I have a finite set $\lbrace r_i \rbrace_{i \in \Lambda}$ with $0 < r_i < 1$ for each $i$. Each $r_i$ has a probability $p_i$ associated to it.
I define $$ r = \prod_{i \in \Lambda}r_i^{p_i} $$ so that $\log r = \sum_{i \in \Lambda} p_i \log r_i$.
Then it is claimed that by the law of large numbers, with probability tending to $1$, the random element $r_I = r_{i_1} \cdot \ldots\cdot r_{i_n}$ chosen with probability $p_I = p_{i_1} \cdot \ldots \cdot p_{i_n}$ satisfies $$ r_I = r^{n(1 + o(1))}. $$
In other words, we claim that when $n$ is large, $r_I = r^n$ almost surely. How can I see this? I can calculate the expected value of $\log r_I$ which is $\log r^n$ but does the law of large numbers not apply into the mean of a sum of random variables? How can I say anything about this single variable?
(The notation $o(1)$ stands for a quantity tending to $0$ as $n \to \infty$.)
EDIT: In which variables am I even using the law on? The conclusion is about $r_I$ which is a singe RV $I \mapsto \mathbb{R}$. Can I present it as a mean of some other variables with expected value $r^n$?
EDIT2: $\log r_I = \sum_{k=1}^n \log r_{i_k}$ and since $\mathbb{E}(\log r_{i_k}) = \log r$, I have $\dfrac{\log r_I}{n} \to \log r$ a.s.
EDIT3: Oh wait I solved it, never mind
$\log r_I = \sum_{k=1}^n \log r_{i_k}$ and since $\mathbb{E}(\log r_{i_k}) = \log r$, I have $\dfrac{\log r_I}{n} \to \log r$ a.s.