Wasserman's AoS, Chapter 7, Question 5
We show that the covariance \(\text{Cov} (\hat F_n(x), \hat F_n(y))\) is \[ \frac{1}{n} F(x) \left( 1 + F(y) \right) , \] for \(x \le y\). For \(x \ge y\), we have \(\frac{1}{n} F(y) \left( 1 + F(x) \right)\).
Using linearity of expectation and the fact that \(\mathbb E \hat F_n(y) = F(y)\), we obtain
\[ \mathbb E (\hat F_n(x) - F(x)) (\hat F_n(y) - F(y)) = \mathbb E (\hat F_n(x)\hat F_n(y)) - F(x)F(y) . \]
By definition of the empirical cumulative distribution function, the first term is equal to
\[ \frac{1}{n^2} \sum_{i, j} \mathbb E (I(X_i \le x)I(X_j \le y)) = \frac{1}{n^2} \left( \sum_{i = j} \mathbb E (I(X_i \le x)I(X_i \le y)) + \sum_{i \ne j} \mathbb E (I(X_i \le x)I(X_j \le y)) \right) . \]
We now simplify the two summations:
Without loss of generality, we may assume that \(x \le y\). With this assumption, \(I(X_i \le x)I(X_i \le y) = I(X_i \le x)\).
Note that the Bernoulli variables \(I(X_i \le x)\) and \(I(X_j \le y)\) are independent for \(i \ne j\) since \(X_i\) and \(X_j\) are independent. This implies that the expectation of the product is the product of the expectations.
These two facts allow us to rewrite the first term as
\[ \begin{align} \frac{1}{n^2} \left( \sum_{i = j} \mathbb E (I(X_i \le x)) + \sum_{i \ne j} \mathbb E (I(X_i \le x)) \mathbb E (I(X_j \le y)) \right) &= \frac{1}{n^2} \left( \sum_{i = j} F(x) + \sum_{i \ne j} F(x)F(y) \right) \\ &= \frac{1}{n} F(x) ( 1 + (n-1) F(y) ) \end{align} . \]
Putting everything together, we see that the covariance is
\[ \begin{align} \frac{1}{n} F(x) (1 + (n-1) F(y)) - F(x)F(y) &= \frac{1}{n} F(x) (1 - F(y)) \end{align} \]
for \(x \le y\).