TITLE: Derivation of standard error of mean
QUESTION [6 upvotes]: I was going through this wikipedia article on standard error. I could not understand the crucial step here. It goes like this:
This formula may be derived from what we know about the variance of a sum of independent random variables.
If $X_1, X_2 , \ldots, X_n$ are n independent observations from a population that has a mean $\mu$ and standard deviation $\sigma$ , then the variance of the total 
$T = (X_1 + X_2 + \cdots + X_n)$ is $n\sigma^2$. Understood.
The variance of T/n must be $\frac{1}{n^2}n\sigma^2=\frac{\sigma^2}{n}$. Not understood.
And the standard deviation of T/n must be $\sigma/{\sqrt{n}}$ .
Of course, T/n is the sample mean $\bar{x}$ .
I went to some basics:
$\displaystyle Var(X)=\frac{1}{n}\sum_{i=1}^{n}({x_i-\mu})^2$
$\displaystyle Var(X)=\frac{1}{n}\sum_{i=1}^{n}({x_i^2+\mu^2-2x_i\mu})$
$\displaystyle Var(X)=\frac{1}{n}\sum_{i=1}^{n}x_i^2+\mu^2-\frac{2}{n}\sum_{i=1}^{n}x_i\mu$
As Sample mean is an unbiased estimate of population mean, we get 
$\displaystyle Var(X)=\frac{1}{n}\sum_{i=1}^{n}x_i^2-\mu^2$
$\displaystyle Var(X)=E(X^2)-(E(X))^2$ 
Nothing useful found from this.
Why is there a $1/n^2$ in that step to get variance ?

REPLY [3 votes]: The only thing we need to prove here is that for any scalar constant $c$, and for a random variable $X$, $$\mathrm{Var}[cX] = c^2 \mathrm{Var}[X].$$  This follows from the property of expectation $$\mathrm{E}[cX] = c\mathrm{E}[X]$$ as follows:  $$\begin{align*} \mathrm{Var}[cX] &= \mathrm{E}[(cX - \mathrm{E}[cX])^2] \\ &= \mathrm{E}[(cX - c\mathrm{E}[X])^2] \\ &= \mathrm{E}[c^2(X - \mathrm{E}[X])^2] \\ &= c^2 \mathrm{E}[(X - \mathrm{E}[X])^2] \\ &= c^2 \mathrm{Var}[X]. \end{align*}$$