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Consider the basic setup for a real valued random varibel i.e $(\Omega,\mathcal{F},P)$ and some r.v $X$ defined on $\Omega$. We then define $P_{X}$ on $(\mathbb{R},\mathcal{B})$ by $P(\omega ; X(\omega) \in B)$ as the probabilty dist. of $X$. Why do we even have the space $(\Omega,\mathcal{F},P)$ why dont just consider the events in $(\mathbb{R},\mathcal{B})$ with probablities w.r.t $P_{X}$ directly instead?

Operations on r.v's should work fine in this spaces aswell, im stuck in some kind of "chicken and the egg"-loop.

Why do we make this move and define all these objects, which to me contain very similar info. Ultimalty one wants to control $X\in B$, I suppose.

user123124
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While just using the law for $X$ is a perfectly good probability space for a universe consisting of a single random variable, what about situations with two, three, four, countably infinite, uncountably infinite numbers of random variables?

In your situation with a single random variable, you can just take $(\Omega,\mathcal{F}, P) = (\mathbb{R},\mathcal{B},P_X)$ as your probability space. $P_X$ is just a specified probability measure on $(\mathbb{R},\mathcal{B}),$ so there's no circularity here (you can define it in terms of a distribution function if you wish). Similarly, if you have a universe with two random variables, you can take $(\Omega,\mathcal{F}, P) = (\mathbb{R^2},\mathcal{B(\mathbb{R}}^2),P_{X_1,X_2})$ for the appropriate probability measure $P_{X_1,X_2}.$ The pattern goes on through situations where there's an infinite number of RVs or even uncountably (like continuous time stochastic processes) though the functional analysis gets challenging. What doesn't change as the complication increases is that it can be described a probability space ... in the case of a continuous time like Brownian motion the natural one would look like $(\Omega,\mathcal{F}, P)=$ (all functions B(t), some sigma algebra, some measure on the space of functions). Since there are many different possible random systems you want to be able to describe, with probability spaces of varying complexity, it's easier to keep the notation abstract. However if you have a concrete situation with a natural set of random variables, you almost always think in terms of the natural probability spaces described above, perhaps in terms of distribution functions instead of measures (but those two are equivalent).

So why bother with the $P_X(B) = P(\omega | X(\omega)\in B)$ definition? Well, $X$ might not be the only thing in the universe and you might want to lift it out to consider it in isolation. (Of course, they are equivalent if $(\Omega,\mathcal{F}, P) = (\mathbb{R},\mathcal{B},P_X)$.) Or you might have chosen a different representation of the probability space than the natural one. For instance, maybe you take $$(\Omega,\mathcal{F}, P) = ([0,1], \mathcal{B}([0,1]), \mathrm{Lebesgue}).$$ This space can be used to model lots of stuff. You can model a random sequence of zeros and ones (whose "natural" outcome space is not $[0,1]$ but rather $\{0,1\}^{\mathbb{N}}$) by mapping $\omega \in [0,1]$ to its decimal expansion. Or you can model a Gaussian RV by mapping $\omega \in [0,1]$ to $X = \Phi^{-1}(\omega) \in \mathbb{R}$ with $\Phi$ the normal CDF (like in inverse transform sampling). You can even use some trickery to model a sequence of random variables with a given distribution. As you alluded to in your question, the two representations have the same information and are equivalent. Since there are many possible spaces to represent a given random model and it doesn't matter which you choose, it's natural to be abstract and write $(\Omega,\mathcal{F}, P).$

spaceisdarkgreen
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  • these can be regarded as "single" objects aswell? with thier own $P$'s on some Borel algebra. And I guess independece and conditionals could be formulated aswell. – user123124 Dec 30 '16 at 08:12
  • I think you're right that usually in practice one doesn't think about the abstract probability space but works directly with joint distributions of various sets of random variables. However, it's not always trivial that a set of infinitely many random variables obeying prescribed conditions exists, and for this the set theoretical mumbo jumbo can be useful. For instance, the existence of Brownian motion. – spaceisdarkgreen Dec 30 '16 at 08:24
  • I like this kind of mumbo jumbo! Im not very familiar with probabily theory which makes my set of examples limited. I just tried to figure out why we have this setup of mesureable spaces. Do you know some theorem or proof where this is essential? with reference – user123124 Dec 30 '16 at 08:27
  • There's a lot of ways that measure theory is valuable (in addition to putting everything on a rigorous footing). It's very flexible (with a probability space you aren't committed to looking at things from a set of prescribed random variables, other than the indicator functions for the events). The measure theory can be good for when intuition dries up. A good basic example that I would feel hard pressed to convince myself of with informal arguments is the Kolmogorov 0-1 law http://math.stackexchange.com/questions/1303735/questions-on-kolmogorov-zero-one-law-proof-in-williams . – spaceisdarkgreen Dec 30 '16 at 08:42
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    I guess there are two different questions here: 1. Why not make your probability space the joint law of a set of random variables? (answer: you are totally free to if that describes your set up. It's just that people like to be more abstract than that and not commit to a concrete representation of this sort). 2. What are the benefits of the measure theory foundation in the first place? (partially answered last comment) – spaceisdarkgreen Dec 30 '16 at 09:09
  • in some sense, I would say my question was "What are the benefits of the more abstact setting". But it is a rather entagled questioned, hence the chicken and egg situation and your formulation as "why not?" in 1 followed by "why?" in 2. A good exmple seattles it tho. – user123124 Dec 30 '16 at 09:16
  • I think I understand now. I thought a little more and added to my answer. – spaceisdarkgreen Dec 30 '16 at 11:25
  • isnt it better to think about it as we have some space $(\Omega, \mathcal{F},P)$ and on this space we can introduce any number of random variables that we want and then let them induce different new probability spaces? i.e $(\mathbb{R},B(\mathbb{R}),P_{X}), (\mathbb{R}^{2},B(\mathbb{R}^{2}),P_{X,Y}),(\mathbb{R}^{T},some algebra, Wiener measure)$ all being object originating from a collection of RV's on the first probability space. But maybe that is what you said? – user123124 Feb 13 '18 at 14:47
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    @user32423 But it's not immediately clear what random variables one can introduce. You need to actually pick a probability space and see what kind of RVs can live there. As I mentioned in my answer, the probability space $([0,1], \mathcal B, Leb)$ is (perhaps somewhat surprisingly) capable of supporting a great many sets of random variables and joint distributions for them, but seeing this requires constructing some elaborate encodings. However in practice we do normally just assume our space is "big enough" to have all the random variables we are interested in. – spaceisdarkgreen Feb 13 '18 at 16:40
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    @user32423 in fact, we usually effectively work (or imagine we work) in the probability space defined by some tuple of random variables and their distribution. But we don't need to be committed to that precise representation: any space and definition of the RVs on it is fine provided the push forward measure of the RVs is what we want it to be. – spaceisdarkgreen Feb 13 '18 at 16:58
  • right. So when we construct Browninan motion we can think of it as we have the sequence of RVs on some(rich enuf) abstract $(\Omega, \mathcal{F},P)$ which allows us to construct the push forward measure (Wiener Measure) on $(\mathbb{R}^{T}, some algebra, Wienermeasure)$? – user123124 Feb 14 '18 at 05:08
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    @user32423 Yes. – spaceisdarkgreen Feb 15 '18 at 00:51