# Mathematics Prelims

## October 26, 2008

### Random Variables

Filed under: Measure Theory,Probability Theory,Topology — cjohnson @ 8:32 pm

If $(\Omega, \mathcal{F}, P)$ is a probability space and $(Y, \Sigma)$ is a measurable space (i.e., a set $Y$ along with a sigma-algebra $\Sigma$ on $Y$), then a random variable is a measurable function $X : \Omega \to Y$.  That is, for each $A \in \Sigma$, we have $X^{-1}(A) = \{ \omega \in \Omega : X(\omega) \in A \} \in \mathcal{F}$.  Generally speaking, we’ll be taking $(Y, \Sigma)$ to be $(\mathbb{R}, \mathcal{B})$ where $\mathcal{B}$ is the Borel algebra on $\mathbb{R}$.

Given a topological space $(S, \tau)$, there exists a sigma-algebra $\mathcal{B}(S)$, called the Borel algebra on $S$, that contains all open sets (members of $\tau$) and is the smallest such sigma-algebra.  This means that for each $\mathcal{O} \in \tau$ we have $\mathcal{O} \in \mathcal{B}(S)$ and also if $\Sigma$ is another sigma-algebra on $S$ with this property, then $\mathcal{B}(S) \subseteq \Sigma$.  In general, given a collection $C$ of subsets of $S$, there exists a sigma-algebra, which we’ll call $\mathcal{F}_C$, that is the smallest sigma-algebra on $S$ (smallest using $\subseteq$ as the ordering relation) containing each $c \in C$; we say that $\mathcal{F}_C$ is the sigma-algebra generated by $C$.  In this sense, the Borel algebra on $S$ is the sigma-algebra generated by the topology $\tau$.  We will usually just write $\mathcal{B}$ to mean $\mathcal{B}(\mathbb{R})$.

In the special case of $(\mathbb{R}, \mathcal{B})$, there’s an easy way to check to see that a function $X : \Omega \to \mathbb{R}$ is a random variable (or a measurable function in general): we just look at the pre-images of intervals.  Since the pre-images of functions are well-behaved with respect to set operations like union, intersection, and complement, it in fact suffices to only consider pre-images of the form ${}[a, \infty)$.  That is, if we show that $X^{-1}([a, \infty)) \in \mathcal{F}$ for every $a \in \mathbb{R}$, we will have shown that $X$ is measurable, and so a random variable.  (Actually, we can look at all intervals of the form $(a, \infty)$, ${}[a, \infty)$, $(-\infty, a)$ or $(-\infty, a]$.  Using properties of sigma-algebras we can easily show that if we have all intervals of any of these forms, we have all intervals of any other form.  Again, using properties of sigma-algebras it’s easy to take that and show that we have all countable unions of intervals — namely all countable unions of open intervals, i.e., all open sets.)

To see this, suppose for every $a \in \mathbb{R}$ we have $X^{-1}([a, \infty)) \in \mathcal{F}$.

$\displaystyle X^{-1}([a, \infty)) \in \mathcal{F}$

$\displaystyle \implies \{ \omega \in \Omega : X(\omega) \geq a \} \in \mathcal{F}$

$\displaystyle \implies \{ \omega \in \Omega : X(\omega) \geq a \}^\complement \in \mathcal{F}$

$\displaystyle \implies \{ \omega \in \Omega : X(\omega) < a \} \in \mathcal{F}$

$\displaystyle \implies X^{-1}((-\infty, a)) \in \mathcal{F}$

So now we have the pre-images of all intervals of the form ${}[a, \infty)$ and $(-\infty, a)$.  If we can also get pre-images for the form $(a, \infty)$, it’ll be an easy jump to countable unions of open intervals.  Note that

$\displaystyle (a, \infty) = \bigcup_{n=1}^\infty \left[ a + \frac{1}{n}, \infty \right)$

Now we easily see that

$\displaystyle X^{-1}((a, \infty)) = X^{-1} \left( \bigcup_{n=1}^\infty \left[ a + \frac{1}{n}, \infty \right)\right)$

$\displaystyle = \bigcup_{n=1}^\infty X^{-1}\left(\left[ a + \frac{1}{n}, \infty \right)\right)$

And we have pre-images of intervals of the form $(a, \infty)$.  Combined with the fact we have intervals of the form $(-\infty, b)$, it’s easy to see that we also have intervals of the form $(a, b)$.  Using properties of sigma-algebras, it’s easy to show now that we have the pre-image of any open set.  This tells us that if $X^{-1}([a, \infty)) \in \mathcal{F}$ for every $a \in \mathbb{R}$, then we have that $X$ is measurable, and so a random variable.

In the case that $\Omega$ is a countable set, we take $\mathcal{F}$ to be $2^\Omega$ (the powerset of $\Omega$), and so any function $X : \Omega \to \mathbb{R}$ is a random variable.  This is because the pre-image of ${}[a, \infty)$ must be something (even if it’s empty); we have for every $a \in \mathbb{R}$ that $X^{-1}([a, \infty)) \in 2^\Omega$, and so every function on a countable sample-space is a random variable.