$\bullet$ Prove: $\mathrm E(\bar x)= \mu$
Answer:
Let $x_1,x_2,x_3\ldots,x_n$ denote the sample observations. The sample mean is $$\bar x= \frac{(x_1+x_2+x_3+\ldots+x_n)}{n}= \frac{1}{n}\sum x_i$$ where $x_i$ is the $i$-th member of of the sample.
Note, in simple random sampling(with or without replacement), the sample members has the same probability distribution as in the variable $x$ in the population.
Therefore, $\mathrm E( x_i)= \mu$
And $$\mathrm E(\bar x)= \frac{1}{n}[\mathrm E(x_1)+ \mathrm E(x_2)+\ldots+\mathrm E(x_n)]= \mu.$$
What I'm not getting is the blocked part that the author wanted to highlight.
Can anyone tell me why actually $x_i$ has the same probability distribution as $x$ in the population especially even when the random sampling is done without replacement?
For $x_i$ and $x_j$ are not independent any-more when the sampling is done without replacement; so can then also the probability distribution of $x_i$ and $x_j$ remain the same as that of $x$ in the population?
How can $\mathrm E(x_i)= \mu$ when the sampling is done without replacement?
Can anyone please explain that?