Nonnegative Bounded Strategy on Supermartingale is a Supermartingale

Last updated Nov 1, 2022

Let $(\Omega, \mathcal{B}, \mathbb{P})$ be a Probability Space and let $(\mathcal{F}{n}){n=0}^{\infty}$ be a Discrete-Time Filtration on $\mathcal{B}$. Let $(X_{n}){n=0}^{\infty}$ be a $(\mathcal{F}{n}){n=0}^{\infty}$-Supermartingale on $\Omega, \mathcal{B}$ and let $(H{n}){n=1}^{\infty}$ be a Strategy for $(X{n}){n=0}^{\infty}$. Then the winnings $(H \cdot X){n}$ for $n \geq 1$ forms a Supermartingale.

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

1. We use the boundedness of $H_{n}$ in the first step to pull it out of the Expectation and give us that $(H \cdot X){n} \in L^{1}(\mathcal{B})$. However, there are other constraints that get use the same result. For example, if we had $H{n}, X_{n-1} \in L^{2}(\mathcal{B})$ $\forall n \in \mathbb{N}$, then we could apply Cauchy-Schwarz Inequality to get that $(H \cdot X)_{n} \in L^{1}(\mathcal{B})$.
2. However, simply requiring $H_{n} \in L^{1}(\mathcal{B})$ is not enough since if $X_{n} = 1/\sqrt{x_{n}}$ for $x_{n} \sim [0,1]$ uniformly iid and we have $H_{n} = \frac{1}{\sqrt{x_{n-1}}}$, then $(H \cdot X){1} = \frac{1}{\sqrt{x{0}}} \cdot \frac{1}{\sqrt{x_{1}}} - \frac{1}{x_{0}}$. Since $$\begin{align*} \mathbb{E}\left(\frac{1}{\sqrt{x_{0}}} \cdot \frac{1}{\sqrt{x_{1}}}\right) &= \mathbb{E}\left(\frac{1}{\sqrt{x_{0}}} \right) \mathbb{E}\left(\frac{1}{\sqrt{x_{1}}}\right) & \text{ independence}\\ &= 4\\ \mathbb{E}\left(\frac{1}{x_{0}}\right) &= \infty \end{align*}$$ we must have that $\mathbb{E}|(H \cdot X)_{1}| = \infty$.