MLE of variance is biased in a Gaussian distribution Referring to: How to understand that MLE of variance is biased in a Gaussian distribution
at some point during calculation the formula of the sum of the expected value becomes a single expected value:

The explanation given is:
With the last step following since due to $E[X^2_n]$
being equal across n due to coming from the same distribution.
I can't understand:


*

*What does the expectation of a value mean, in this case what is even $E[X_n]$

*Why should the expected value of an observation value be equal?

*What's the proof for $Var[1/N *...] = (1/N)^2 Var[...]$
Thanks for the help and sorry for the bad format I don't know how to write formulas, this is my first post xD
 A: Note that $x_n$'s are random variables, not specific numbers. Assuming that $x_n$ is gaussian distributed with density $p$, it's expectation is per definition 
$$ E[x_n] = \int_\mathbb{R} x \cdot p(x) dx. $$
If we apply a function to $x_n$, it's expectation is
$$E[f(x_n)] = \int_\mathbb{R} f(x) \cdot p(x) dx, $$
in your case $f(x) = x^2$.
The calculation you reference, assumes that the $x_n$ all come from the same distribution, i.e. they have the same density $p$ and thus the same expectation. 
For your last question see e.g. here for a more general statement and proof.
A: $X_n$ is a Gaussian RV for each $n$; since iid, they all have the same mean, variance, moments etc. i.e. simply we have $$E[X_1^2]=E[X_2^2]=\dots=E[X_n^2]$$
Therefore, $\frac{1}{N}\sum_{n=1}^N E[X_n^2]=\frac{1}{N}NE[X_1^2]=E[X_2^2]=E[X_n^2]$ for any $n$. This is why we can remove the summation.
The proof for scaling property of variance can be: $$\operatorname{var(cX)}=E[(cX)^2]-E[cX]^2=c^2(E[X^2]-E[X]^2)=c^2\operatorname{var}(X)$$
