 Solution

Write $z^{-}=\max (-z,0)$. For $a<b$, let $U_N[a,b]$ be the upcrossing count obtained by successively buying at an observation at or below $a$ and selling at an observation at or above $b$, with the first purchase allowed at time zero. The Doob upcrossings lemma for an integrable discrete-time martingale isThe weaker last bound is also the usual Doob upcrossing inequality.

To prove it, hold one unit between each purchase and sale, and zero units otherwise. If $H_k$ records the holding over $(k,k+1]$, the decision is ${\mathcal{F}}_k$-measurable, with $H_k\in \{0,1\}$. The resulting martingale transform has gainEach completed upcrossing earns at least $b-a$. An unfinished holding loses at most $(X_N-a{)}^{-}$. If a purchase occurs at zero, its price is $X_0\le a$, giving the additional discount $(X_0-a{)}^{-}$, whether that holding has been sold or remains open. If no purchase occurs at zero, this extra term is zero. Therefore, path by path,Taking expectations proves the bound.

The Martingale convergence theorem, stated without proof, says that a discrete-time martingale satisfying ${\sup }_n\mathbb{E}|X_n|<\infty$ converges almost surely to a finite integrable random variable $X_{\infty }$. In particular every nonnegative martingale converges almost surely to a finite limit. Under uniform integrability, the uniformly integrable martingale convergence theorem additionally gives convergence in L1 and $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$.