8
$\begingroup$

In the comments on my answer to a recent question about the sum of random variables, I came across a link to the Wikipedia article on the ratio distribution, and noticed the following peculiar claim there:

The algebraic rules known with ordinary numbers do not apply for the algebra of random variables. For example, if a product is $C = AB$ and a ratio is $D=C/A$ it does not necessarily mean that the distributions of $D$ and $B$ are the same.

This claim has been in the article since 2007. It was added there by the same seemingly reputable editor who originally created the article and contributed much of its original and current content, and it is seemingly cited to the book The Algebra of Random Variables by Melvin D. Springer, published in 1979 (although it's not 100% clear whether or not the citation marker that appears later in the same paragraph is actually meant to cover this claim as well).


Obviously, the claim seems like nonsense to me. I could just edit it out of the Wikipedia article, but given that it has stood unchallenged there for over 10 years, I figured I should make sure I'm not the one who's mistaken here. Not having Springer's book at hand to check the (possible) citation, I figured I'd ask the experts here for help. In particular, since the claim as stated really consists of two parts, so does my question:

Part 1: Do random variables follow the same algebraic rules as ordinary numbers, or do they (in some sense) not? If they do not, how do the rules differ? Does it depend on what (generally accepted) formalism one adopts?

Part 2: It is clear that, even for ordinary numbers, $D = \frac{AB}{A}$ is not always equal to $B$, since $D$ is not even defined when $A = 0$. Is this trivial difference the only way in which $D$ and $B$ can fail to be equal, even when they are random variables? In particular, does the following statement always hold for (real- or complex-valued) random variables: $$A \ne 0 \implies \frac{AB}{A} = B.$$

Part 3 (bonus): What does Springer's book actually say about this, and is there anything in there that could in some sense be taken to support the claim quoted above? Is it, as I would presume, actually regarded as a reliable source for claims about mainstream mathematics and statistics?

$\endgroup$
  • 1
    $\begingroup$ A math comment on Part 2: It is always the case that $ab/b = a$ when it is defined, i.e. not the equation is the problem but the mere definition! In that sense, assume that $B$ is a RV such that $B(\omega) \neq 0$ for all $\omega$. Then $AB/B$ and $B$ have the same distribution simply because they are the same random variable. A random variable is simply a function from some probability space $\Omega$ into whatever set (the reals if you want to do this kind of algebra in a natural setting with it)... Also: What exactly do you mean by $A \neq 0$? for all $\omega$? Just for some?... $\endgroup$ – Fabian Werner Mar 7 '18 at 15:58
  • 3
    $\begingroup$ (+1) The language in that Wikipedia article is poor, but its intent is clear: it means to discuss the algebra of distributions, not random variables per se. If you choose to edit it, then consider clarifying the language without changing the idea it is trying to convey. $\endgroup$ – whuber Mar 7 '18 at 16:03
  • 1
    $\begingroup$ @FabianWerner: I'm using the (AFAIK fairly standard) convention that we may omit the $(\omega)$ when writing the r.v. $A(\omega)$, but of course that may not be what the Wikipedia article is doing. You may be onto something there. $\endgroup$ – Ilmari Karonen Mar 7 '18 at 16:04
  • 2
    $\begingroup$ @Carl: Why do you think they would be vectors of any kind? Division by a vector is generally not defined anyway, so for the expression to make any sense, $A$ pretty much has to be a scalar (or more precisely a scalar-valued function of the probability space, if you want to strictly follow the standard formalism as noted by Fabian Werner above). I suppose $B$ could be a vector, although I see no particular reason to assume that it is. $\endgroup$ – Ilmari Karonen Mar 7 '18 at 19:47
  • 2
    $\begingroup$ @Carl: Uh, no. At least not unless you're working in a three-element probability space (usually, the elements or even the size of the probability space $\Omega$ are not explicitly specified, since it's really just a formal tool for working with variables whose values are uncertain) and are using a funny notation for functions over that space. $\endgroup$ – Ilmari Karonen Mar 7 '18 at 20:19
0
$\begingroup$

The algebra of random variables (ARV) is an extension of the usual algebra of numbers "high school algebra". This must be so because numbers can be embedded in the ARV as rv equal to a constant with probability 1. So there cannot be any inconsistency, but it could well be new properties which doesn't say anything about numbers. In the ARV equality is equality in distribution, so it is really an algebra of distributions. But for rv's constant with probability 1, this is an extension of equality of numbers in the usual sense.

About the given example from Wikipedia, there is no inconsistency there, only a (maybe for someone) surprising possibility that arises because there are many random variables such that $X$ and $X^{-1}$ have the same distribution, while there are only two numbers with this property, $-1$ and 1. The Cauchy distribution have this property, see What can we say about distributions of random variables $X$ such that $X$ and its inverse $1/X$ have the same distribution?.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.