1
$\begingroup$

I am going through Introduction to Probability by Grinstead and Snell and apparently the book's definition of random variable is not conventional as discovered in another question I had here and I would like to understand their differences.

The book describes random variables and its distribution function as (page 27 of the PDF):

Let X be a random variable which denotes the value of the outcome of a certain experiment, and assume that this experiment has only finitely many possible outcomes. Let $\Omega$ be the sample space of the experiment (i.e., the set of all possible values of X, or equivalently, the set of all possible outcomes of the experiment.) A distribution function for X is a real-valued function m whose domain is $\Omega$ and which satisfies: $m(\omega) \geq 0$, for all $\omega \in\Omega$ and $\sum_{\omega \in \Omega } m(\omega)=1$

This seems to not line up with the Wikipedia definition here that describes random variables as a function $X:\Omega -> E$ where $E$ is usually $R$.

  1. In the Wiki definition, it is correct to say that $X$ is not an outcome in $\Omega$ but instead maps $\omega$ to, usually, a real number? And is the probability mass function $p_X(x)=P(X = x)$ equivalent to my book's distribution function $m(\omega)$, except the domain of the probability mass function is all possible values of $X$, not $\Omega$?

  2. Using the Wiki definition, is the event (say $A$) described by $X = j$ the set $A= \{ \omega_1, \omega_2, .. \} $ for all $\omega \in\Omega$ where $X = j$, such that $P(X = j) = P(A)$ (and $P(X = j | E) = P(A |E)$ if dealing with conditional probability conditioned on event $E$)?

  3. In the continuous case, is there any practical difference between the two definitions? It seems if $X$ is viewed as a function, then $X(x)=x\in\Omega$ so would be equivalent as taking on an outcome from a continuous sample space?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$
  1. Yes, it is correct to say that $X$ maps $\omega$ to, usually, a real number. Grinstead and Snell's $m$ is actually the probability measure of the probability space. (Minor quibble: Wikipedia misuses the verb "equip." It is $\Omega$ that is equipped with $P$, not the other way around.)
  2. Yes, $X=j$ denotes the event $\{\, \omega \in \Omega \mid X(\omega) = j \,\}$.
  3. In the continuous case, Grinstead and Snell abandon the idea that $X$ gives the outcome of the experiment. They regard $\Omega$ as a subset of $\mathbb{R}^n$, while $X$ takes values in $\mathbb{R}$. While I cannot speak for the authors, I suspect these choices were primarily driven by pedagogical expediency.
$\endgroup$
5
  • $\begingroup$ Is the pmf of $X(\omega)$ functionally the same as $m$ in that both denote the probability of $X$, except that the domain of the two are different ($\mathbb{R}$ vs. $\Omega$)? Also, is the Wikipedia definition the preferred way of looking at random variables? I'm curious what the advantages of Grinstead and Snell's definition is, especially since starting from Chapter 4 they also start using the random variable as a function of ω. $\endgroup$
    – Yandle
    Commented May 15, 2019 at 4:12
  • $\begingroup$ The Wikipedia approach is the one taken in more advanced treatments of probability. Most introductory treatments, including most undergraduate probability textbooks and virtually all undergraduate discrete math textbooks, however, depart in a number of ways from the standard treatment to achieve significant simplifications. Of course, at some point, students of probability realize that there's more to it than they've been told so far, as they move to more general definitions, but this is not uncommon in math and other disciplines, and is often worth doing. $\endgroup$
    – Fabio Somenzi
    Commented May 15, 2019 at 5:23
  • $\begingroup$ Concerning the pmf vs. $m$ issue, note that the pmf of $X$ is defined on the range of $X$, which for a discrete random variable is not $\mathbb{R}$ by definition of discrete rv. $\endgroup$
    – Fabio Somenzi
    Commented May 15, 2019 at 5:30
  • $\begingroup$ Had another question that just came to me, for the continuous case (which I think also works for the discrete case?), per the Wiki definition, is the proper notation for the event where $X_1(\omega)<x_1,X_2(\omega) <x_2\dots,X_n(\omega)<x_n$ written as a set of n-tuple $\{\omega\in\Omega \space|\space (X_1(\omega)<x_1,X_2(\omega) <x_2\dots,X_n(\omega)<x_n)\}$? Each $X$ would still be a function of an outcome $\omega$ whose domain is now a subset of $R^n$? $\endgroup$
    – Yandle
    Commented May 21, 2019 at 0:18
  • $\begingroup$ Notation varies with authors. The one you used is rather common, but it helps understanding, in my opinion, to explicitly write $\{\omega \in \Omega \mid X_1(\omega) < x_1 \wedge X_2(\omega) < x_2 \wedge \cdots \wedge X_n(\omega) < x_n \}$. Each random variable $X_i$ is a function of the outcome $\omega$, which ranges over $\Omega$, whatever that is. That is, $X_i : \Omega \to \mathbb{R}$. $\endgroup$
    – Fabio Somenzi
    Commented May 21, 2019 at 3:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .