Given is a sample $S = {X_1,...,X_n}$ from a target population $P$.
The sampling is done by Simple Random Sampling (random sampling without replacement); n is the sample size.
The sample mean is defined as $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$
The following two theorems are given:
(1) $E[\bar{X_n}] = \mu$, with $\mu$ the population mean; that is, the estimator of the mean is unbiased
(2) $V[\bar{X_n}]$ = $\frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n Cov(X_i, X_j)$
My question:
When we calculate the variance of $\bar{X_n}$ we assume that the $X_i$ are dependent. This is how I explain it to myself: If we have a population of different-colored balls, after drawing a blue ball and not putting it back, the probability of drawing a next ball with some color has changed.
How is this dependence not reflected when we calculate $E[\bar{X_n}]$, where it is assumed that all $X_i$ have the same expectation $E[X_i]=\mu$? That is, how do the expectations not change when the sampling is done without replacement?
The only assumption here is that the $X_i$ are identically distributed; i.e., they are drawn from the same population.
It is important to remember that linearity of expectation does not require independence of the variables; e.g., $$\operatorname{E}[A+B] = \operatorname{E}[A]+\operatorname{E}[B]$$ even if $A$ and $B$ are jointly dependent random variables, provided that their individual expectations exist.
Therefore, the theorems follow even when the $X_i$ are dependent but identically distributed. When they are independent, then the covariance is $$\operatorname{Cov}[X_i X_j] = \begin{cases} \operatorname{Var}[X_i], & i = j \\ 0, & i \ne j \end{cases}$$ but this is not assumed.