TITLE: Statistics: Why do we divide by $\sqrt{n}$ for sample standard deviation
QUESTION [7 upvotes]: Can someone tell me if my explanations/understanding is on the right track? 
Suppose we have a set of variances, each of them identical, where $V_{1}(x) + V_{2}(x) + ...+ V_{j}(x) = \sigma^2$. If we wanted to take $\frac{1}{n}$-th ($n \leq j$) of the variances we would have $\frac{\sigma^2}{n}$, taking the square root this becomes $\frac{\sigma}{\sqrt{n}}$ 
My other idea is to let $V_{1}(x) = V_{2}(x) = ...=V_{j}(x) = \sigma^2$ so summing these all together we would get $j\sigma^2$ so the standard deviation is $\sqrt{j} \sigma$ and taking the average we would get $\frac{\sigma}{\sqrt{j}}$

REPLY [3 votes]: Your second idea is correct if you assume your $V$'s are independent.  Independence is the important property, because variance is then additive, as you show, and then the rest of your reasoning is correct.  If your $V$'s are not independent, then you won't have additivity of variance because there could be covariance between the observations.