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This is within the context of "survey sampling".

Let $P = \{1,...,N\}$ be the target population, and $S = \{s_1,...,s_n\}$ a sample from P. P and S are identified by the corresponding values of the characteristics of interest:

$S = {X_1,...,X_n} \subset {x_1,...,x_N} = P$

The population mean and variance are $\mu$ and $\sigma^2$, respectively. These are defined by $\mu = \frac{1}{N} \sum_i^n x_i$ and $\sigma^2 = \frac{1}{N}\sum_i^n (x_i - \mu)^2$

Sampling is performed by "simple random sampling" without replacement.

Question

A first lemma states that:

  1. $P(X_i=\xi_j) = \frac{m}{N}$, with m the number of times a distinct value $\xi_j$ of $X_i$ is present in the original population.

Two further lemmas state now that:

  1. $E[X_i] = \mu$ and $Var[X_i] = \sigma^2$

This is confusing to me. If $X_i$ is one particular observation, say $X_1 = 5$, drawn from a normal distribution with $\mu = 0, \sigma^2 = 1$, how can the expected value of this particular observation (called a "sample unit" in our lecture notes) be equal to the mean of the entire population?

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    $\begingroup$ Maybe I'm even more dense than usual tonight, but this Question leaves me totally clueless. // Are you dealing with samples from a multivariate population here? Maybe $n$-variate? // You need to provide some context in your question. Can you define symbols? Explain how $(x_1,\dots,x_n)$ is a population? Maybe show what you have tried and why you need help. Maybe say what topic you are studying. Name/author of text where lemmas appear. $\endgroup$ – BruceET May 12 at 4:21
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    $\begingroup$ Thank you for the feedback - these are lecture notes which are unfortunately not publicly available. But I have added more context and made what I wrote more precise. $\endgroup$ – Pugl May 13 at 19:15
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Your notation is quite confusing to me, but about item number two: "random sample" means that the random variables $X_1, \dots, X_n$ are independent and identically distributed. Particularly, they all have the same mean and variance. That is why $\text{E}(X_i) = \mu$ and $\text{Var}(X_i) = \sigma^2$.

I think you are confused about the difference between a random variable $X$ and the realized value $x$ that you get when you draw from $X$. If you drew a value of 5 from a random variable $X$ with a $N(0, 1)$ distribution---which would be extraordinary, by the way, since it would be 5 standard deviations away from the mean---then you don't take the "expected value of this particular observation". The particular observation 5 is a constant. You could take the expected value of $X$, the random variable, and you would find that $\text{E}(X) = 0$, since the expected value of a normally-distributed random variable is equal to the first parameter $(0)$ of the distribution.

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  • $\begingroup$ Thank you, that is helpful, and you are right that I am confused by the notation. I understand as you write that E(X) = 0. What was confusing to me was the subscript "i" in $X_i$. I understand now that $E[X_i]$ refers to the expected value over all $X_i$, with i in 1,..,N (N being the sample size) over infinitely many samples. Does that then not imply that $E[X_i] \rightarrow \infty$, that is, that we have convergence to the population mean when the number of times we sample goes to infinity? $\endgroup$ – Pugl May 13 at 19:19
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    $\begingroup$ I think you may still be confused. $\text{E}(X_i)$ is just the expected value of the random variable $X_i$. It's got nothing to do with infinitely many samples. Typically, when you take a random sample of size $n$, this can be thought of as involving one draw from each of the (independent and identically-distributed) random variables $X_1, X_2, \dots, X_n$. They each have an expected value (which are all equal). So $\text{E}(X_1) = \mu, \text{E}(X_2) = \mu$, and so on. $\endgroup$ – Novice May 15 at 5:08
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    $\begingroup$ If you are still confused, I would suggest either consulting your instructor, and/or asking another question on this site (with more details and explanations). If you are taking an introductory statistics course then I think you probably haven't understood the basic setup of random sampling yet. If on the other hand, you are taking a more advanced or specialized statistics course, then perhaps I am the one who is confused here. $\endgroup$ – Novice May 15 at 5:12

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