Mar 30 '13 at 14:55

It's rather $Y\le t$ instead of $Y\le y$. With that, for an $\omega$ we have $$\omega\in\{\max(X,Y)\le t\} \iff (X(\omega)\le t)\land(Y(\omega)\le t) \\ \iff \omega\in\{X\le t\}\cap\{Y\le t\}\,.$$ Similarly for the $\min$.