Statistics: Why do we divide by $\sqrt{n}$ for sample standard deviation

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 $\frac{\sigma}{\sqrt{j}}$

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Neither description is clear. You have not described what random variables these are variances of, and whether these are independent or not. – André Nicolas Aug 7 '13 at 0:05
Lets say that you are rolling a die so $x$ = the # that shows up when you roll a die. Repeat the experiment $j$ times. Now record only $\frac{1}{n}$-th of the experiments. This is what I was thinking of for the first explanation. If anyone has an alternative explanation, I wouldn't mind hearing it. – Person Aug 7 '13 at 0:13
Should use $X_1,X_2,\dots,X_n$. You probably want to assume independence, and want the sample mean $(X_1+\cdots+X_n)/n$. – André Nicolas Aug 7 '13 at 0:25

1 Answer

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.

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