Explanation on the different forms of proportion standard error and population variance of proportion (Stats)

Question

Let $\sigma^2$ be the population variance, $p$ be the population proportion, and $\hat{p}$ be the sample proportion. My understanding is that $\sigma^2_p = p(1-p)$ (bernoulli trial) and the standard error is $$S_\hat{p} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ It doesn't make sense to me why do the terms standard deviation and standard error have $\sqrt{\frac{1}{n}}$ inside them while the variance doesn't have one. Is there anything I'm missing? Perhaps some similar behavior with the variance of the mean or some properties of binomial?

Answer 1

A sum of iid Bernouilli variables $X_i$ is a Binomial. The variance of the Binomial distribution is $Var[\sum X_i] = n p (1-p)$. Hence it is immediate that the variance of the mean of iid Bernouillis is $Var[\sum X_i / n] = Var[\sum X_i] / n^2 = n p (1-p) / n^2 = p (1-p) / n$. And the standard deviation is the square root of the variance, $\sqrt{p (1-p) / n}$