Understanding Cochran 1977 proof of variance of the sample mean in sampling without replacement

Question

Notation

$\mathcal{X}$ is the multiset of size $N$ of all possible values in our population. These values are fixed.

$\mathcal{S}$ is the set of all $\binom{N}{n}$ possible subsets of $\mathcal{X}$ of fixed size $n$.

Let $S\in\mathcal{S}$ be a random variable representing a simple random sample from $\mathcal{X}$ without replacement.

$x_i \in \mathcal{X}$ is the $i^{th}$ member of the population, $i\in\{1,...,N\}$.

$x_{si}$ is the $i^{th}$ member of a given sample $s\in\mathcal{S}$, $i\in\{1,...n\}$

Claim

$\frac{1}{n}\sum_{i=1}^{n} x_{Si}$ is an unbiased estimate of $\frac{1}{N}\sum_{i=1}^{N} x_i$

Proof (Cochran 1977)

Since every unit $x_i$ appears in the same number of samples, it is clear that:

$\mathbb{E} \sum_{i=1}^n x_{Si}$ must be some multiple of $\sum_{i=1}^{N} x_i$

The multiplier must be $n/N$, since the expression on the left has $n$ terms and that on the right has $N$ terms. This leads to the result.

Question

I'm familiar with the other proofs of the unbiasedness of this estimator using indicator variables, linearity of expectation, and combinatorics, but I'm having a hard time wrapping my head around the argument being made here.

The argument that every unit appears in the same number of samples is clear to me. Everything else that follows is not. In particular:

• Why does the left-hand expression have to be a multiple of the right?
• How do we get the multiplier from the number of terms?

Answer 1

The Cochran proof is a slicker variant of the indicator proof. (It might be the same one you're familiar with.) If $S$ is a randomly selected subset with $n$ elements, then the following holds as an identity between random variables: $$ \sum_{i=1}^n x_{{\cal S}i} = \sum_{j=1}^N x_jI(x_j\in S)\tag 1 $$ Since all subsets of $n$ elements are equally likely, the expectation of (1) is then $$ E\sum_{i=1}^n x_{{\cal S}i} \stackrel{(1)}= \frac1{|{\cal S}|}\sum_{X\in{\cal S}}\left[\sum_{j=1}^N x_jI(x_j\in X)\right] \stackrel{(2)}= \sum_{j=1}^N x_j\left[\frac1{|{\cal S}|}\sum_{X\in{\cal S}}I(x_j\in X)\right] \stackrel{(3)}= \sum_{j=1}^N x_jP(x_j\in S) $$ where $\sum_{X\in \cal S}$ denotes summation over every subset (nonrandom) of $n$ elements; there are $|\cal S|$ possible values for $X$. Step (2) is an interchange of summation. In step (3) we realize that each $x_j$ is seen in the same number of subsets, so all the $P(x_j\in S)$ are the same, and equal to $n/N$ since that's the probability that a given element from a set of $N$ items will be chosen to appear in a subset of $n$.