TITLE: series of random variables
QUESTION [0 upvotes]: 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 they'll all have the same value?!?
of course that's 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 don't see how that's 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...

REPLY [3 votes]: 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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