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:
- $P(X_i=\xi_j) = \frac{m}{N}$, 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:
- $E[X_i] = \mu$ 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?