Let $X_1,...,X_n$ be i.i.d. R.V. with variance $\sigma^2$ Is the $$\text{Var}{\frac{X_1+...+X_n}{n}}=\frac{\sigma^2}{n}$$ or rather $$\text{Var}{\frac{X_1+...+X_n}{n}}=\frac{\sigma^2}{n-1}$$ ?
I just remember that sometimes $n$ is decreased to $n-1$ by loosing one degree of freedom when computing moment's, is this that case ?
It's the first. When you compute the variance you get with $\bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i$ that $$ var[\bar{X}_n] = \frac{1}{n^2} \sum_{i=1}^n var[X_i] = \frac{n}{n^2} var[X_1] = \frac{\sigma^2}{n}, $$ where $\sigma^2$ is the variance of the $X_i$. The reduced $n-1$ in the denominator usually comes into play when estimating the variance of the $X_i$. In this case $$ \widehat{\sigma}^2 = \frac{1}{n-1} \sum_{i=1}^n(X_i - \bar{X}_n)^2 $$ is an unbiased estimator for $\sigma^2$, i.e., $\mathbb{E}[\widehat{\sigma}^2] = \sigma^2$. Another place where $n-1$ can be found is in the construction of confidence intervals for the mean when using a t-statistic.