Let $X_1,\ldots X_n\sim Bern(p)$ be Bernoulli random variables and $x_1\ldots x_n$ their realizations. Suppose I want to find the variance of the estimator $\overline X=\frac{X_1+\ldots X_n}{n}$. If we don't know the parameter $p$, one way to do this is using the formula:

$var(\bar X) = \frac{s^2}{n}$, where $s=\frac{\sum_{i=1}^n (x_i-\bar X)^2}{n}$

However, if I understood well, in this slide of this class the professor uses Slutsky theorem to find this formula for the variance replacing the true parameter $p$ by the estimate $\hat p$:

$var(\bar X)=\frac{\hat p(1-\hat p)}{n}$

where $\hat p = \frac{x_1+\ldots x_n}{n}$.

I want to know which formula I should use in this case


1 Answer 1


You can use both because they're the same:

$$\begin{align}\hat{\operatorname{var}(\bar X)}&=\frac{\sum (x_i-\bar x)^2}{n^2}=\frac{\sum x_i^2-2\sum x_i\bar x+\sum\bar x^2}{n^2}\\&=\frac{\sum x_i-2\hat p\sum x_i+n\hat p^2}{n^2}=\frac{n\hat p-2\hat p n \hat p+n\hat p^2}{n^2}\\&=\frac{\hat p-\hat p^2}{n}=\frac{\hat p (1-\hat p)}{n}\end{align}$$

Note that, the key ideas here are $\hat p=\bar x$ and $x_i^2=x_i$ for a Bernoulli random variable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.