Why is this process bounded?

Let $$X$$ be a random variable such that $$P(X=1) = p$$ and $$P(X = -1) =1-p$$. Let $$(X_j)_{j \in \mathbb{N}}$$ be a sequence of independent copies of $$X$$. We define a process $$M = (M_n)_{n \in \mathbb{N}}$$ by $$M_0 = 0$$ and $$M_n = \sum_{j=1}^n X_j$$ for $$n > 0$$. Pick integers $$k < 0 < m$$ and let

$$T( \omega) = \inf \{ n | M_n( \omega) =k \text{ or } M_n( \omega) = m \}.$$

We know that $$T$$ is a stopping time and that $$T$$ is almost surely finite. We want to use the Optional Stopping for Bounded Processes Theorem, so we define a new process $$Y$$ by $$Y_0 = M_0$$ and $$Y_n = M_{n \wedge T}$$ for $$n > 0$$. We know that $$Y$$ is a (sub/super)-martingale exactly when $$M$$ is. At this point it is supposedly obvious that $$Y$$ is bounded, but I struggle to see why. If we knew that $$T$$ were bounded by for example $$K$$, we would have

$$|Y_n| \leq \sum_{j=1}^K |X_j| \leq K,$$

but we only know that $$T$$ is finite, so this would not work. How to we show that $$Y$$ is bounded?

To clarify, I want $$Y$$ to be uniformly bounded in $$n$$. So that I can use the following theorem:

If $$Y$$ is a bounded submartingale and $$S \leq T < \infty$$ are finite stopping times, $$(Y_S, Y_T)$$ is a $$( \mathcal{F}_S, \mathcal{F}_T)$$-submartingale.

• Bounded in what sense? Why do you want $Y$ to be bounded? Nov 25, 2021 at 10:17
• I have now edited the question to specify this point. Sorry, I should have been clearer about that. Nov 25, 2021 at 10:26
• I just looked it up. It seems that your type of random variable $X$ is only called Rademacher distributed for $p=1/2$. Maybe just let the terminology of the distribution away. It is enough to say $X \sim p\delta_1 + (1-p)\delta_{-1}$. Nov 25, 2021 at 10:46

To your question: $$\vert Y_n \vert = \vert M_{n\wedge T}\vert = \vert M_n \vert 1_{\{T > n\}} + \vert M_T \vert 1_{\{T \leq n\}}$$
We have $$\{T > n \} = \{ k< M_\ell < m \;\forall \ell \in\{0, \ldots , n\}\}$$
and $$M_T \in \{k,m\}$$ almost surely. Thus
\begin{aligned}&\vert M_n \vert 1_{\{T > n\}} + \vert M_T \vert 1_{\{T \leq n\}}\\ &\leq \max ( - k , m) 1_{\{T > n\}} + \max(-k , m)1_{\{T \leq n\}} \\&\leq \max(-k , m) \end{aligned}