Prove $Y_n:=\sup|\hat{F}_n(x)-F(x)|$ is a reverse submartingale

Question

Suppose $\{X_j,j \ge 1\}$ are iid with common distribution $F$ and let $\hat{F}_n$ be the empirical distribution based on $X_1,\dots,X_n$. Show $$Y_n:=\sup|\hat{F}_n(x)-F(x)|$$ is a reverse submartingale.

Hint: Consider first $\{\hat{F_n}-F(x), n \ge 1\}$, then take absolute values, and then take the supremum over a countable set.

Above is the problem. I think we should use $$\hat{F}_n(x)=\frac{1}{n}\sum 1_{\{X_i \le x \}}.$$ But I could not find a proper $B_n$ s.t. $E(\hat{F}_n(x)|B_{n+1})=\hat{F}_{n+1}(x)$.

Answer 1

Lemma Let $x \in \mathbb{R}$ be fixed. $(\hat{F}_n(x))_{n \in \mathbb{N}}$ is a reverse martingale with respect to the reverse filtration $(\mathcal{B}_n)_{n \in \mathbb{N}}$ defined by $$\mathcal{B}_n := \sigma(\hat{F}_n(y), y \in \mathbb{R}; X_{n+1},X_{n+2},\ldots).$$

Proof: First of all, note that $(\mathcal{B}_n)_{n \in \mathbb{N}}$ defines a reverse filtration, i.e. $\mathcal{B}_n \supseteq \mathcal{B}_{n+1}$. Indeed: The $\mathcal{B}_n$-measurability of $\hat{F}_n(x)$, $x \in \mathbb{R}$, and $X_{n+1}$ entails the $\mathcal{B}_n$-measurability of $\hat{F}_{n+1}(x)$; hence, $\mathcal{B}_{n+1} \subseteq \mathcal{B}_n$.

Let $1 \leq j \leq n$. Since the random variables $X_1,\ldots,X_n$ are iid and independent of $(X_{i})_{i \geq n+1}$, we find

$$\mathbb{E}(1_{X_j \leq x} \mid \mathcal{B}_n) = \mathbb{E}\big(1_{X_j \leq x} \mid \sigma(\hat{F}_n(y); y \in \mathbb{R})\big) = \mathbb{E}\big(1_{X_1 \leq x} \mid \sigma(\hat{F}_n(y); y \in \mathbb{R})\big). \tag{1}$$

Consequently,

$$\begin{align*} \hat{F}_n(x) &= \mathbb{E}\big(\hat{F}_n(x) \mid \sigma(\hat{F}_n(y); y \in \mathbb{R})\big) \\ &= \frac{1}{n} \sum_{j=1}^n \mathbb{E}\big(1_{X_j \leq x} \mid \sigma(\hat{F}_n(y); y \in \mathbb{R})\big) \\ &\stackrel{(1)}{=} \mathbb{E}\big(1_{X_1 \leq x} \mid \sigma(\hat{F}_n(y); y \in \mathbb{R})\big) = \mathbb{E}(1_{X_1 \leq x} \mid \mathcal{B}_n). \end{align*}$$

Therefore, we see from the tower property that $(\hat{F}_n(x),\mathcal{B}_n)_{n \in \mathbb{N}}$ is a reverse martingale.

Theorem The sequence $$Y_n := \sup_{x \in \mathbb{R}} |\hat{F}_n(x)-F(x)|$$ defines a reverse submartingale with respect to the filtration $(\mathcal{B}_n)_{n \in \mathbb{N}}$.

Proof We have already shown that $(\hat{F}_n(x)-F(x))_{n \in \mathbb{N}}$ is a reverse martingale with respect to $(\mathcal{B}_n)_{n \in \mathbb{N}}$. By Jensen's inequality, this implies that $(|\hat{F}_n(x)-F(x)|)_{n \in \mathbb{N}}$ is a reverse submartingale. As $\hat{F}_n$, $F$ are left-continuous, we have

$$Y_n = \sup_{x \in \mathbb{Q}} |\hat{F}_n(x)-F(x)|$$

and therefore $Y_n$ is $\mathcal{B}_n$-measurable (since it is the supremum of countably many $\mathcal{B}_n$-measurable random variables). Moreover,

$$\begin{align*} \mathbb{E}(Y_n \mid \mathcal{B}_{n+1}) &= \mathbb{E} \left( \sup_{x \in \mathbb{R}} |\hat{F}_n(x)-F(x)| \, \bigg| \, \mathcal{B}_{n+1} \right) \\ &\geq \mathbb{E} \left( |\hat{F}_n(x)-F(x)| \mid \mathcal{B}_{n+1} \right) \\ &\geq |\hat{F}_{n+1}(x)-F(x)|. \end{align*}$$

Taking the supremum on the right-hand side yields

$$\mathbb{E}(Y_n \mid \mathcal{B}_{n+1}) \geq \sup_{x \in \mathbb{R}} |\hat{F}_{n+1}(x)-F(x)| = Y_{n+1}.$$