Martingale transform ($(A\cdot M{)}_n$)

If $M$ is a martingale and $A$ is predictable, then, subject to integrability,is a martingale.

For $Y_t={\sum }_{s\le t}{K}_s(M_s-M_{s-1})$ and fixed finite deterministic $T$, bounded events in ${\mathcal{F}}_{t-1}$ restricting both $Y_{t-1}$ and $K_t$ show that nonnegativity of $Y_t$ forces nonnegativity of $Y_{t-1}$. Induction makes the entire stopped process nonnegative. The nonnegative discrete-time local martingale is a martingale criterion then gives $\mathbb{E}{Y}_T=Y_0=0$. Nonnegative terminal gain is therefore zero almost surely, even without an intermediate wealth bound.

For a finite-valued predictable process $K$, this is a stopping time and increases to infinity. The stopped martingale transform uses coefficients $K_s{1}_{\{s\le {\sigma }_j\}}$, which are predictable and bounded by $j$. Thus a transform of a discrete-time martingale is a local martingale without a global boundedness assumption. The stop is before, not after, the first large coefficient is used.

For a square-integrable martingale transform and positive conditional increment variance,

For a stopping time $T$,The indicator is predictable, so a stopped integrable martingale is a martingale.

For a simple symmetric random walk, the infinite mean first passage of a simple symmetric random walk gives a finite almost sure first passage $T_1$ to $1$. The stopped martingale $M_n=1-S_{n\wedge T_1}$ is nonnegative and eventually zero almost surely. Nevertheless $\mathbb{E}{M}_n=1$ for every $n$, so it has almost sure convergence but no convergence in L1. It is therefore not uniformly integrable. Before absorption the nearest-neighbour walk is at most zero, ensuring nonnegativity.

Let $S_0=0$ be a simple symmetric random walk, let $T_a$ be its first hitting time of the positive integer $a$, and reverse every increment after $T_a$. The resulting path is again a simple symmetric random walk. Consequently,

For $M_n={\max }_{0\le k\le n}{S}_k$ and a positive integer $a$,Exactly one term can be nonzero because $S_n$ has the parity of $n$.

If ${\mathcal{F}}_n$ is generated by independent symmetric signs ${\xi }_1,\dots ,{\xi }_n$, every martingale has predictable coefficients $B_n$ such thatOne may take $B_n=\mathbb{E}[M_n{\xi }_n\mid {\mathcal{F}}_{n-1}]$.

For a bounded stopping time $T$ and a square-integrable representation $M_n=M_0+{\sum }_{k\le n}{B}_k{\xi }_k$,The cross terms vanish because predictable multiples of distinct independent signs are orthogonal martingale differences.