# Expected value of a "sample unit"

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?

• 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. May 12, 2020 at 4:21
• 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.
– Pugl
May 13, 2020 at 19:15

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.
• 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?
• 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. May 15, 2020 at 5:08