# sampling: question about variance estimation

I'm new in this field, I'm studying montecarlo methods for sampling; I understand that,with a basic approach, we have an estimation for the mean $$\hat{\mu}=\frac{1}{n}\sum f(x)$$ and for variance $$\hat{\sigma}^2=\frac{\sigma^2}{n}$$ so I read that, while $n\to +\infty,\hat{\sigma}^2\to 0$ and this I don't understand... For example, we can imagine that I have a population that follows a Normal distribution as $$N(10,3)$$ Now, I want to use MC to estimate this distribution; more sample I take from this distribution, more little becomes my estimated variance, going to $0$...but my variance estimation should go to $3$ to be a good estiamation of the real $\sigma^2$... where is my error or my misunderstanding?

• $\dfrac{\sigma^2}{n}$ is the variance of the mean of $n$ i.i.d. samples from a distribution with variance $\sigma^2$ (and similarly $n \sigma^2$ is the variance of the sum). As $n$ increases, you increasingly expect the sample mean to be close to the population mean – Henry Oct 9 '17 at 9:53
• so, $\frac{\sigma^2}{n}$ is not $Var(f(x))$ but $Var(\hat{\mu})$? this is my error, right? – volperossa Oct 9 '17 at 10:00
• @volpeross this is closely related to standard error of the mean. en.wikipedia.org/wiki/Standard_error (but variance instead of standard deviation) – German Demidov Oct 9 '17 at 10:54
• @volperossa. I would prefer to say that for a sample you have $\text{Var}\left(\frac1n \sum_i X_i \right) = \frac{\sigma^2}{n}$, as I am not totally clear what you mean by $f(x)$ – Henry Oct 9 '17 at 11:23