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?


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:

$$Var(\hat{F}_n(x)) = \frac{F(x)(1 - F(x))}{n}$$

  • 3
    $\begingroup$ Recall the definition of the ecdf, which is an average of Bernoulli r.v.s $\endgroup$ Jan 3, 2023 at 10:43
  • $\begingroup$ Stating that "$\hat{F}_n(x)$ is an estimator of the population CDF $F(x)$" is not enough to conclude that "we can substitute $F(x)$ for $E[\hat{F}_n(x)]$". $\endgroup$
    – statmerkur
    Jan 3, 2023 at 11:00
  • $\begingroup$ @ChristophHanck So what should I add to my solution? I'm missing the point. $\endgroup$
    – CORy
    Jan 3, 2023 at 11:28
  • 2
    $\begingroup$ Basically what the answers that have come in by now say - recall that $Var(aX)=a^2Var(X)$. $\endgroup$ Jan 3, 2023 at 13:11
  • 1
    $\begingroup$ Because $n\hat F(x)$ counts how many independent copies of $X$ are in the interval $(-\infty, x],$ it has a Binomial distribution with parameters $n$ and $F(x)$ and you're done. $\endgroup$
    – whuber
    Jan 3, 2023 at 16:51

2 Answers 2


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)). $

  • 1
    $\begingroup$ concise and straight to the point (+1). Trivial note: the numbering (2) seems redundant, it is not actually used. $\endgroup$
    – utobi
    Jan 3, 2023 at 21:10
  • 1
    $\begingroup$ I numbered that for if OP needed, I would have expanded. $\endgroup$ Jan 3, 2023 at 22:16

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 $\mathop{R_n}\left(x\right) \sim \mathop{\text{Binomial}}\left(n, \mathop{F}\left(x\right)\right)$.


$\mathop{R_n}\left(x\right) \mathrel{:=}\sum_{i=1}^n \mathop{\mathbf{1}_{\left(-\infty,\, x\right]}}\left(X_i\right)$ 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.

  • $\begingroup$ That is $R_n(x)$ here? $\endgroup$
    – CORy
    Jan 3, 2023 at 11:26
  • 1
    $\begingroup$ @CORy Hint: It's a sum of $n$ i.i.d. indicator random variables. $\endgroup$
    – statmerkur
    Jan 3, 2023 at 11:30
  • $\begingroup$ So maybe $\hat{F}_n(x) = 1/n \sum i$ where i is the random variable? $\endgroup$
    – CORy
    Jan 3, 2023 at 11:58
  • 2
    $\begingroup$ @CORy what does your random variable $i$ stand for? $\endgroup$
    – statmerkur
    Jan 3, 2023 at 12:03
  • 1
    $\begingroup$ @CORy You already use $\text{Var}(R_n(x)) = nF(x)(1 - F(x))$ in your edited derivation. What does this tell you about $\text{Var}(R_n(x)/n)$? Do you need the other variance decomposition steps? $\endgroup$
    – statmerkur
    Jan 3, 2023 at 12:40

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