If $X_1,X_2,…,X_n∼Bernoulli(p)$
Variance of the average of $X$ is
$Var[S_x/n]=\frac{p(1−p)}{n}$
But if we have sample, where all $X$ are equal, $\hat{p}=1$ (or zero), and estimation of var of mean became equal to 0. Is it correct?
If $X_1,X_2,…,X_n∼Bernoulli(p)$
Variance of the average of $X$ is
$Var[S_x/n]=\frac{p(1−p)}{n}$
But if we have sample, where all $X$ are equal, $\hat{p}=1$ (or zero), and estimation of var of mean became equal to 0. Is it correct?
When $p=1$ or $p=0$, variance of each individual $X_i$, mean of them, sum of them, or any deterministic function of them will be $0$ because now $X_i$'s are certain (they become constant), i.e. you know what's going to happen. For example, if $p=1$, then all $X$ will be $1$, and the average will still be $1$. One can tell you the result without uncertainty as I did.