On Wikipedia, for Bessels Correction, Proof of correctness – Alternate 1, https://en.wikipedia.org/wiki/Bessel%27s_correction
It says at the beginning of the proof:
I don't understand what it means to have an expected value of an independent sample such as $E[x_1]$ and $E[x_2]$.
Isn't the the whole definition of expected value the mean of the probability of all the samples, so how could there be an expected value for specific independent samples?
Also, how could $E[x_1^2] =E[x_2^2]$ since they are different independent samples?
Also, if $E[x_1^2] =E[x_2^2]$, shouldn't it follow that $E[x_1] =E[x_2]$ and therefore why doesn't $E[x_1x_2] =E[x_1^2]=E[x_2^2]$?
I am very confused...
Imagine that I have two distinct coins, $C_1$ and $C_2$, both of which are perfectly fair, and $x_i$ takes value $0$ if $C_i$ comes up tails, and $5$ if it comes up heads. Then both $x_1, x_2$ are drawn from the same distribution, but that does not mean that they are identical -- it could very well be that one coin turned up tails and one turned up heads, so that $x_0 \neq x_1$.
The expected value $\mathbb E[x_1]$ is the "true, theoretical average" of the values of $x_1$: in this case it is $$ \frac12 \times 0 + \frac12 \times 5 = \frac52. $$ The same, of course, is true for $\mathbb E[x_2]$. This is why $\mathbb E[x_1] = \mathbb E[x_2]$, and similarly $\mathbb E[x_1^2] = \mathbb E[x_2^2] = \frac{25}2$, even though it need not be that $x_1 = x_2$.
You can approximate the value $\mathbb E[x_1]$ in practice by repeating the experiment of tossing coin $C_1$ a lot of times, and averaging the results, but that does not necessarily directly relate to $x_2$.
However, the fact that $\mathbb E[x_1] = \mathbb E[x_2]$ does not imply that $\mathbb E[x_1x_2] = \mathbb E[x_1^2]$: you cannot just change the $x_1$ or $x_2$ around within the $\mathbb E$, since it is not in general true that $x_1 = x_2$. In fact, you can calculate directly: $x_1^2$ has value 0 or 25, both with probability $1/2$, so $\mathbb E[x_1^2] = \frac{25}2$. The number $x_1 \times x_2$ has value 0 with probability 3/4 (at least one tails) and value 25 with probability 1/4 (both are heads), so it has expected value $\frac{25}4$.
I think most of the confusion here is caused by $x_1, x_2$ really referring to two closely related but distinct things: both to a random variable which is distributed a certain way, and to an actual sample from that random variable. That's why it seems to change around whether $x_1, x_2$ get referred to as numbers, or to something more complex (which you can, for example, apply the operator $\mathbb E$ to).