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 Apr 6 awarded Supporter Sep 24 awarded Autobiographer Jan 9 comment is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” Concerning the "full of": One quote, that you may view as arrogant (although it was a comment out of frustration - you shouldn't confuse arrogance with frustration) doesn't make my question full of arrogance. Besides the quote, the rest of it is a neutral mathematical enquiry. Also, a lot of other questions here on this site are uninformed (no need to hyperbolize it), so either you should downvote all of them or restrict to downvote because of my "arrogance" - which would again make me ask the above question. Jan 9 comment is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” @did I can see your viewpoint, but I have to object to "quite typically" and "full of": I only posted two questions, so you can't possibly know me that well to say "typically" (and if a lot of people downvote I think it's natural I respond to that). Jan 7 comment What does it mean for a random variable to describe an experiment? @leonboy haha, that somehow made me laugh; I was just trying to say a really dumb different rant, after Sarwate dismissed my enquiry, and I'm fairly sure I succeeded in doing that. take it with a grain of salt :) Jan 7 comment What does it mean for a random variable to describe an experiment? @Alex that book is really great, thanks a lot Jan 7 accepted What does it mean for a random variable to describe an experiment? Jan 7 comment is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” @did But doing "revenge-downvotes" based on "moral-jduge" decisions and not mathematics is a consequence of a [see comment above]-type behavior ;) Is arrogant mathematics worse mathematics ? Jan 7 comment What does it mean for a random variable to describe an experiment? @jay-sun My problem is not that I do not consider the mathematical description appropriate - it is that it seems to me that there are tacit assumption made when someone says for example "Let $X:\Omega\rightarrow \mathbb{R}$ describe the tossing of two coins", since probabilities are missing, nothing about $\Omega$ is said etc. Jan 7 awarded Commentator Jan 7 comment is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” @gnometorule Well, true, though if adults whose area of expertise gets criticised do a "revenge-downvote", act a bit like they were kindergarten I think. Since the question itself, it seems to me, was up to the standards more or less (at least, I've see worse questions with a positive vote count). Jan 7 awarded Scholar Jan 7 comment is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” the variables on it make sense (the sample space for example has to have more elements in it than the values the random variable can take). This "physicists approach" of adding assumptions to our objects of consideration during our reasoning is just terrible (for me at least). Jan 7 accepted is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” Jan 7 comment is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” Great! And another example where the books are sloppy, since - as you say - the assumption $P'(v_I=\omega)=P(\omega)$ should also have been made. (upvoting takes 15 rep, and since a lot of people downvote me, I'll probably have to wait until I can upvote). Jan 7 comment is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” @RahulNarain Well, I didn't say that probability is stupid and it sucks. I said it more elegantly with a quote :D (And why are uninterested people always downvoting ?) But jokes aside, what I'm criticising is that after having browsed through dozens of web pages - and this forum too!! - I always find the same sloppy way of handling matters in probability theory: Usually a sample isn't given - but we aren't told either that we should assume one fixed space exists. But if we do assume an arbitrary, in the next line, we see that this space actually does have to fulfill certain criteria such that Jan 7 comment is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” @uncookedfalcon Ok, I tried to be more precise. In the link you provided indeed it's explained how to construct a product measure, but the connection with independent variables isn't treated (which is the thing I don't quite understand - edited my question). Also - good to know there are also other people who find this latter approach more natural :). Were you also so frustrated with probability and the extremely sloppy way of handling the formalism ? (Do you perhaps know an undergraduate book - i.e. without hard measure theory - that is formally more sound ?) Jan 7 revised is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$” added 178 characters in body Jan 7 comment What does it mean for a random variable to describe an experiment? @RahulNarain Well the question in the title isn't disconnected from the question in the body, so it should be clear (to the readers) what isn't clear (to me). Necessarily having a question mark in the body seems rather burocratic, but since this is the first reasonable remark as to what isn't ok with my question, I'm happy to comply with it. Jan 7 revised What does it mean for a random variable to describe an experiment? added 305 characters in body