# 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 $m = \frac{1}{|\mathcal{S}|} \sum_{s_j \in \mathcal{S}} m_j$

\begin{aligned} m \binom{N}{n} &= \sum_{s_j \in \mathcal{S}} m_j \\ nm \binom{N}{n} &= \underbrace{\sum_{s_j \in \mathcal{S}}}_{\text{fix sample}} \underbrace{\sum_{x \in s_j}}_{\text{find item}} x \\ mN \binom{N-1}{n-1} &= \underbrace{\sum_{x \in \Omega}}_{\text{fix item}} \underbrace{\sum_{s \in \mathcal{S}_x}}_{\text{find sample}} x \end{aligned}

The transition of the double summation on the RHS can be explained by the bijection

$$\{(s_j, x) \mid s_j \in \mathcal{S}, x \in s_j \} \leftrightarrow \{(x, s) \mid x \in \Omega, s \in \mathcal{S}_x \}.$$

Then the RHS of the last equality is

$$\sum_{x \in \Omega} \sum_{s \in \mathcal{S}_x} x = \sum_{x \in \Omega} |\mathcal{S}_x| x = \binom{N-1}{n-1} \sum_{x \in \Omega} x.$$

Take out $\binom{N-1}{n-1}$ on both sides to get $mN = \sum_{x\in\Omega} x$. Hence $m = 0$.

# Second moment

Population variance: $\sigma^2 = \frac1N \sum_{x \in \Omega} x^2$
Variance of $n$-samples: $\sigma_m^2 = \frac{1}{\binom{N}{n}} \sum_{s_j \in \mathcal{S}} m_j^2$

\begin{aligned} & \binom{N}{n} \sigma_m^2 \\ =& \frac{1}{n^2} \sum_{s_j\in\mathcal{S}} \left(\sum_{x\in s_j} x \right)^2 \\ =& \frac{1}{n^2} \left[ \sum_{s_j\in\mathcal{S}} \left( \sum_{x \in s_j} x^2 + \sum_{\small \begin{array}{l} & x,y \in s_j \\ & x \ne y\end{array}} xy \right) \right] \\ =& \frac{1}{n^2} \left[ \binom{N-1}{n-1} \sum_{x \in \Omega} x^2 + \binom{N-2}{n-2} \sum_{\small \begin{array}{l} & x,y \in \Omega \\ & x \ne y\end{array}} xy \right] \\ =& \frac{1}{n^2} \left[ \binom{N-1}{n-1} - \binom{N-2}{n-2} \right] \sum_{x \in \Omega} x^2 \\ =& \frac{1}{n^2} \frac{N-n}{n-1} \binom{N-2}{n-2} N \sigma^2 \\ =& \frac{\sigma^2}{n} \frac{N-n}{N-1} \binom{N}{n} \xrightarrow[N \to \infty]{} \frac{\sigma^2}{n}, \end{aligned}

which is the standard error of sample mean.