Finite Population Sampling without Replacement

Personal note of finite population sampling

First moment

Population: Ω={x1,,xN} \Omega = \{ x_1, \dots, x_N \}
Collection of nn-samples: S={sΩni,js,ij}\mathcal{S} = \{ s \in \Omega^n \mid \forall i,j \in s, i \ne j \}
Collection of nn-samples containing xx: Sx={sSxs} \mathcal{S}_x = \{ s \in \mathcal{S} \mid x \in s \}
Observe that Sx=(N1n1) |\mathcal{S}_x| = \binom{N-1}{n-1} .
Let population mean be zero. μ=0\mu = 0, i.e. i=1Nxi=0 \sum_{i = 1}^N x_i = 0
Fix an order for S\mathcal{S}: S={s1,s2,,sS} \mathcal{S} = \{ s_1, s_2, \dots, s_{|\mathcal{S}|} \} .
jj-th nn-sample mean mj=1nxSjx m_j = \frac1n \sum_{x \in \mathcal{S}_j} x
Remark: I don’t use sj \sum s_j as in T \cup \mathcal{T} in topology to avoid misreading the nn-sample sj s_j as an element.
mean of nn-sample mean m=1SsjSmj m = \frac{1}{|\mathcal{S}|} \sum_{s_j \in \mathcal{S}} m_j

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