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A random variable: is it a function or an equivalence class of functions?

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closed as too localized by Pete L. Clark, Loop Space, Qiaochu Yuan, Anton Geraschenko Jan 22 '10 at 2:42

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Could this not quickly be answered by Google? It is customary to be sloppy in talking about "functions" on a probability space (or more general measure space) when one really identifies functions that agree except on a set of measure zero. –  Jonas Meyer Jan 21 '10 at 9:04
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Commment: I voted to close this question. To explain: it's essentially asking about a definition, hence is a very straightforward question. That's fine with me. However, I am trying out the philosophy that when a straightforward question gets an adequate answer, we close it to encourage people to spend their energy answering the tougher / more interesting questions. Comments welcome of course. –  Pete L. Clark Jan 21 '10 at 9:47
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I'm voting to close as well, but for a different reason to Pete! Particularly in light of gowers' answer, I think that the question is too vague. The question does not give any context for what would constitute a good answer, and thus I worry that gowers' excellent answer might go completely over the head of the questioner - thus being a bit of a waste of gowers' time. If the question is edited to make this more precise, I will rescind my vote! (Unfortunately, 'too vague' isn't one of the options for closing) –  Loop Space Jan 21 '10 at 11:19
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@Andrew: Your reason is also sensible. –  Pete L. Clark Jan 21 '10 at 11:21
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@Andrew, even though Gowers' answer may be over the head of the questioner, there are still many of us who would benefit from it. Had you closed the thread three hours ago when you made that comment, it may have fallen off the main page and many of us (including me) would never have seen Gowers' response. In that case, you really would have made a waste of his time! My opinion is that on MathOverflow, garbage in doesn't automatically mean garbage out. Even an awfully-worded question should stay open when its answers are worth reading. –  Tom LaGatta Jan 21 '10 at 15:19

2 Answers 2

I think this question may be slightly deeper than some people are giving it credit for being. I lectured a course in probability to first-year undergraduates at Cambridge recently, and a previous lecturer, who was a genuine probabilist, was very keen to impress on me the importance of talking "correctly" about random variables. It took me a while to understand what he meant, but basically his concern was that the notion of a sample space should be very much in the background. It's tempting to define a random variable as a function on a probability measure space (not that this particular course used measure theory -- but some more elementary substitute for the definition would have been needed), but his view was that this was absolutely not how probabilists think about random variables.

The practical point was not so much to come up with a better formal definition of random variables, but rather to try, whenever possible, to prove results about random variables without referring to the sample space. It's surprising how little you need to mention them (or not surprising if you're a real probabilist). I seem to remember that the one place where I found I really wanted sample spaces was when it came to proving linearity of expectations.

The compromise I reached in that particular course was to define random variables using sample spaces (which makes them seem fairly straightforward objects) but then to tell people to prove as much as they could just with reference to the distribution of the random variable itself. In other words, I gave the "wrong" definition and immediately admitted that it was wrong.

Added very slightly later: I see that I am interpreting the question differently from everyone else. I am not talking about two functions on a measure space being equivalent if they agree outside a set of measure zero -- which is indeed not a very interesting issue. I am talking about two functions on different measure spaces being the same random variable if you can find a nice map between the measure spaces such that one function is (up to a set of measure zero) the obvious transformation of the other. One of the big advantages of not specifying a sample space is when you start talking about several random variables. For instance, if you start by discussing the tossing of a coin, it's sort of clear that your sample space is $\{0,1\}$, but if another coin enters the picture does that mean you have to go back and prove everything for a more complicated function defined on $\{0,1\}^2$? Not if you didn't mention the sample space in the first place but just the random variable.

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@gowers: When I read the question, the point that probabilists like to think of two random variables as being "essentially the same" if they have the same distribution functions flickered across my mind. (I know just enough probability to have seen that in texts.) But I don't think they go as far as saying "A random variable is an equivalence class..." or even "Two random variables are equivalent if..." So IMO the less interesting equivalence relation -- i.e. almost everywhere equality -- is probably what is meant. (This may be an example of a too vague question wasting people's time.) –  Pete L. Clark Jan 21 '10 at 11:11
    
This notion of sameness is a morphism, not an equality. If you can flip two fair coins, you may want to say there is a symmetry switching them, but not that they are equal just because there is an isomorphism from an ideal fair coin to either. I think we really want the whole probability space to be replaced by an equivalence class of spaces with some subsets describable. –  Douglas Zare Jan 21 '10 at 14:41
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Relevant: terrytao.wordpress.com/2010/01/01/… –  Qiaochu Yuan Jan 21 '10 at 14:47
    
@Clark: [I don't think they go as far as saying ... "Two random variables are equivalent if..."] They do say, "Two random variables are identically distributed if...", or things like that. –  Gerald Edgar Jan 21 '10 at 14:54
    
@gowers: I think, you are underestimating the probabilists. They do think about r.v.'s as functions on prob. spaces. However, often it is just distributional properties that matter and the probabilist naturally and smoothly switches to the distributional point of view. It's true that the space of realizations of a stochastic process is often a natural choice of the prob. space, but there are numerous examples in modern probability of other clever choices of probability spaces. E.g.: Skorokhod's method of one probability space, or the notion of a weak solution of a stochastic equation. –  Yuri Bakhtin Jan 21 '10 at 15:51

Hi,

"stricto sensu" it is a measurable function with respect to a $\sigma$-field (notice that no probability measure is really needed here)

Now as you always work with a probability measure in this context, it is almost always implicit that properties of random variables are defined up to the class defined by the equivalence relation "equalility with probablility 1".

This might explain the ambiguity that you perceive in the definition of a random variable.

Regards

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