Nonnegative Bounded Strategy on Submartingale is a Submartingale

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 $(\mathcal{F}{n}){n=0}^{\infty}$-Submartingale 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 Submartingale.

Recall that An Adapted Process is a Supermartingale iff its negative is a Submartingale, so $({-X}{n}){n=0}^{\infty}$ is a Supermartingale. Then by Nonnegative Bounded Strategy on Supermartingale is a Supermartingale, we have that $(H \cdot -X){n}$ is a Supermartingale for $n \geq 1$. But for $n \geq 1$: $$\begin{align*} (H \cdot -X){n} &= \sum\limits_{m=1}^{n} H_{m} (-X_{m} + X_{m-1})\\ &=-\sum\limits_{m=1}^{n} H_{m} (X_{m} - X_{m-1})\\ &=-(H \cdot X){n}. \end{align*}$$ Again, because An Adapted Process is a Supermartingale iff its negative is a Submartingale we have that $(H \cdot X){n}$ for $n \geq 1$ is a Submartingale $\blacksquare$