Doob decomposition for bounded submartingales

Question

Assume $X_n$ is a uniformly bounded submartingale, i.e. there is a constant $K>0$ so that $|X_n|<K$ for all $n, \omega$. Let $X_n=M_n+A_n$ be the Doob decomposition. Can we conclude $M_n$ and $A_n$ are also uniformly bounded?

Answer 1

No, it's in general not true.

Let $S_n = \sum_{j=1}^n \xi_j$ be a simple random walk, i.e. $(\xi_j)_j$ are iid random variables with $$\mathbb{P}(\xi_j=1) = \mathbb{P}(\xi_j=-1)=\frac{1}{2}.$$ Clearly, $S_n^2$ is a submartingale and its compensator is given by $\langle S \rangle_n = n$. Now define

$$\tau := \inf\{n \in \mathbb{N}; |S_n|=2\},$$

then $X_n := S_{n \wedge \tau}^2$ is a submartingale which is uniformly bounded and its compensator equals

$$A_n =(n \wedge \tau).$$

If $(A_n)_{n \in \mathbb{N}}$ was uniformly bounded, say $A_n \leq M$ for some $M>0$ (not depending on $n$ and $\omega$), then this would imply $\tau \leq M$ almost surely... which is absurd because

$$\begin{align*} \mathbb{P}(\tau \geq 2n) &\geq \mathbb{P}(|S_0| \leq 1,\ldots,|S_{2n}| \leq 1) \\ &\geq \mathbb{P}(\xi_1=1, \xi_2 = -1, \xi_3 =1, \xi_4=-1,\ldots,\xi_{2n-1}=1,\xi_{2n}=-1) \\ &= \frac{1}{2^n} > 0 \end{align*}$$

for any $n \in \mathbb{N}$.