# 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?

• 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.

• 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
• @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
• @cardinal, in @Leo's comment it was asked if $[0,1]$ is 'preferred' in all circumstances. I am just saying that IMO there is no such $\Omega$ and that it is beneficial not to require anything about $\Omega$ in general. When you want to work with a specific example, there may be reasons to choose one specific $\Omega$. Note, however, that the 'tautology' is sweeping under the rug that the existence of Brownian motion as a probability measure on $\mathcal{C}$ needs to be established. – NRH May 14 '11 at 19:27
• @NRH, sorry for my slowness of mind today. I had failed to connect the preferred reference to @Leo's previous comment. Thanks. Regarding the "tautology" remark, my point was that in other constructions, the a.s. continuity of sample paths is a theorem, whereas, under the $\mathcal{C}$-based construction with identity map, it is tautological. Of course, the fact that B.M. can be constructed this way must first be shown. But, that is a little bit beside the point. – cardinal May 14 '11 at 19:45

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.

What seems to be the issue is that the arithmetic transformations involved in the CLT are quite clearly expressed in terms of random variables but they do not translate so well in terms of the measures.

• (+1) Good description. I think the other reason the former notation is so popular is that it more naturally translates to intuitive notions in applications. (Voted up several hours ago.) – cardinal May 14 '11 at 23:39
• @cardinal, thanks for making that point more clear. It seems more natural to think and argue in terms of a sum of variables, not a convolution of probability measures, and we would like the mathematics to reflect that. – NRH May 15 '11 at 0:06

I only recently stumbled over this new way to think about the Random Variable $X$ as well as about the background space $\Omega$. I am not sure whether this is the question you were looking for, as it is not a mathematical reason, but I think it provides a very neat way to think of RVs.

Imagine a situation in which we throw a coin. This experimental setup consists of a Set of possible initial conditions that include the physical description of how the coin is tossed. The background space consists of all those possible initial conditions. For simplicities sake we might assume that the coin tosses only vary in velocity, then we would set $\Omega = [0,v_{max}]$

The random variable $X$ can then be thought of as a function that maps every initial state $\omega \in \Omega$ with the corresponding outcome of the experiment, i.e. whether it is tails or head.

For the RV: $X:([0,v_{max}], B\cap [0,v_{max}], Q)\to (\{0,1\}, 2^{\{0,1\}})$ the measure $Q$ would then correspond to the probability measure over the initial conditions, which together with the dynamics of the experiment represented by $X$ determines the probability distribution over the outcomes.

For reference of this idea you can look at Tim Maudlin`s or Micheal Strevens chapters in "Probabilties in Physics" (2011)