Trying to understand Bienaymé formula In Bienaymé formula, it states that $var(\bar X) = \large\frac{\sigma^2}{n}$.
However, when I was going through the proof here, it says the variances of $X_1,X_2,X_3......X_n$ are the same(assuming they are all independent). Can anyone explain the reason behind it? 
I am confused about how different random variables can have the same variance.
 A: Consider the random variable $$X = \begin{cases}1 & \text{with probability } 1/2\\
-1 & \text{with probability } 1/2 \end{cases}$$
this random variable has variance $1$. Now consider $$Y = X + 3$$
this random variable has also variance $1$.
The reason we make such an hypothesis is that this make explicit the properties used in the proof (independence and equal variances). Often  we deal with $\textit{i.i.d.}$ random variables which satisfy the conditions of the theorem (Bienaymé formula).
A: You can certainly have random variables that are independent and identically distributed (iid).  For example, say I have five numbered coins, each of which is fair.  I am interested in whether the coins land heads when I flip them.  If I flip coin number $i$ once, let the random variable $X_i = 1$ if I get a head, and $X_i = 0$ if I get a tail.  Then $$\Pr[X_i = 1] = 1/2, \quad \Pr[X_i = 0] = 1/2,$$ and because all the coins are fair and independent of each other, the random variables $X_1, X_2, X_3, X_4, X_5$ representing the outcome of flips of each coin, are iid.  But that does NOT mean that $X_1 = X_2 = X_3 = X_4 = X_5$, because random variables are random, and represent outcomes from a random process.  You could certainly have the first coin land heads but the fifth coin land tails.  The fact that they are identically distributed is a reflection of the fact that we are modeling their random behavior in the same fashion, not that their outcomes are linked to another.
Now that this is cleared up, we can consider the expected value:  clearly, $$\operatorname{E}[X_i] = 0 \Pr[X_i = 0] + 1 \Pr[X_i = 1] = 0(1/2) + 1(1/2) = 1/2,$$ and $$\begin{align*} \operatorname{Var}[X_i] &= \operatorname{E}[(X_i - \operatorname{E}[X_i])^2] \\ &= (0 - 1/2)^2 \Pr[X_i = 0] + (1 - 1/2)^2 \Pr[X_i = 1] \\ &= (1/4)(1/2) + (1/4)(1/2) \\ &= 1/4. \end{align*}$$
So we see that each coin has the same expectation and variance as the others (after all, they are iid).  The expectation of the sum is, by linearity of expectation, is simply $$\operatorname{E}[X_1 + X_2 + X_3 + X_4 + X_5] = 5(1/2) = 5/2.$$  Although the variance is not in general a linear operator, when the variables are independent, variance is also linear.  Therefore, $$\operatorname{Var}[X_1 + X_2 + X_3 + X_4 + X_5] = 5(1/4) = 5/4.$$  Now, the expectation of the sample mean $\bar X = (X_1 + X_2 + X_3 + X_4 + X_5)/5$ is simply $$\operatorname{E}[\bar X] = (5/2)/5 = 1/2,$$ as we would expect, but the variance of the sample mean is $$\operatorname{Var}[\bar X] = \operatorname{Var}[(X_1 + X_2 + X_3 + X_4 + X_5)/5] = \frac{1}{5^2} \operatorname{Var}[X_1 + X_2 + X_3 + X_4 + X_5] = \frac{5/4}{25} = \frac{1}{20}.$$  This is because if $c$ is a scalar constant, $$\begin{align*} \operatorname{Var}[cX] &= \operatorname{E}[(cX - \operatorname{E}[cX])^2] \\ &= \operatorname{E}[(cX - c \operatorname{E}[X])^2] \\ &= \operatorname{E}[c^2(X - \operatorname{E}[X])^2] \\ &= c^2 \operatorname{E}[(X - \operatorname{E}[X])^2] \\ &= c^2 \operatorname{Var}[X]. \end{align*}$$  So when we scale a random variable by some constant $c$, then the variance is scaled by $c^2$.
A: Taking a random sample $X_1, X_2, \dots, X_n$ from a population with mean $\mu$ and
variance $\sigma^2$ means that the $X_i$ are independent and that
$E(X_i) = \mu$ and $V(X_i) = \sigma^2.$ All of these random variables
have the same variance because they represent observations from
the same population.
Consequently, defining 
$$\bar X = \frac{1}{n}\sum_{i=1}^n X_i = \frac{X_1 + X_2 + \cdots + X_n}{n},$$
one has $E(\bar X) = \mu$ and $V(\bar X) = \sigma^2/n.$
It seems you are trying to understand the proof for $V(\bar X)$, which uses the assumption of independence.
This proof is given in your link. (If there is a step in that
you don't understand, please leave a Comment.)
I see that two other Answers have been posted while I was typing
this. The Answer by @ConradoCosta shows two RVs with the same variance (but not because of random sampling); I up-voted it and left a Comment there with yet another one. I hope one of the three Answers is helpful.
