 Solution

For a discrete-time martingale $(M_n)$ and real $a<b$, let $U_N[a,b]$ be the upcrossing count: buy on the first observation at or below $a$, sell on the next observation at or above $b$, and repeat up to time $N$, counting only completed pairs. Writing $z^{-}=max(-z,0)$, the Doob upcrossing inequality isThis convention permits the first purchase at time zero. Equivalently, the right side is $\mathbb{E}(M_N-a{)}^{+}-\mathbb{E}(M_0-a)$, because the martingale has constant expected value.

For completeness, let $H_k\in \{0,1\}$ indicate whether the trading rule holds one unit during $(k-1,k]$. This is a predictable process, so the martingale transform $G_N={\sum }_{k=1}^N{H}_k(M_k-M_{k-1})$ has zero expected value. Every completed trade gains at least $b-a$, and an unfinished trade loses at most $(M_N-a{)}^{-}$. Thus $G_N\ge (b-a)U_N[a,b]-(M_N-a{)}^{-}$, giving the inequality.