# is “$X_1,\ldots,X_n$ independent” equivalent to “$P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$”

Sometimes it seems to me, that when we are considering independent variables on a sample space of the form $\Omega^n$, we use the assumed independence of variables to show that $P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$, where $P'$ is a yet unknown measure on $\Omega^n$.
Whereas I would find it natural (supposing we know of a probability distribution $P$ on $\Omega$) to instead define a distribution on $\Omega^n$, derived from $P$ (as the multiplication of individual probabilities), and then show that our $X_i$'s are indeed independent.
So why is the first way of doing it more commonly used and hot do I show using the independence that $P'(\omega_1\ldots,\omega_n)=P(\omega_1)\cdot \ldots \dots P(\omega_n)$ ? (I managed to do it the other way around and show independence using the above product measure)

To end it with Russell: "My mathematical lecturers never showed me any reason to suppose probability theory was anything but a tissue of fallacies".

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This is actually wildly amusing: It seems that the math.SE community reacts allergic to quotes or to criticism concerning the bad pedagogy (in notation) of probability theory. Or is it forbidden to let the frustration created the bad formalism of a subfield of maths to transpire ? Anyway, keep them downvotes comming :) If someone wants to downvote my other question, let me make it easy for him, by providing a link: math.stackexchange.com/questions/272181/… – nus Jan 7 '13 at 15:59
I think as a rule of thumb, it's helpful for readers if you clearly pin down a question. For example, if you were to ask, is the latter approach, which I find more natural (me too!), commonly used? (the answer is yes: en.wikipedia.org/wiki/Product_measure ) – uncookedfalcon Jan 7 '13 at 16:52
If I go to, say, StackOverflow and ask a programming question and end it with "C++ is a stupid language and it makes no sense and it sucks", I shouldn't be surprised if it makes people less interested in helping me understand C++. You should consider whether your "criticism" has any more substance than this. – Rahul Jan 7 '13 at 17:15
-1. Ignorance is not an excuse for arrogance. – Did Jan 7 '13 at 18:04
@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 ? – nus Jan 7 '13 at 18:38

As to why it's more commonly used, I can't say...for example, when I was learning this stuff I saw these in the opposite order.

To be clear, for what follows, we are in the case that $\Omega$ is finite (or countably infinite), so that you can recover what $P$ does to arbitrary subsets of $\Omega$ from what it does to singletons (by $P(S) = \sum_{s \in S} P(s)$). (otherwise the implication $P'(\omega_1, \ldots, \omega_n) = P(\omega_1) \ldots P(\omega_n) \Rightarrow X_i$ are independent doesn't hold: if say $n = 2, \Omega = \mathbb{R}$ with $P$ given by integrating a cdf, the probability of any point is 0, so all we know about $P'$ is that it gives no weight to individual points)

For the desired implication, what does it mean that the "variables" are independent? Well, the $i^{th}$ variable $X_i: \Omega^n \rightarrow \Omega$ is simply projection onto the $i^{th}$ coordinate.

Now then, to see what $P'$ of $(\omega_1, \ldots, \omega_n)$ is, we note that this point is exactly the subset of $\Omega^n$ where $X_1 = \omega_1 \ldots X_n = \omega_n$. So by independence $$P'(X_1 = \omega_1 \cap \ldots \cap X_n = \omega_n) = \prod_i P'(X_i = \omega_i)$$

So it looks we want not just the independence of the $X_i$, but also that $P'(X_i = \omega) = P(\omega)$ for all $\omega$ and $i$ to get the desired result.

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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). – nus Jan 7 '13 at 17:30
regarding this and your earlier comment (and my experience learning probability), I'm thinking this isn't the fault of "probability theory" or "books" in general, but rather just the book you're using :)! maybe try checking out grimmett & stirzaker's probability and random processes - I recall it being pretty good, while avoiding measure theory. no worries about the upvote lol – uncookedfalcon Jan 7 '13 at 17:38