3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What are the tools you know, to prove that something is a random variable? $\endgroup$ – Did Mar 30 '13 at 14:54
  • 1
    $\begingroup$ If you have several random variables, then practically any combination of them will be a random variable. :) $\endgroup$ – Caran-d'Ache Mar 30 '13 at 14:55
4
$\begingroup$

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 $\min$.

$\endgroup$
  • $\begingroup$ Woops, typo. Thanks, I'll fix that. $\endgroup$ – Gil Mar 30 '13 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.