# What is $\bar\mu_n$?

Am I right in thinking that it is the average of the sum of $n$ different populations means?

Here it is used in the context that confused me. It's the Chebychev WLLN, apparently.

"If $x_i, i = 1, . . ., n$ is a sample of $n$ observations such that $E[x_i] = \mu_i < \infty$ and Var[$x_i] = \sigma_i^2$ such that $\bar\sigma_i^2/n = (1/n^2)\Sigma_i \sigma_i^2 \rightarrow 0$ as $n \rightarrow \infty$ then $plim(\bar x_n - \bar\mu_n$) = 0."

Is this saying that each sample of $i$, corresponds to it's own population of $i$ (from above, $E[x_i] = \mu_i$) and as the sample get bigger we have to average over populations?

So if I were to draw the random variable 1 from a population of {1,2,3} and the random variable 4 from the population {4,5,6} then the population of my sample 1,4 is {1,2,3,4,5,6}?

## 1 Answer

You are right.

"If $x_i, i = 1, . . ., n$ is a sample of $n$ observations such that $E[x_i] = \mu_i < \infty$ and Var[$x_i] = \sigma_i^2$ such that $\bar\sigma_i^2/n = (1/n^2)\Sigma_i \sigma_i^2 \rightarrow 0$ as $n \rightarrow \infty$ then

$$\lim_{n\rightarrow \infty}P\left(\left|\frac 1n\sum_{i=1}^nX_i - \frac 1n\sum_{i=1}^nE(X_i)\right|<\epsilon\right) =1$$

I guess you can make the notational mapping.

Since by design we assume different moments for each $X_i$, each comes from a different population. So if by $\{1,2,3\}$ you mean values of the index $i$, then $\{1,2,3\}$ is not a population, but a set including three values of the index with each value representing a different population.

If you consider the random variables $\{X_1,X_4\}$, it is a pair coming from two different populations -you do not "unite" the two populations "into one" because, being different with respect to the object of study (convergence of sample moments), how could they form a single population (for the purposes of the specific study)? Have you contemplated how is the abstract concept of "statistical population" defined?

• No I didn't intend for {1,2,3} to be the index values of $i$, I meant for {1,2,3} to be the population that the sample $x_1$ is drawn from. For example, the sample $x_1$ has one element {1} but $\mu_1 = 2$ (1+2+3/3) Jul 26, 2014 at 21:55
• You mean the values that the random variable can take? This is never called a "population". It is the set from which the members of the population take their values, usually called the "support". In the example you state in the question, the joint support is the cartesian product $\{1,2,3\} \times \{4,5,6\}$ You have a two-dimensional vector here - the support should also be two-dimensional. Jul 26, 2014 at 21:59
• Continuing with the index notation for a moment, in this example do you think that $x_1$ and $x_n$ are draws from 2 different population but the number of observations in sample $n$ is greater than the number of observations in sample 1. If this is the case though(and I'm not saying it is, because I don't fully understand it), then we don't need to average across populations because $plim(\bar x_n - \mu_n) = 0$. I guess I'm wondering what is happening when the index is going to $n$. Are we drawing larger numbers from the same population or from different populations? Jul 26, 2014 at 22:05
• With respect to the most recent comment, I had actually never heard the word "support" used like that (though I had often used the phrase "common support" when doing PSM. So all the values that a population mean are calculated from are called the "support"? Jul 26, 2014 at 22:15
• As per the terminology, the "support" of the distribution of a random variable, is standard at least in mathematical statistics, for the range of the random variable, which in turn is the domain of the probability functions (and from where we "draw" realizations of random variables). Jul 26, 2014 at 23:01