# series of random variables

3:30am and i was just looking at the law of large numbers and thought about the idea of a series of random variables and got myself very stuck.

when we consider the series X'_n = X_1 + ... X_n and say it (divided by n) converges to E(X_1), one of the definitions of a.s. convergence is based on the probability space. specifically, the theorem can be stated:

P({ w in omega : 1/n * X'_n(w) --> E(X_1) }) = 1

which then led me to the question, what is X'_n(w)?

indeed w is an element in our sample space. lets say our random variables are i.i.d. (assumption of the theorem) with known distribution (lets say, exponential, even though we wont use this fact). then a single w is but a number on the interval [0,+infinite).
so X'_n(w) = (X_1 + ... + X_n)(w)
= X_1(w) + ... X_n(w) = n * X_1(w)

since we have n random variables that are all looking at the same event from the sample space, i.e. the same event (in this case real number), and since theyre identically distributed theyll all have the same value?!?

of course thats not the intention of the statement of the law. the theorem says that if we sample n times for large enough n, blah blah blah. which of course means that we sample n different values, different w's, i just dont see how thats expressed in the formulation of the definition.

im missing something very basic - maybe the fact that im way too tired to be thinking about this at 3:30am. staying awake when im too tired to function is one of those mistakes that always seems tempting to make at the time, always turns out to be a mistake afterwards, and is the perpetual lesson that i will never learn from...

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The variables $X_1,\ldots,X_n$ are intended to be \emph{independent}. That means that given a sample space for one of them $\Omega$, the appropriate sample space for all of them is $\Omega^n$. For example, if $X_i$ is a Bernoulli random variable, then you can take $\Omega = \{0,1\}$ and $\Omega^n$ is just the set of all possible outcomes.

Of course, a measure space comes with a $\sigma$-field and a probability measures, so these also have to be extended; when you're dealing with finitely many variables this is straightforward. For infinite (countable) sequences it might be less obvious, but you can look it up in any good introductory book.

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yeh i figured as much, although it seemed strange that the sample space changes. ive never encountered a limit where the sample space is dependant on the limit-variable. so for every different n we're actually considering a different sample space altogether. thats not problematic, although its still foreign to me. – davin Jan 2 '11 at 10:52
In limit theorem you'd like all variables to share a sample space (otherwise you can't reason about their interrelations), so you need to take the limit up front. – Yuval Filmus Jan 2 '11 at 17:14

First of all, note that a sample point $\omega$ (unlike $X(\omega)$) need not necessarily be a number (or vector). But that's not the point here.

As in your example, let the $X_i$ be i.i.d. exponential rv's. For our purposes, let's consider only the finite case of $n$ rv's, $X_1,\ldots,X_n$. Then, to see why a sample point $\omega$ should correspond to a vector $\omega = (\omega_1,\ldots,\omega_n) \in [0,\infty)^n$ (rather than to a number in $[0,\infty)$), consider the following observation: $${\rm P}\bigg(\sum\limits_{i = 1}^n {X_i } \in (a,b]\bigg) = {\rm P}\bigg[(X_1 , \ldots ,X_n ) \in \bigg \{ (x_1 , \ldots ,x_n ):\sum\limits_{i = 1}^n {x_i } \in (a,b]\bigg \}\bigg].$$ Hence, we can actually consider vectors $(X_1 (\omega),\ldots, X_n (\omega))$, and, in turn, vectors $\omega = (\omega_1,\ldots,\omega_n) \in [0,\infty)^n$.

EDIT: Since, in our example, the $X_i$ are i.i.d. exponentials, then given any $n$-dimensional rectangle $R_n=[a_1,b_1] \times \cdots \times [a_n,b_n] \subset [0,\infty)^n$, we have $${\rm P}[(X_1 , \ldots ,X_n ) \in R_n] = \prod\limits_{i = 1}^n {{\rm P}(X_i \in [a_i ,b_i ])} = \prod\limits_{i = 1}^n {\int_{a_i }^{b_i } {f(x_i )\,{\rm d}x_i } } ,$$ where $f(x_i)=\lambda e^{-\lambda x_i}$, $x_i \geq 0$, is the exponential$(\lambda)$ density function. The basic idea here is to consider "product measure space". Another key concept here is "convolution".

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