# Why are random variables defined as functions?

I'm having problems understanding the concept of a random variable as a function. I understand the mechanics (I think) but I do not understand the motivation...

Say $(\Omega, B, P)$ is a probability triple, where $\Omega = [0,1]$, $B$ is the borel-$\sigma$-algebra on that interval and $P$ is the regular lebesgue measure. Let $X$ be a random variable from $B$ to $\{1,2,3,4,5,6\}$ such that $X([0,1/6)) = 1$, $X([1/6,2/6)) = 2$, ..., $X([5/6,1]) = 6$, so $X$ has a discrete uniform distribution on the values 1 through 6.

That's all good, but I do not understand the necessity of the original probability triple... we could have directly constructed something equivalent as $(\{1,2,3,4,5,6\}, S, P_x)$ where $S$ is all the appropriate $\sigma$-algebra of the space, and $P_x$ is a measure that assigns to each subset the measure (# of elements)/6. Also, the choice of $\Omega=[0,1]$ was arbitrary-- it could've been $[0,2]$, or any other set.

So my question is, why bother constructing an arbitrary $\Omega$ with a $\sigma$-algebra and a measure, and define a random variable as a map from the $\sigma$-algebra to the real line?

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Note that the random variable is the function from $\Omega$ to $\mathbb{R}$, not from $\mathcal{B}$ to $R$. The requirement is that random variable is measurable with respect to $\mathcal{B}$. –  mpiktas May 14 '11 at 12:03

If you are wondering why all this machinery is used when something much simpler could suffice--you are right, for most common situations. However, the measure-theoretic version of probability was developed by Kolmogorov for the purpose of establishing a theory of such generality that it could handle, in some cases, very abstract and complicated probability spaces. In fact, Kolmogorov's measure theoretic foundations for probability ultimately allowed probabilistic tools to be applied far beyond their original intended domain of application into areas such as harmonic analysis.

At first it does seem more straightforward to skip any "underlying" $\sigma$-algebra $\Omega$, and to simply assign probability masses to the events comprising the sample space directly, as you have proposed. Indeed, probabilists effectively do the same thing whenever they choose to work with the "induced-measure" on the sample space defined by $P \circ X^{-1}$. However, things start getting tricky when you start getting into infinite dimensional spaces. Suppose you want to prove the Strong Law of Large Numbers for the specific case of flipping fair coins (that is, that the proportion of heads tends arbitrarily closely to 1/2 as the number of coin flips goes to infinity). You could attempt to construct a $\sigma$-algebra on the set of infinite sequences of the form $(H,T,H,...)$. But here can find that it is much more convenient to take the underlying space to be $\Omega = [0,1)$; and then use the binary representations of real numbers (e.g. $0.10100...$) to represent sequences of coin flips (1 being heads, 0 being tails.) An illustration of this very example can be found in the first few chapters of Billingsley's Probability and Measure.

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Thanks! I'll check that book out. However, since the $\Omega$ is still arbitrary (it could just as well been $[0,2)$ in your example, is the unit interval $[0,1]$ or $[0,1)$ the 'prefered' space which will work in all circumstances? Or are there situations where a more complicated $\Omega$, like $R^2$ would be beneficial? –  Leo Vasquez May 13 '11 at 22:57
@Leo: Yes. Continuous-time stochastic processes provide an example. The canonical example is Brownian motion, where the sample space $\Omega$ is taken to be $\mathcal{C}$, the space of all continuous real-valued functions. –  cardinal May 14 '11 at 13:35
@cardinal, the sample space is (usually) $\mathcal{C}$. The background space $\Omega$ can be $\mathcal{C}$ with $X$ the identity map. This is the canonical background space. The Kahunen-Loève expansion gives a series expansion of the sample path of Brownian motion based on a sequence of i.i.d. normal random variables, which, in turn, could be obtained as transformations of the Lebesgue measure on $[0, 1]$. Thus it is possible to get Brownian motion using $\Omega = [0,1]$. –  NRH May 14 '11 at 18:43
@NRH, Yes, I should have said can be taken instead of is taken. I was (somewhat purposefully) trying to brush that under the rug. –  cardinal May 14 '11 at 18:48
My point with the comment above is that there may be many choices of $\Omega$ but none are preferred. My experience is that any specific requirements on the structure of $\Omega$ is more of a nuisance than a benefit. –  NRH May 14 '11 at 18:50
The issues regarding $\sigma$-algebras are mathematical subtleties, that do not really explain why or if we need a background space. Indeed, I would say that there is no compelling evidence that the background space is a necessity. For any probabilistic setup $(E, \mathbb{E}, \mu)$ where $E$ is the sample space, $\mathbb{E}$ the $\sigma$-algebra and $\mu$ a probability measure, the interest is in $\mu$, and there is no abstract reason that we want $\mu$ to be the image measure of a measurable map $X : (\Omega, \mathbb{B}) \to (E, \mathbb{E})$.
However, the use of an abstract background space gives mathematical convenience that makes many results appear more natural and intuitive. The objective is always to say something about $\mu$, the distribution of $X$, but it may be easier and more clearly expressed in terms of $X$.
An example is given by the central limit theorem. If $X_1, \ldots, X_n$ are i.i.d. real valued with mean $\mu$ and variance $\sigma^2$ the CLT says that $$P\left(\frac{\sqrt{n}}{\sigma} \left(\frac{1}{n}\sum_{i=1}^n X_i - \xi\right) \leq x \right) \to \Phi(x)$$ where $\Phi$ is the distribution function for the standard normal distribution. If the distribution of $X_i$ is $\mu$ the corresponding result in terms of the measure reads $$\rho_{\sqrt{n}/\sigma} \circ \tau_{\xi} \circ \rho_{1/n}(\mu^{*n})((-\infty, x]) \to \Phi(x)$$ Some explanation of the terminology is needed. By $\mu^{*n}$ we mean the $n$-times convolution of $\mu$ (the distribution of the sum). The functions $\rho_c$ are the linear functions $\rho_c(x) = cx$ and $\tau_{\xi}$ is the translation $\tau_{\xi}(x) = x - \xi$. One could probably get used to the second formulation, but it does a good job at hiding what it is all about.