Bounded Strategy on Martingale is a Martingale

Last updated Nov 1, 2022

a

First note that for $n \geq 1$ because $H_{n}$ is bounded, say by $M \in \mathbb{R}{\geq 0}$ $$\begin{align*} \mathbb{E}|(H \cdot X){n}| &= \mathbb{E}\left|\sum\limits_{m=1}^{n}H_{m}(X_{m} - X_{m-1})\right|\\ &\leq M \sum\limits_{m=1}^{n} \mathbb{E}|X_{m} - X_{m-1}| & \text{(Triangle Inequality)}\\ &\leq M \sum\limits_{m=1}^{n} \mathbb{E}|X_{m}| + \mathbb{E}| X_{m-1}| & \text{(Triangle Inequality)}\\\ &< \infty \end{align*}$$ since $X_{m} \in L^{1}(\mathcal{B})$ $\forall m \geq 0$. So $(H \cdot X)_{n} \in L^{1}(\mathcal{B})$ as well. $\checkmark$

Next, for $n \geq 1$, we have that $X_{m}, H_{m+1} \in \mathcal{F}{n}$ for all $0 \leq m \leq n$, so $$(H \cdot X){n} = \sum\limits_{m=1}^{n} H_{m} (X_{m} - X_{m-1}) \in \mathcal{F}_{n}$$ $\checkmark$

Observe that for $n \geq 0$ $$\begin{align*} \mathbb{E}((H \cdot X){n+1} | \mathcal{F}{n}) &= \mathbb{E}(\sum\limits_{m=1}^{n+1} H_{m}(X_{m} - X_{m-1}) | \mathcal{F}{n})\\ &=\sum\limits{m=1}^{n+1}\mathbb{E}(H_{m}(X_{m} - X_{m-1}) | \mathcal{F}{n})\\ &=\sum\limits{m=1}^{n}H_{m}(X_{m} - X_{m-1}) - \mathbb{E}(H_{n+1} (X_{n+1} - X_{n}) | \mathcal{F}{n})\\ &= (H \cdot X){n} \end{align*}$$ since $$\begin{align*} \mathbb{E}(H_{n+1} (X_{n+1} - X_{n}) | \mathcal{F}{n}) &= H{n+1} (\mathbb{E}(X_{n+1} | \mathcal{F}{n}) -X{n})\\ &= 0. \end{align*}$$ $\checkmark$ $\blacksquare$

1. This theorem nicely states that any bounded gambling strategy cannot make an unfavorable game favorable.