# Deriving the sampling probability of a unit for any sampling method

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.

• Does this answer your question? Probability of a unit being selected in sample Jul 23, 2021 at 16:48
• @DavidLukeThiessen not quite, this talks about specific sampling methods - not the general case Jul 23, 2021 at 16:51

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 $$E[I_j] = q$$,

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

$$q = n/N$$

• Great proof. super clear and understandable. thanks! Jul 23, 2021 at 17:36