Central limit theorem notation Can you tell me if my understanding of the CLT is correct? Maybe it's just a matter of notation.
The classical CLT states:
Let $X_1,...,X_i,...,X_n$ be a sequence of iid random variables drawn from a distribution with mean $\mu$ and variance $\sigma^2 < \infty$.
Then, let the sample avarage be equal to $\bar{X}_n = \frac{X_1+...+X_i+...+X_n}{n}$.
The CLT says, when $n$ becomes large then $\sqrt{n}(\bar{X}_n-\mu)$ is approx. normal with mean $0$ and variance $\sigma^{2}$.
What this theorem is saying is: take a random sample from a population, then compute the sample average and keep it aside, then you keep sampling let's say for additional 100,000 times, and you get additional (possibly different, at least most of them) 100,000 sample averages. Then, if you look at the sampling distribution of all these sample averages, it would be approximately normal.
Now, the generic $X_i$ is the sample average obtained from the $i$ random sampling or is it simply a random observation obtained from the $i$ random sampling? I mean, is it $\bar{X}_i$ or a single observation?
What I'm asking is: should I can consider $X_i$ either as a single observation (in this case each draw means to draw a single value at each trial, and the number of random draws is $n$)  or as a sample mean (and in this case I'm drawing samples fixed in size, and in this case, $n$ is the number of random samples drawn)?
Shall the same interpretation for $X_i$ hold when we present the law of large numbers?

So I think you have a common misunderstanding. each $X_i$ is a single observation.
Just as 1 throw of a dice has uniform distribution on 1-6 ( you don't need to consider throwing 1000s of dice), a single sample mean is approximately distributed normally ( with the approximation improving as the sample size increases).
So I can calculate probabilities on my single sample mean, just as I can calculate that eg the probability of getting a number of 4 and above with a single dice throw is 1/2. and I know that after many dice throws the proportion of times I get 4 and above should approach 1/2.  But this long run effect is a consequence of saying that the probability of throwing 4-6 is 1/2.
So with one sample mean and variance, I can calculate the approximate distribution of my single sample mean, and discuss the probability that my single sample mean was greater than a given value etc.
