Help with a proof regarding empirical CDF Given $X_1,X_2...X_n \sim F$ and $(\hat{F}_n(x))$ is an empirical CDF
I need to prove that $$Var(\hat{F}_n(x)) = \frac{F(x)(1 - F(x))}{n}$$
So what I did is:
The variance of an estimator $\hat{F}_n(x)$ of a population CDF $F(x)$ is this: (if I'm not mistaken)
$$Var(\hat{F}_n(x)) = E[(\hat{F}_n(x) - E[\hat{F}_n(x)])^2]$$
Since $\hat{F}_n(x)$ is an estimator of the population CDF $F(x)$, we can substitute $F(x)$ for $E[\hat{F}_n(x)]$ in the above equation to get:
$$Var(\hat{F}_n(x)) = E[(\hat{F}_n(x) - F(x))^2]$$
We can then use the fact that $\hat{F}_n(x)$ is a Bernoulli random variable with probability of success equal to $F(x)$ to expand the square and simplify the expression:
$$Var(\hat{F}_n(x)) = E[\hat{F}_n(x)^2 - 2\hat{F}_n(x)F(x) + F(x)^2]$$
Since it's a bernoulli we get that $\hat{F}_n(x)^2 = \hat{F}_n(x)$ and we know that $E[\hat{F}_n(x)] = F(x)$, therefore we plug the new values:
$$Var(\hat{F}_n(x)) = F(x) - 2F(x)^2 + F(x)^2 = F(x)(1 - F(x))$$
And this is where I get to a dead end, I get this formula but what I need to prove is this formula but with the right side divided by $n$, and I didn't get this. What did I do wrong?
EDIT-
After reading your comments I tried adjusting the approach a little bit, I did the following, I hope I'm right:
First we find the variance of $\mathop{\hat{F}_n}\left(x\right)$.
The variance of a random variable is defined as $Var(X) = E[(X - E[X])^2]$.
Substituting $\mathop{\hat{F}_n}\left(x\right)$ for $X$, we have:
$$Var(\hat{F}_n(x)) = E[(\hat{F}_n(x) - E[\hat{F}_n(x)])^2]$$
We can rewrite $\hat{F}_n(x)$ as $\frac{R_n(x)}{n}$, where $R_n(x) \sim \text{Binomial}(n, F(x))$, and that is because empirical CDF is the average of Bernoulli random variables. Therefore,
$$Var(\hat{F}_n(x)) = E\left[\left(\frac{R_n(x)}{n} - E\left[\frac{R_n(x)}{n}\right]\right)^2\right]$$
$$Var(\hat{F}_n(x)) = E\left[\frac{R_n(x)^2}{n^2}\right] - \left(E\left[\frac{R_n(x)}{n}\right]\right)^2$$
Since $R_n(x)$ is a binomial random variable, we can use the formula for the variance of a binomial random variable, which is $Var(R_n(x)) = nF(x)(1 - F(x))$. Therefore,
$$Var(\hat{F}_n(x)) = \frac{nF(x)(1 - F(x))}{n^2} - \left(\frac{nF(x)}{n}\right)^2$$
After simplifying we get:
$$Var(\hat{F}_n(x)) = \frac{F(x)(1 - F(x))}{n}$$
 A: Define $$Y_i(x)=\mathbb I_{\{X_i\leq x\}}$$ $\forall i\in\{1, 2,\ldots, n\}.$
Notice $$Y_i(x) \overset{\text{iid}}{\sim}\mathcal{Ber}(\theta)\tag 1\label 1$$ where $\theta := F(x) . $
Now express (how?) $$n \hat F_n(x) =\sum_{i=1}^n Y_i(x) ;\tag 2$$
Use $\eqref 1$ above to yield $\operatorname{Var}(F_n(x)). $
A: Note that you can write $\mathop{\hat{F}_n}\left(x\right)$ as $\mathop{\hat{F}_n}\left(x\right) = \mathop{R_n}\left(x\right)/n$, where $\mathop{R_n}\left(x\right) \sim \mathop{\text{Binomial}}\left(n, \mathop{F}\left(x\right)\right)$.
Proof.

$\mathop{R_n}\left(x\right) \mathrel{:=}\sum_{i=1}^n \mathop{\mathbf{1}_{\left(-\infty,\, x\right]}}\left(X_i\right)$ counts the number of successes, meaning the number of $X_i$s in $\left(-\infty, x\right]$, in $n$ independent Bernoulli trials with success probability $\mathop{\mathbb{P}}\left(X_i \in \left(-\infty, x\right] \right)  = \mathop{F}\left(x\right)$ each. Hence, $\mathop{R_n}\left(x\right) \sim \mathop{\text{Binomial}}\left(n, \mathop{F}\left(x\right)\right)$. By definition, $\mathop{\hat{F}_n}\left(x\right) = n^{-1}\sum_{i=1}^n \mathop{\mathbf{1}_{\left(-\infty,\, x\right]}}\left(X_i\right)$ and the statement follows.

