What we defined: Suppose we have a (discrete) probability model $\left(\Omega,P\right)$, where $P$ is the probability function (at least, that was the way it was introduced in a course I took; that means only that $\Omega$ is at most countable, that $P\left(\bigcup_{i}A_{i}\right)=\sum P\left(A_{i}\right)$, for all (at most countable and disjoint) events $A_{i}$ and that of course $P\left(\Omega\right)=1$). We defined a random variable (rd from now on) $X$ to be a mapping $X:\Omega\rightarrow\mathbb{R}$ and then discussed some aspects of $P\left(X=k\right)$ for some $k\in\mathbb{R}$ .

What bothers me: To me, this definition seems rather artificial: Why define a mapping like that, if there is no apparent need for it (at least in this probability model) ?

Since if we have an event $A\subseteq\Omega$, that depends on some parameter $k\in\mathbb{R}$, (for example the sum of the faces of dice be $k$) then we could just as easily define a collection of events $A_{k}$ - one for each parameter - and discuss aspects of $P\left(A_{k}\right)$, instead of the above way by using $X$. Of course one could now argue, that $A$ does not always need to depend on some parameter $k$, so one in some cases really has to "convert" probabilities to numbers via $X$, but of all examples that I have seen until now, even in $\Omega$ isn't made up out of numbers, somewhere a parameter $k$ does sneak in, so we could equivalently work with $P\left(A_{k}\right)$ instead of $P\left(X=k\right)$, since defining the subsets of $\Omega$ whose elements we want to count (to establish the probability of the subset) always amounts to using some $k$ in the definition of those subsets (this reasoning extends of course also to other cases like when we consider $P\left(X\leq k\right)$, since this can also be circumvented by considering an appropriate $P\left(\cup_{j\leq k}A_{j}\right)$).

Thus, introducing rd's seems to me to be a superfluous definition of events without specifying $k$"

  • $\begingroup$ First of all $P(\bigcup_i A_i)\leq \sum P(A_i)$ not equal as you said it was. Secondly, I do not see a difference in using an $A_k$ and $X=k$, they are both keeping track of they same info. My guess is that is better notation and that it will be handy if you move on in probability, I only took a first course in it, so I'm a newbie. $\endgroup$ Mar 22, 2012 at 6:34
  • $\begingroup$ The events were supposed to be disjoint, so that there is equality. I fixed it now. $\endgroup$
    – MyCatsHat
    Mar 22, 2012 at 7:07
  • $\begingroup$ Your suggestion breaks down as soon as one consider ones result with uncountably many possible values and/or as soon as one needs to combine several random results. $\endgroup$
    – Did
    Mar 22, 2012 at 7:25
  • $\begingroup$ @DidierPiau Could you please elaborate why/were it breaks down (especially in the case of combining rd's since I don't know yet anything about results with uncountably many possible values) ? $\endgroup$
    – MyCatsHat
    Mar 22, 2012 at 7:32
  • 3
    $\begingroup$ About combining random variables: try to show that E(2X+5Y)=2E(X)+5E(Y), using only events. This is doable, but awfully cumbersome. $\endgroup$
    – Did
    Mar 22, 2012 at 9:07

3 Answers 3


In short, the abstract concepts are for blurring out the details, i.e. the abstract concept of random variables is useful because it allows us to work at higher level than ordinary atoms.

To give some more motivation:

  • You can have more than one random variable on single probability space, but doing it with atoms you would need to care for every possible combination.
  • The notion of conditional expected value $\mathbb{E}(X | Y)$ is very useful, but cumbersome to define in terms of atoms.
  • You can nicely characterize the random variable in terms of characteristic function.

Finally, there's one more thing to add: humans have a very strong intuition about randomness, but in mathematics there is no randomness at all. All you have is just functions (I will skip other entities for simplicity), and every time you apply the same arguments to a function you get the same result. I will repeat: in mathematics there is no randomness at all.

How one deal with that? By introducing something like "the state of the world" or what we often denote by $\Omega$, the universe. So if the world happens to be in some state $\omega_1 \in \Omega$, then the outcome of every experiment is precisely determined, we can predict perfectly every outcome of every action (tossing a coin or whatever it is). However, we don't know in which state the world is, and this way we hacked the randomness into math. And this interpretation gives natural rise to the definition of random variables -- they are only random, because we do not know in which state of the world we live in.

To conclude, it does not matter whether $\Omega$ is finite or not, it does not matter if we prefer to deal with raw sets and atoms, or theorems about abstract concepts, the notion of a random variable is very useful, and I would even say that is the intuitive connection between pure mathematics and the imperfect (and therefore beautiful) world we live in.

Edit: To answer one of the comments of OP (to long for another comment), random variables aren't theoretically necessary, as we could inline their definitions into our proofs, etc., but in practice any such proof would be unreadable and unmanageable, so random variables are necessary. Consider real numbers: one could always work with sequences of rational numbers or continued fractions, or whatever, so the question is, are they essential? I think yes, and there are many similar examples.

  • 5
    $\begingroup$ Indeed. Why do we need $2,3,4,\ldots$ when we could use $1+1,1+1+1,1+1+1+1,\ldots$? $\endgroup$ Mar 22, 2012 at 16:17
  • $\begingroup$ @RobertIsrael In fact logicans do that, i.e. $0 = Zero$, $1 = Succ(Zero)$, $2 = Succ(Succ(Zero))$ and so on, but we know they can use abstraction (at least the logicans I know) ;-) $\endgroup$
    – dtldarek
    Mar 22, 2012 at 16:21
  • $\begingroup$ Excellent answer, @dtldarek! $\endgroup$ Sep 23, 2016 at 23:27
  • $\begingroup$ Cantor rolls in his grave whenever you claim the continuum hypothesis is stated as: $$S(S(0))^{\aleph_0}=\aleph_{S(0)}$$ $\endgroup$
    – Asaf Karagila
    Oct 22, 2016 at 11:02

If everything was about computing probabilities of events $A_k$, you might be able to get away with avoiding random variables. However, there's a lot more to probability than that. When, for example, you want to calculate means and variances, or talk about the relations between many "parametrized families", it's very useful to have random variables.

  • 4
    $\begingroup$ But is it just useful or is it necessary to use rd's ? While I agree, that are a useful formal tool, which enables writing various probabilities in a concise way, I'm unsure about their necessity. $\endgroup$
    – MyCatsHat
    Mar 22, 2012 at 15:03
  • 1
    $\begingroup$ @user26698 To long for a comment, so I posted an answer, see the edit. $\endgroup$
    – dtldarek
    Mar 22, 2012 at 15:40

In principle, you can get away with joint distribution laws instead of random variables but the things get extremely messy and formulations ridiculously tangled (try to reformulate the strong law of large numbers this way and you'll see what I mean). In addition, you lose all intuition coming from classical analysis and the Lebesgue integration theory. Even such a simple operation as the truncation of a random variable at some level will become the projection of the distribution measure, etc. So, why to have a headache expressing everything in a fancy way when a nice and powerful language is available?

Of course, the Kolmogorov formalism may lead to some confusion too. For instance, if you have a random variable on $\Omega$, and want to introduce an independent copy (a standard trick in many proofs), it may be just impossible to do it on the same $\Omega$, so, technically speaking, a random variable is not quite the same as a function on a measure space; you should rather say that it can be realized as a function on a measure space and to check that the properties you define do not depend on the realization and are determined by the (joint) distribution only. Still, after proving it three times or so, it becomes fairly obvious what is legitimate and what is not and you can safely forget about fine issues like this one (There are more of them; the next one comes when you need conditional probabilities in the continuous setting, which requires the underlying $\Omega$ to be a Polish space, etc.).


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