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The usual definition of a random variable (or random element) is that of a measurable function $X : (\Omega, \mathcal{F}, P) \rightarrow (\Omega', \mathcal{F}')$. Now I am not aware of any property/theorem that depends on the specific values of $X$ for every $\omega \in \Omega$. In particular any other $P$-almost surely equal random variable $X'$ is generally considered as equivalent to $X$ for all practical purposes.

So is there a good reason not to define random variables as equivalent classes rather than laboriously precising each time that such or such statement is true almost surely, that such or such sequence converges almost surely, that such or such object is unique almost surely, etc ? As a comparison defining $L^p$ spaces as spaces of equivalent classes of almost everywhere equal functions helps a lot in simplifying the phrasing of the theory.

So are there some interesting/complex cases where we would really need to keep the distinction between almost surely equal random variables?

Edit

In agreement with @Pedro Tamaroff's comment I'm removing the last addendum to this question and opening a new one.

Contactomorph
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    First examples coming to mind concern the almost sure properties of realizations of random processes indexed by uncountable sets, say the almost sure Hölder continuity of the paths of Brownian motion $(B_t)$. If one allows to modify each random variable $B_t$ on a null set, the resulting path $t\mapsto B_t(\omega)$ may become ugly on an event of positive probability. – Did Nov 21 '16 at 23:49
  • Indeed, if $B_t$ were defined as a one-parameter family of equivalence classes, then a question like "is $t \mapsto B_t(\omega)$ continuous for almost every $\omega$" becomes ill defined, since the answer may depend on the choice of representatives. This is the idea behind a modification of a stochastic process. – Nate Eldredge Nov 22 '16 at 00:01
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    @Did: I think this should be an answer. My knowledge of post-undergrad probability is nil, but that is a very convincing example (unlike the only answer so far, which I don't even understand). – tomasz Nov 22 '16 at 00:05
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    OP: So, in the end, you modified substantially the question after you received answers and after you have been signalled such modifications are to be avoided because they have the undesirable effect of making previous answers off-topic. A reversal to the previous version of the question seems desirable. – Did Dec 05 '16 at 12:44
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    @Burakumin Please, do not modify a post after answers have been provided with edits that render such answers incomplete, incoherent or altogether incorrect. Ask a new question instead. Regards, – Pedro Dec 05 '16 at 17:11
  • @Did Your help is appreciated, but please stick to a civil tone. Regards, – Pedro Dec 05 '16 at 17:12
  • @Pedro Any specific example of uncivil tone? – Did Dec 05 '16 at 17:51
  • @Did It seems at least unkind to accuse someone of plagiarism when trying to better their post by looking at answers given. – Pedro Dec 05 '16 at 17:57
  • @Pedro I suggest to check a precise definition of plagiarism. If this is all that you find uncivil in my comments, I stand by them. For the record, the presentation you just gave ("trying to better one's text", one also finds "being inspired by", even "giving an hommage", conscious or unconscious...) is one of the classic ways that plagiarists use to justify their acts. The irony. – Did Dec 05 '16 at 18:06
  • @Did I believe you shouldn't be taking this matter too seriously. I think the OP has no reason whatsoever to plagiarise any of what you wrote. It seems counterproductive to find conflict where there probably is none. – Pedro Dec 05 '16 at 18:08
  • @Pedro OK, they did but they had no reason to so it does not matter? Right. Other examples of uncivility? – Did Dec 05 '16 at 18:12
  • @Did I have nothing more to add. It just preoccupies me that you should quarrel so often with users. Regards, – Pedro Dec 05 '16 at 18:15
  • @Pedro "To quarrel" (def.): 1. To mention that the comments to one's own answer are unclear. 2. To suggest to read the literature instead of reinventing the wheel. 3. To explain that some kind of exchanges explicitely wished for by the OP are in fact not suited to the present forum. 4. To be explained that one's contribution is "not that relevant". 5. To signal that modifying significantly one's question after one received answers is not kosher because it nullifies said answers. 6. To realize with delight that the OP nevertheless significantly modifies their question and that, to do so, ... – Did Dec 05 '16 at 18:36
  • ... they, in effect, plagiarize one's own contribution. Enough? (If you think that being fair is being symmetrical, please reconsider.) – Did Dec 05 '16 at 18:36

2 Answers2

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One of the first classes of examples coming to mind where this matters concerns the almost sure properties of realizations of random processes indexed by uncountable sets, say the almost sure Hölder continuity of the paths of Brownian motion $(B_t)$. If one allows to modify each random variable $B_t$ on a null set, the resulting paths $t\mapsto B_t(\omega)$ may become ugly for every $\omega$ in an event of positive probability.

Edit: Regarding "ugly" above, user @tomasz mentioned a useful point in a comment below, which I now reproduce: if one allows to modify each random variable on a null set, the supremum of an arbitrary (uncountable) family of measurable functions need not be measurable, not even if the functions are almost everywhere zero (say, indicators of points).

Did
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  • What "ugly" means ? Wouldn't we have something like: « For almost every $\omega$, $t \mapsto B_t(\omega)$ is almost everywhere continuous. » ? If this is the case that does not sound as a big loss. – Contactomorph Nov 22 '16 at 09:09
  • This is indeed "a big loss" if one insists that the paths are continuous, almost surely. And one often does. (Did you vote to delete this, by any chance?) – Did Nov 22 '16 at 10:28
  • I think a big issue here is that the supremum of an arbitrary (uncountable) family of measurable functions need not be measurable, not even if the functions are almost everywhere zero (say, indicators of points). – tomasz Nov 23 '16 at 11:13
  • @tomasz Yes. Point added to my answer (if you disagree, just yell). :-) – Did Nov 23 '16 at 11:23
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    @Burakumin Why are you staying silent? You were asked at least one precise question, if only for good manners you could see fit to address it... – Did Nov 23 '16 at 11:24
  • @did: I didn't vote to delete anything. I just mishandled the addition of my comment (from my phone) and accidentally replaced your answer. Then I had to find how to revert it to the original post. Sorry for the mess! – Contactomorph Nov 24 '16 at 10:53
  • @Did Rethinking about it I still does not feel convinced by the Brownian motion example. Unless I misunderstand, you think of the Brownian motion as an assignement of a random variable to every $t$. But it seems the valid extension would here be to consider the dual description: an assignement of a function to every $\omega$. An equivalence class now contains $B_1$ and $B_2$ if they share all their trajectories except a neglibible subset of them. Those trajectories themselves can still be required to be continuous (except again a neglible subset of them). – Contactomorph Nov 24 '16 at 18:18
  • Not sure I am following your point but you can be assured that indeed I, as everybody else interested in the subject, very much considers simultaneously Brownian motion as a collection of functions $t\mapsto B_t(\omega)$ and as a collection of random variables $\omega\mapsto B_t(\omega)$. (If I may add, everything I indicated on the present page is arch classical, there is not a whiff of originality or personal point of view from my part in it.) // What are $B_1$ and $B_2$ in your comment supposed to be exactly? – Did Nov 24 '16 at 18:33
  • Sure both descriptions are possible. My point is this one : if one now identifies random elements that are a. s. equal (let's call here the resulting equivalent class a reduced random element) there are now two possible definitions of a reduced stochatic process that are not equivalent anymore : either an assignement of a reduced real random variable to each $t$ or an equivalent class of stochatic processes that share almost all their trajectories. $B_1$ and $B_2$ are two standard brownian motions that belong to the same reduced brownian motion, following the latter definition. – Contactomorph Nov 24 '16 at 19:50
  • So in case that was not clear, it seems to me that you and @tomaz's point about changing random variables $B_t$ on a null set only exists because you're using the former definition. But in the end it would make more sense to choose the latter, and according to this one, changing $t$-wise values of the (reduced) brownian motion is not a valid operation. – Contactomorph Nov 25 '16 at 10:20
  • Your reading of our contribution seems to be rather orthogonal to what we actually wrote. I say "seems" because I am not sure what your point actually is, which would not be the exact point we stressed ourselves, only formulated as if we had missed it and in your own, non canonical, lingo. Unless you move to more constructive exchanges, I am sorry but I feel I cannot bring you much more than what is already on the page... Maybe reading some good sources on processes in continuous time could help? – Did Nov 25 '16 at 11:43
  • First I don't appreciate very much the fact of being implicitely accused of steering the discussion around "non-constructive exchanges". My objections may be incorrect, but they are certainly sincere. I do think that my remarks are not orthogonal to what you and @tomasz wrote and that your counter-example is not that relevant. Now I may have misunderstood it. Or maybe you may have misunderstood my point. So if you are interested in "constructive exchanges" I propose we move to a math/scientific forum (whichever you want) as I believe StackExchange is not an appropriate tool for discussions. – Contactomorph Nov 30 '16 at 20:39
  • Your message has disappeared once again. Does that mean you are not interested in clarifying the misunderstanding and/or the conversation? – Contactomorph Dec 02 '16 at 09:36
  • No, that means that my comment, rather mysteriously, was flagged and deleted (apparently, simply mentioning that the kind of exchange you seem to wish for is not suited to the present forum, is seen as offensive by some users...). Re what I am interested in, please refer to my previous comment. I can only reiterate that my post fully answers your question as it is stated and that your insistence that it does not might be due to some preconceived notions you wish to stick to at all cost. From this point, sorry if I do not think I should feel overly concerned... – Did Dec 02 '16 at 09:49
  • I still believe that discussing what may be confusing in the phrasing of my question or wrong with my objections could have helped a lot. Now I plan to edit my question, hoping other people might react. – Contactomorph Dec 02 '16 at 17:52
  • The phrasing of your question is not confusing and the content of your question as it is written has been addressed. On the other hand, your objections to the answers you receive are slightly mysterious, probably because you do not really listen to what is said to you. In this context, that you complain afterwards for a lack of discussion is slightly paradoxical. Anyway, best of luck with the hundreds of other users who will, no doubt, fully understand what you really mean (but excuse me if I leave your future comments unanswered, several of them are already quite offtopic). – Did Dec 02 '16 at 19:41
  • ((Oh, and modifying a question after it received answers is a big no-no on the site, for obvious reasons. That is, if ever you care.)) – Did Dec 02 '16 at 19:42
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Although every finite moment of a variate $X$ will be equal to the corresponding moment of the two "equivalent" variate $X^\star$, one might consider the range of a distribution as an interesting property. If you consider the range to be interesting, then consider the following two distributions, both derived from an underlying uniform random $U$ on $[0,1]$: $$ X: \begin{array}{lc} X = & \left\{ \begin{array}{cl} U & U \mbox{ irrational}\\ -U & U \mbox{ rational} \end{array}\right.\\ X^\star = &U \end{array} $$ The variate $X$ is almost surely equal to $X^\star$ but the range of $X$ is $[-1,1)$ whilst the range of $X^\star$ is $(0,1)$.

Mark Fischler
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    I don't know if that's a compelling example; to many it would just suggest that range is not in fact interesting, and one should instead look at essential range, which is well-defined with respect to a.s. equality, or even equality of distribution. In this case both random variables have $[0,1]$ as their essential range. – Nate Eldredge Nov 22 '16 at 00:04