# Why is the expected value of a data point equal to population mean?

For example, I draw samples until I have I have the random variables $X_1,...,X_n$. I'm told that if I select any of these random variables, say $X_4$, its expected value will equal the population mean, i.e. $E[X_4]=\mu$. This is what I do not understand.

I understand that if my sample included every single possible observation from the population, i.e. $n=N$, then $E[X_4]=\mu$ makes sense to me. But if $n<N$, why do we still expect it to be $\mu$?

If my population is $\{1,2,3,4,5\}$, each value equally likely. Clearly, $\mu=3$. If I take a sample, say, $X_4 = \{1,2,3\}$, then $E[X_4]=(1/3)+(2/3)+(3/3)=2\ne \mu$. So I seem to have calculated a sample mean if anything. So something is wrong with this calculation. Therefore, I do not understand why $E[X_4]=\mu$.

• It would be better to use uppercase $X_4$ for the random variable and $x_4$ for a particular observation of $X_4$. You would not than say $X_4 = \{1,2,3\}$ and probably not $E[x_4]$ though you could still talk about $E[X_4]$ – Henry Apr 20 '18 at 9:57

A sample takes the form $\{X_1,\dots,X_n\}$ and the $X_i$ are random variables.
Then it is true that: $$\mathsf EX_i=\mu\text{ for every }i$$ but stating that is something different as stating that: $$\frac1n(X_1+\cdots+X_n)=\mu$$ which is in general not true.
By writing $X_4=\{1,2,3\}$ as a possible sample you suggest that $X_4$ is a set with $3$ random elements (that all have taken some value). In that context $\mathsf EX_4$ has no meaning at all.
• So $X_1$, for example, is not a sample in itself, but rather, is a single value within a sample that has the potential to take on various values, each with a particular probability? Hence this is why it is called a random variable? Does $X_1$ have the potential to take on all possible values contained within the population? – HumptyDumpty Apr 20 '18 at 9:48
• Yes. For instance if $\{1,2,3,4,5\}$ and you pick a sample of $3$ then $X_1$ corresponds with the first pick. It can be any of the elements $1,2,3,4,5$ and if the distribution is uniform then this with equal probabilities. Then $X_1=\frac15(1+2+3+4+5)=3=\mu$. This also works for $X_2$ and $X_3$. – drhab Apr 20 '18 at 9:52
• So if we are taking multiple samples. What is the notation for this? Is it something like $S_1=\{X_1, X_2, X_3\}$, $S_2=\{X_1, X_2, X_3\}$, $S_3=\{X_1, X_2, X_3\}$? – HumptyDumpty Apr 20 '18 at 10:11
• No, because like this $S_1=S_2=S_3$. I would go for $S_k=\langle X_1^{(k)},X_2^{(k)},X_3^{(k)}\rangle$ for $k=1,2,\dots$. Or, if the order does not matter $S_k=\{X_1^{(k)},X_2^{(k)},X_3^{(k)}\}$. – drhab Apr 20 '18 at 10:16