# What are the $X_1, X_2, …, X_n$ in the Sampling Distribution of $\bar{X}=\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ X_{ j } }$?

The sampling distribution of $\bar{X}$ defined in a book that I am reading is $\bar{X}=\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ X_{ j } }$.

I know $X_1, X_2, ..., X_n$ are random variables. But what is confusing to me is that should $n$ here be seen as the number of observations or the number of trials in each observation?

If the $n$ here is the number of observations, then, does that mean that the $X_1, X_2, ..., X_n$ are mean values of their own individual trials? For example, $X_1$ is the average of say 10 trials. So $X_1$ itself is $X_1=\frac { 1 }{ 10 } \sum _{ i=1 }^{ 10 }{ Y_{ i } }$, where $Y_{1...10}$ are the trials made in the observation set of $X_1$. In this case, however, $X_1, X_2, ..., X_n$ are more of like constant instead of random variables any more. Then it shouldn't be just $X_1, X_2, ..., X_n$ but $\bar{X_1}, \bar{X_2}, ..., \bar{X_n}$ and $\bar{X}=\bar{X_1}, \bar{X_2}, ..., \bar{X_n}$. But this doesn't look like how it is defined.

Then, if the $n$ here is the number of trials in each observation, then $\bar{X}$ is just the mean value of the trials in this single observation, which again doesn't make a lot of sense because this is just average of one set of observation. From my understanding, Sampling Distribution is the "average of the averages of $n$ sets of observations" and so this interpretation doesn't align with my understanding too.

Which of my interpretations is right? What is the right way to look at the definition of the sampling distribution of $\bar{X}$ as $\bar{X}=\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ X_{ j } }$ and what are the $X_1, X_2, ..., X_n$?

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What is your book? –  Did Mar 18 '12 at 10:01
Maybe I shouldn't say that it is a book because it's like a set of bind notes but it feels like a book when reading it. I think I'm having the idea right about the sampling distribution being the "average of the averages of $n$ set of observations". But I'm not very confident either and the notations are very confusing to me. –  xenon Mar 18 '12 at 14:37
So this is not a book. Where does the set of notes come from? –  Did Mar 18 '12 at 14:42
School? But why? –  xenon Mar 18 '12 at 14:55
To determine whether the set of notes is awfully badly formulated, or you just misread it. –  Did Mar 18 '12 at 15:09

Each $X_i$ is a random variable, not an average of random variables. The book calls $\frac{1}{n}\sum X_i$ the sample average because you can think of the sample value, or outcome, $X_i(\omega) = x_i$ as the value that shows up on the $i$th trial of an experiment when $\omega$ occurs.
Think of it this way: you have an outcome space $\Omega$ that represents all the possible "realities" or "states of the world" that might arise. If, in the course of your experiment, or random sample, the world is observed to be in a given state $\omega \in \Omega$, then the random variable $X_i$ (a function) will be observed to have the value $X_i(\omega) = x_i$ (a number).
When we write capital letters in probability, like $X$, we usually mean a random variable, which is a function on the sample space $\Omega$. This function will take on a specific value $X(\omega) = x$ when we observe a specific state of nature $\omega \in \Omega$. (The lower case letters, like $x$, are just numbers.)
Thanks! So say if $n=10$, for every $X_i$ trial, is a randomly sampled occurrence. Then, $X=X_1+X_2+...+X_{10}$ and I can say that $\bar{X}=\frac { 1 }{ 10 } \sum _{ i=1 }^{ 10 }{ X_{ i } }$. Is this right? Then in central limit theorem, we have to make several $X$'s and the average of the several attempts of $X$ converge to the true mean of $\mu$. –  xenon Mar 18 '12 at 14:53
Thanks. I re-read your explanation and rethought through it and I think I'm clearer now. Just to make sure that I am totally right, am I right to say that the Law of Large Numbers says that when $n$ gets large, $\bar{X}$ will converge to the true mean, which is $E(X)=\mu$? Then for the CLT, it says that collecting many many samples of $\bar{X}$ builds a distribution of $\bar{X}$ which follows the normal distribution? –  xenon Mar 20 '12 at 14:56