Finite Population Sampling without Replacement

Personal note of finite population sampling

First moment

Population: $ \Omega = \{ x_1, \dots, x_N \} $
Collection of $n$-samples: $\mathcal{S} = \{ s \in \Omega^n \mid \forall i,j \in s, i \ne j \} $
Collection of $n$-samples containing $x$: $ \mathcal{S}_x = \{ s \in \mathcal{S} \mid x \in s \} $
Observe that $ |\mathcal{S}_x| = \binom{N-1}{n-1} $.
Let population mean be zero. $\mu = 0$, i.e. $ \sum_{i = 1}^N x_i = 0 $
Fix an order for $\mathcal{S}$: $ \mathcal{S} = \{ s_1, s_2, \dots, s_{|\mathcal{S}|} \} $.
$j$-th $n$-sample mean $ m_j = \frac1n \sum_{x \in \mathcal{S}_j} x $
Remark: I don’t use $ \sum s_j $ as in $ \cup \mathcal{T} $ in topology to avoid misreading the $n$-sample $ s_j $ as an element.
mean of $n$-sample mean $ m = \frac{1}{|\mathcal{S}|} \sum_{s_j \in \mathcal{S}} m_j $

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