Expected value of a "sample unit"

Question

This is within the context of "survey sampling".

Let $P = \{1,...,N\}$ be the target population, and $S = \{s_1,...,s_n\}$ a sample from P. P and S are identified by the corresponding values of the characteristics of interest:

$S = {X_1,...,X_n} \subset {x_1,...,x_N} = P$

The population mean and variance are $\mu$ and $\sigma^2$, respectively. These are defined by $\mu = \frac{1}{N} \sum_i^n x_i$ and $\sigma^2 = \frac{1}{N}\sum_i^n (x_i - \mu)^2$

Sampling is performed by "simple random sampling" without replacement.

Question

A first lemma states that:

1. $P(X_i=\xi_j) = \frac{m}{N}$, with m the number of times a distinct value $\xi_j$ of $X_i$ is present in the original population.

Two further lemmas state now that:

2. $E[X_i] = \mu$ and $Var[X_i] = \sigma^2$

This is confusing to me. If $X_i$ is one particular observation, say $X_1 = 5$, drawn from a normal distribution with $\mu = 0, \sigma^2 = 1$, how can the expected value of this particular observation (called a "sample unit" in our lecture notes) be equal to the mean of the entire population?

Answer 1

Your notation is quite confusing to me, but about item number two: "random sample" means that the random variables $X_1, \dots, X_n$ are independent and identically distributed. Particularly, they all have the same mean and variance. That is why $\text{E}(X_i) = \mu$ and $\text{Var}(X_i) = \sigma^2$.

I think you are confused about the difference between a random variable $X$ and the realized value $x$ that you get when you draw from $X$. If you drew a value of 5 from a random variable $X$ with a $N(0, 1)$ distribution---which would be extraordinary, by the way, since it would be 5 standard deviations away from the mean---then you don't take the "expected value of this particular observation". The particular observation 5 is a constant. You could take the expected value of $X$, the random variable, and you would find that $\text{E}(X) = 0$, since the expected value of a normally-distributed random variable is equal to the first parameter $(0)$ of the distribution.