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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?

Ilmari Karonen
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    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?... – Fabian Werner Mar 07 '18 at 15:58
  • In that sense I think that in the wikipedi article as well the definition is the key point and not the actual equation, i.e. they may define $AB/A$ in a weird way or allow random variables to take the value $\infty$ or so... – Fabian Werner Mar 07 '18 at 16:01
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    (+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. – whuber Mar 07 '18 at 16:03
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    @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. – Ilmari Karonen Mar 07 '18 at 16:04
  • Well, if you chose to do so then what *is* the random variable $AB/A$? You need to define a function here and the natural definition is $(AB/A)(\omega) = A(\omega)B(\omega)/A(\omega)$ and this definition only makes sense for those $\omega$ s.t. $A(\omega) \neq 0$, hence, the $\omega$ is actually very important here. As stated above: what does $A\neq 0$ actually mean? – Fabian Werner Mar 07 '18 at 16:06
  • @whuber: I suspected something like that might be the case, but even then, I'd be curious to know how much and in what ways the rules of this algebra of distributions differ from ordinary algebra on real/complex numbers, and in particular how exactly one would describe the difference between $B$ and $D = (AB)/A$ in that algebra. Maybe I just need to see if I can find Springer's book in a library and read it. – Ilmari Karonen Mar 07 '18 at 16:08
  • @FabianWerner: Writing random variables as explicit functions of $\omega$, I suppose my statement would become $A(\omega) \ne 0 \implies \left( \frac{AB}{A} \right)(\omega) = B(\omega)$. Obviously this then depends on how one defines division of functions, but I believe any sensible definition that allows $\frac{AB}{A}$ to be defined at all should satisfy this statement. – Ilmari Karonen Mar 07 '18 at 16:16
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    One defines the division of functions as the divisions of their evaluations. The problem is that this is not always possible, namely when the evaluation of the denominator is zero... How to continue the definition in that case? The answer whether or not $AB/A$ and $B$ have the same distribution depends on that question and the article does not answer it. Hence, we can neither know nor not know whether or not $AB/A=A$ or not. Still: What do you mean by $A(\omega) \neq 0$? For all $\omega$? For all but a null set? For some? ... You need to decide. – Fabian Werner Mar 07 '18 at 16:20
  • @FabianWerner: The usual convention is that assertions with free variables are implicitly assumed to be assertions about all possible valuations of the variables. So $\forall A \in S^\Omega, B \in S^\Omega, \omega \in \Omega: \left( A(\omega) \ne 0 \implies \left( \frac{AB}{A} \right)(\omega) = B(\omega) \right)$, where $S$ is the domain (e.g. $S = \mathbb R$ or $S = \mathbb C$) the random variables are defined over. – Ilmari Karonen Mar 07 '18 at 16:27
  • ... But this is getting kind of sidetracked; my original question was more about whether the statement in my question could be reasonably said to hold (or fail) in the usual implicit shorthand language of random variables *without* these explicit qualifications, not about whether or not I *could* write more explicitly in a way that does make it hold. – Ilmari Karonen Mar 07 '18 at 16:29
  • Since ''whether or not the equation holds'' does indeed depend on these 'sidetracked' assumptions, the answer is clearly: neither yes nor no. If you ask an alien (or a little child that has never heard of mathematics) whether or not 1+1=2 then it will ask you: tell me what 1 and 2 mean and I can tell you whether that statement is true or not... Let us assume that for me 'usual convention' means that we allow $\infty$ as value and if $A(\omega) = 0$ then $*/A(\omega) = \infty$ then for $A=0$ we have that $AB/A$ is constantly $\infty$ and is entirely different from $A=0$... – Fabian Werner Mar 07 '18 at 16:35
  • What exactly are A and B? Please choose something: A and B are 1) single paired numbers, 2) || vectors of number pairs 3) $\bot $ vectors of number pairs 3) vectors of unpaired numbers of unequal length. – Carl Mar 07 '18 at 19:43
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    @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. – Ilmari Karonen Mar 07 '18 at 19:47
  • I am trying to nail down what is being done with numbers. Is this correct or not: Let $A=\{a,b,c\},\,B=\{d,e,f\}$. Then is $A/B=\{a/d,b/e,c/f\}$ and is $A+B$ then $\{a,b,c\}+\{d,e,f\}=\{a+d,b+e,c+f\}$ – Carl Mar 07 '18 at 20:13
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    @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. – Ilmari Karonen Mar 07 '18 at 20:19
  • Then I have no idea how to plot $A/B$. Note, if you make an association like $A/B$ there have to be numbers divided somehow somewhen or the division is undefined. I think this is being done as operations on paired lists, but I have not gotten a straight answer yet. – Carl Mar 07 '18 at 20:38

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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?.

kjetil b halvorsen
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