Why do we need to define a random variable as a function?

Question

I recently learned the mathematical definition of a random variable, namely:

A random variable is a measurable function $X: \Omega \rightarrow \mathbb{R}$ whose domain $\Omega$ is equipped with a $\sigma$-algebra $\mathcal{F}$ and a probability measure $P$.

This definition seems unintuitive to me, so I wanted to see how some of the famous classes of random variables are defined using the definition (e.g. Gaussian, Uniform). However, from what I understand in these cases there is no clear sample space $\Omega$ but rather a pdf which is used to define the function. That got me thinking why can't we just define a random variable as a probability measure on the Borel sets instead of using another measure on another $\sigma$-algebra and then requiring the function to be measurable?

Answer 1

Suppose I toss $n$ coins. It's natural to model this probabilistically in terms of a sample space $\{ H, T \}^n$ constructed as the product of $n$ copies of the sample space of possible outcomes of a coin toss, with the uniform probability measure. Furthermore the individual coin tosses themselves are naturally functions on this sample space, namely the $n$ functions $C_i : \{ H, T \}^n \to \{ H, T \}$ given by the $n$ projections. These functions are random variables! More precisely they are $\{ H, T \}$-valued random variables, where I haven't chosen any inclusion into $\mathbb{R}$. We can consider, very generally, random variables with values in any set, e.g. random functions, random graphs….

Now suppose we want to ask a question like: what's the expected number of heads? It's natural to model this in terms of a sum of random variables, namely the sum of the $n$ random variables $X_i : \{ H, T \}^n \to \mathbb{R}$ which is $1$ if the $i^\text{th}$ coin is heads and $0$ otherwise. It is not possible to even discuss the sum $\sum_{i=1}^n X_i$ of random variables, which gives the number of heads, if you insist on identifying each $X_i$ with the corresponding pushforward of the probability measure to $\mathbb{R}$, since this is not enough information to determine what the distribution of the sum is; to determine the sum of random variables in general requires that you know the joint distribution of the $X_i$, or equivalently that you know the pushforward of the corresponding map to $\mathbb{R}^n$.

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Generally it just turns out that a powerful and flexible answer to the question "what is a random element of a (measurable) set $S$?" is that it's a (measurable) function $X : \Omega \to S$, because this definition allows you to flexibly apply operations to random elements of $S$ as if they were ordinary elements of $S$: for example if $S$ is an abelian group you can take sums, if $S$ is a ring you can take sums and products, etc. So this is a philosophy in which functions into an object are "generalized elements" of that object, which is for example very common in category theory and algebraic geometry.

You could instead decide that a random element of $S$ is a probability measure on $S$. This has the virtue of not depending on an auxiliary choice of probability space $\Omega$ (which is not unique in the above definition, and this is a subtlety that needs to be addressed; I really like Terence Tao's discussion here) but makes it awkward to talk about operations on random variables. As above, to talk about the sum $X + Y$ of two random variables you need to specify their joint distribution, which is a measure on $S^2$, not a pair of measures on $S$. Meanwhile in the usual formalism, to talk about the sum $X + Y$ of two random variables they only need to be functions on the same sample space $\Omega$, which is what equips them with a joint distribution.

You could do everything purely in terms of measures if you really wanted to but it just seems inconvenient to me. It won't suffice to consider measures on $\mathbb{R}$ only.

Answer 2

$\newcommand\Om\Omega\newcommand\om\omega\newcommand\R{\Bbb R}$It is actually more natural to define a random variable (r.v.) $X$ as a function from a set $\Om$ -- which is usually referred to as the sample space -- to (say) $\R$ than a probability measure over $\R$. Indeed, one can think of the sample space $\Om$ as a population, so that then $X(\om)$ will be the value of the function/characteristic $X$ for the member $\om$ of the population $\Om$.

The probability measure that is the distribution of a r.v. $X$ is just one (even if very important) feature of the r.v. $X$, as there are usually many r.v.'s with the same distribution, even on the same probability space.

The convenience of dealing with r.v.'s rather than just with their distributions is even much greater if one has to deal with several (and possibly infinitely many) r.v.'s at once. Then one can easily deal with sums, products, and other functions of several r.v.'s, which may be rather inconvenient to do in terms of just distributions, even joint ones.

Answer 3

Throw a die twice. The set of all possible outcomes is $$ \Omega = \left\{ \, \begin{array}{cccccc} 11 & 12 & 13 & 14 & 15 & 16 \\ 21 & 22 & 23 & 24 & 25 & 26 \\ 31 & 32 & 33 & 34 & 35 & 36 \\ 41 & 42 & 43 & 44 & 45 & 46 \\ 51 & 52 & 53 & 54 & 55 & 56 \\ 61 & 62 & 63 & 64 & 65 & 66 \end{array} \, \right\}. $$

• The sum of the results of the two dice is a random variable.
• The maximum of the two is a random variable.
• The result shown by the second die is a random variable.
• The function whose value is $1$ if the sum is${}\le4$ and $0$ otherwise is a random variable.

If you just have a probability measure on Borel sets of reals, then the real line is $\Omega.$