Let $(\Omega, \mathcal{F}, P)$ be a probability space and $X_1(\omega), \dots, X_n(\omega)$ be random variables defined on the space. Suppose we are concerned with the joint behavior of the variables. Then one usually considers the variables as a single multivariate random variable defined as a vector of (one-dimensional) random variables: $\mathbf{X}(\omega) = (X_1(\omega), \dots, X_n(\omega))^T$.

My first set of questions is: Is it necessary to pack random variables into a random vector to have a legitimate (mathematically correct) discussion about their joint distribution? Can one consider them as just a set $\{ X_1(\omega), \dots, X_n(\omega) \}$?

A vector induces a particular order on its elements while a set does not. It is quite common to denote, for example, the joint PDF as $f_{X_1, \dots, X_n}(x_1, \dots, x_n)$ where, again, a particular order is implied even though the vector notation is not used explicitly. So, is it all about ordering?

My second set of questions is: Is it correct to denote a set of random variables as $\text{(a-regular-weight-letter-goes-here)}(\omega)$, for example, $X(\omega) = \{ X_1(\omega), \dots, X_2(\omega) \}$? (The question is mainly concerned with the use of $\omega$ next to a set of random variables.) Then is it acceptable to denote, for instance, the corresponding PDF by $f_X(x)$ meaning that $x = \{ x_1, \dots, x_n \}$? Since the order is not specified, it is presumably wrong.

Thank you.

Regards, Ivan

  • $\begingroup$ In my experience, I have not heard from theoretical probabilists much use of the phrases "pdf" or "cdf" or the like, but rather "density/density function" or "distribution/distribution function" respectively. (Distribution also sometimes refers to the stieltjes measure) On your first question, order matters. Often times people talk about invariance of the joint distribution under permutation. This would be lost without keeping track of the ordering. $\endgroup$
    – Jeff
    Commented Dec 21, 2013 at 8:05
  • $\begingroup$ Order matters. If the $X_k$ are measurements that are, for example, real numbers, then you need to track which number is which, not just the set of numbers. $\endgroup$
    – copper.hat
    Commented Dec 21, 2013 at 8:35
  • $\begingroup$ @copper.hat, I guess one way to get away with using sets could be to equip them with totally ordered index sets, as it is done for random processes. For instance, $\{ X_i \}_{i \in I}$, $I \subset \mathbf{N}$. $\endgroup$
    – Ivan
    Commented Dec 21, 2013 at 9:13
  • $\begingroup$ @Jeff, regarding the first part of your comment, you wanted to draw my attention to the fact that the terminology I am using is not very welcomed, and it is better to avoid it? Such abbreviations are not common? Thank you. $\endgroup$
    – Ivan
    Commented Dec 21, 2013 at 9:30
  • $\begingroup$ @Ivan your terminology has been in my experience very common among some of the applied probability classes, but for some reason probabilists just never seem to say it. Just my personal observation. I cannot say anything more than that about what the "actual convention" is. $\endgroup$
    – Jeff
    Commented Dec 21, 2013 at 15:56

1 Answer 1


To be concrete, assume every $X_i$ is a real valued random variable. Then a first observation is that you confuse $X_1(\omega)$, say, which is a number, and $X_1$, which is a function. For example, the random vector you consider is $X=(X_1,\ldots,X_n)$ (with or without transpose), which is a function from $\Omega$ to $\mathbb R^n$, certainly not $X(\omega)=(X_1(\omega),\ldots,X_n(\omega))$, which, for each $\omega$ in $\Omega$, is a point in $\mathbb R^n$.

Now to the main part of your question. Consider some real numbers $x_k$. There is definitely a difference between $x=(x_1,\ldots,x_n)$, which is a point in $\mathbb R^n$, and $u=\{x_1,\ldots,x_n\}$, which is a subset (of size at most $n$) of $\mathbb R$. The former determines the latter but the latter carries strictly less information than the former.

For example, several points $(x_1,x_2,x_3)$ in $\mathbb R^3$ correspond to $u=\{3,17,42\}$ and several points $(x_1,x_2,x_3)$ in $\mathbb R^3$ correspond to $u=\{17,42\}$ (but only one point corresponds to $u=\{42\}$). The same applies to random variables.

To sum up, the object quasi universally considered in such a setting is the vector $X=(X_1,\ldots,X_n)$ (and its distribution on $\mathbb R^n$, say). One can translate these, with some loss of information, in terms of the random set $U=\{X_1,\ldots,X_n\}$ (and in terms of its distribution on the space of nonempty finite subsets of $\mathbb R$, endowed with its natural sigma-algebra) but this is almost never done a priori.

  • $\begingroup$ Thank you, Did. I appreciate very much your first observation as well. $\endgroup$
    – Ivan
    Commented Dec 22, 2013 at 11:41
  • 1
    $\begingroup$ Regarding "infinitely many $(x_1, x_2, x_3)$ correspond to $u = \{3, 17, 42\}$," you meant six, i.e., the number of all permutations of three objects? $\endgroup$
    – Ivan
    Commented Dec 22, 2013 at 11:47
  • $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$
    – Did
    Commented Dec 22, 2013 at 14:43

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