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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.

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  • $\begingroup$ Bounded in what sense? Why do you want $Y$ to be bounded? $\endgroup$ Nov 25, 2021 at 10:17
  • $\begingroup$ I have now edited the question to specify this point. Sorry, I should have been clearer about that. $\endgroup$ Nov 25, 2021 at 10:26
  • $\begingroup$ 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 $X \sim p\delta_1 + (1-p)\delta_{-1}$. $\endgroup$
    – Falrach
    Nov 25, 2021 at 10:46

1 Answer 1

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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

$$\begin{aligned}&\vert M_n \vert 1_{\{T > n\}} + \vert M_T \vert 1_{\{T \leq n\}}\\ &\leq \max ( - k , m) 1_{\{T > n\}} + \max(-k , m)1_{\{T \leq n\}} \\&\leq \max(-k , m) \end{aligned}$$

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