Upcrossing Inequality

Last updated Nov 1, 2022

Let $(\Omega, \mathcal{B}, \mathbb{P})$ be a Probability Space. Let $(X_{n}){n \geq 1}$ be a discrete-time Submartingale wrt Discrete-Time Filtration $\mathcal{F}{*} := (\mathcal{F}{n} \subset \mathcal{B}){n \in \mathbb{N}}$. Then $\forall a < b \in \mathbb{R}$, $$(b-a)\mathbb{E}(U_{n}[a,b]) \leq \mathbb{E}(X_{n} - a)^{+} - \mathbb{E}(X_{1} - a)^{+}$$

Let $(H_{n}){n \geq 1}$ be the Upcrossing Strategy for $(X{n})_{n \geq 1}$.

We let $Y_{n} = a + (X_{n} - a)^{+}$ for $n \geq 1$. Because of Submartingale Relations from Jensen and convexity of $(\cdot)^{+}$, we have that $(Y_{n}){n \geq 1}$ is a Submartingale. The only difference between $X{n}$ and $Y_{n}$ is that $Y_{n} = a$ when $X_{n} \leq a$ for all $n \geq 1$. Therefore $(Y_{n}){n \geq 1}$ has a the same number of Complete Upcrossings as $(X{n}){n \in \mathbb{N}}$. Since $(Y{n}){n \geq 1}$ cannot dip below $a$, we have that $$(b-a)U{n}[a,b] \leq (H \cdot Y){n} = \sum\limits{m=1}^{n} H_{m}(Y_{m} - Y_{m-1})$$ with $Y_{0} = 0$ as in the definition of winnings, since $H_{n} = 1$ If and Only If $Y_{n}$ is rising up to $b$ after a fall from above $b$ down to $a$ for $n \geq 2$. That is, we will include a $(b-a)$ for each upcrossing in the sum along with any extra that comes from breaking above $b$ and any remaining incomplete upcrossing. Furthermore, for $n \geq 1$ $$Y_{n} = \sum\limits_{m=1}^{n} Y_{m} - Y_{m-1} = \sum\limits_{m=1}^{n} (H_{m} - (1 - H_{m}))(Y_{m} - Y_{m-1}) = (H \cdot Y){n} + ((1-H) \cdot Y){n}$$ denote $K := 1-H$. Thus, $\mathbb{E}(Y_{n})= \mathbb{E}((H \cdot Y){n}) + \mathbb{E}((K \cdot Y){n})$. Since $K$ is just another Strategy, Nonnegative Bounded Strategy on Submartingale is a Submartingale tells us that $((K \cdot Y){n}){n \geq 1}$ is another Submartingale. Then, we have that $\mathbb{E}((K \cdot Y){n}) \geq \mathbb{E}((K \cdot Y){1})$ since Submartingales have Non-Decreasing Expectation. Note that $H_{1} = 0$ since all $N_{2k-1} \geq 1$ for $k \geq 1$. Thus $$(K \cdot Y){1} = Y{1} - Y_{0} = Y_{1}.$$ Therefore, $\mathbb{E}(Y_{n}) - \mathbb{E}(Y_{1}) \geq \mathbb{E}(H \cdot Y){n}$. Putting it together with above, we get $$(b-a)\mathbb{E}(U{n}[a,b]) \leq \mathbb{E}((H \cdot Y){n}) \leq \mathbb{E}(Y{n}) - \mathbb{E}(Y_{1})$$ Substituting our original expression for $Y_{n}$, $$(b-a)\mathbb{E}(U_{n}[a,b]) \leq a + \mathbb{E}(X_{n} - a)^{+} - a - \mathbb{E}(X_{1} - a)^{+} = \mathbb{E}(X_{n} - a)^{+} - \mathbb{E}(X_{1} - a)^{+}$$ $\blacksquare$