# Nomenclature of random variables $\{X=0, Y=0\}$ same as $\{X=0\}\cap \{Y=0\}$?

just a small doubt. My exercises keep oscillating their nomenclature on this small detail and I always have the other version.

Let $X,Y$ be random variables. Is $\{X=0, Y=0\}$ the same as $\{X=0\}\cap \{Y=0\}$?

Another example. Let $N$ be the number of Users on a webpage. Two files are available for download, one with 200 kb and another with 400 kb size.

$$\begin{align} X_n(w) := w_n = \{ & 0:=\text{user downloads no file}, \\ & 1:=\text{user downloads the first file (200 kb)}, \\ & 2 :=\text{user downloads the second file (400 kb)}, \\ & 3:=\text{user downloads both files (600 kb)}\} \end{align}$$

I want to express, at least one user downloaded the 200 kb file. Here's how I expressed it $\{X_1 + X_2 + \cdots + X_n \geq 1\}$. Would this be ok? The book expressed it as $\{X_1=1\}\cup\{X_1=3\}\cup \cdots \cup\{X_n=1\}\cup\{X_n=3\}$.

Another thing to express: no user downloaded the 200 kb file. I expressed it as $|\{X_k=1, 1 \leq k \leq N\}|=0$. The book as $\{X_1 \neq 1\}\cap \cdots \cap \{X_n \neq 1\}$. Would my solution be ok?

I'm always in doubt when I'm allowed to use symbols like $+$ and $|\mathrm{modulo}|$ (to get the number of elements). Is this generally always allowed? Many thanks in advance!

Thanks in advance guys!

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## 2 Answers

$\{X=0,Y=0\}$ and $\{X=0\}\cap\{Y=0\}$ are the same thing. Both notations refer to $$\{\omega\in\Omega : X(\omega)=0\ \ \&\ \ Y(\omega)=0\} = \{\omega\in\Omega : X(\omega)=0\}\cap\{\omega\in\Omega : Y(\omega)=0\}.$$

Your notation saying $$\begin{align} X_n(w) := w_n = \{ & 0:=\text{user downloads no file}, \\ & 1:=\text{user downloads the first file (200 kb)}, \\ & 2 :=\text{user downloads the second file (400 kb)}, \\ & 3:=\text{user downloads both files (600 kb)}\} \end{align}$$ seems confused. I suspect maybe you meant $$\begin{align} \Omega = \{ & 0:=\text{user downloads no file}, \\ & 1:=\text{user downloads the first file (200 kb)}, \\ & 2 :=\text{user downloads the second file (400 kb)}, \\ & 3:=\text{user downloads both files (600 kb)}\}, \end{align}$$ although even that may differ from what's appropriate if you're bringing in $n$ different random variables. Your later notation makes it look as if what the author of the book had in mind is that $X_k$ is the number of kb downloaded by the $k$th user, for $k=1,\ldots,n$. Just what $w$ is, you're not clear about, and at this point I'm wondering if you're confusing $w$ with $\omega$. Probably what is needed is this:

$$\begin{align} \{ & 0:=\text{user downloads no file}, \\ & 1:=\text{user downloads the first file (200 kb)}, \\ & 2 :=\text{user downloads the second file (400 kb)}, \\ & 3:=\text{user downloads both files (600 kb)}\}^n \end{align}$$ i.e. then $n$th power of that set of four elements. This is the set of all $n$-tuples where each component of an $n$-tuple is one of these four elements. Then, when $\omega$ is any such $n$-tuple, $X_k(\omega)$ is its $k$th component, which is one of those four elements.

For example, if $n=3$, so there are three users, then $$\begin{align} \Omega = \{ & (0,0,0), (0,0,1), (0,0,2), (0,0,3), (0,1,0), (0,1,1), (0,1,2), (0,1,3),\ldots\ldots\ldots \\ \\ & \ldots\ldots\ldots, (3,3,3) \}, \end{align}$$ with $64$ elements. If, for example, $\omega=(2,3,0,1)$, then $X_2(\omega)=3$.

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Your second example is incorrect: if no user downloaded the 200K file, but at least one user downloaded the 400K file, we will still have $X_1 + \dots + X_n \ge 1$.

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thanks, well observed! but the notation I used would be ok? – Clash Jan 28 '12 at 22:33
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