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in my lecture our professor said the following: Given a population of N, where each unit has the same probability to get chosen, and a sampling method where the population is divided to samples of size n for all possible samples, the sampling probability for each unit in the population will be n/N.

It sounds rather counter intuitive - I know it's true for a simple random sample, but how can I derive the general case? Thanks.

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I'm not sure this is the best way to prove this, but I think it does the job.

Suppose $\{M\}_i \subset \{1,\ldots, N\}, i \in 1,\ldots,K, |\{M\}| = n$ is a collection of $K$ possible samples of size $n$ from the population, not necessarily containing all possible subsets of size $n$. Let $\{m\}$ be sampled from $\{M\}_i$ with probabilities $p_1, \ldots, p_K$, not necessarily equal. But assume that this entire construction is such that the probability that individual $j$ is in $\{m\}$, say $q_j$, is equal for all $j$, $q_j = q, j \in \{1, \ldots, N\}$

Define $I_j, j \in \{1,\ldots,N\}$ to be an indicator function, taking value 1 if $j$ is in $\{m\}$ and 0 otherwise. Then the expected value of $I_j$ is equal to $q_j = q$.

Sum the $N$ indicator functions, $\sum_{j=1}^{N} I_j$. The sum now counts how many individuals were in $\{m\}$. But $|\{m\}| = n$ for all possible samples. Therefore $\sum_{j=1}^{N} I_j = n$. Take expectations of both sides. As $n$ is constant, $E[n] = n$.

$E[\sum_{j=1}^{N} I_j] = n$

By linearity of expectation,

$\sum_{j=1}^{N} E[I_j] = n$

Since $E[I_j] = q$,

$\sum_{j=1}^{N} q = n$

$ q = n/N$

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  • $\begingroup$ Great proof. super clear and understandable. thanks! $\endgroup$
    – Omri. B
    Jul 23, 2021 at 17:36

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