= Solution
Stop just before a large <predictable> coefficient would be used. Set
$$
\sigma_j=\inf\{t\geq0:|K_{t+1}|>j\},
$$
with the infimum of the empty set equal to infinity. Because $K_{t+1}$ is $\mathcal F_t$-measurable, $\sigma_j$ is a <stopping time>. The increment of the stopped <martingale transform> is
$$
Y_{t\wedge\sigma_j}-Y_{(t-1)\wedge\sigma_j}
=\mathbf1_{\{t\leq\sigma_j\}}K_t(M_t-M_{t-1}).
$$
Its coefficient is <predictable> and bounded by $j$: the first coefficient exceeding $j$ occurs at the step after stopping, and is never included. Part (c) makes $Y^{\sigma_j}$ a <martingale>. Since the finitely many $K_s$ on any fixed finite horizon are finite almost surely, $\sigma_j\uparrow\infty$ almost surely. Thus
$$
\boxed{Y\text{ is a discrete-time local martingale}.}
$$
This is <predictable-coefficient localization of a martingale transform>. Stopping after taking the large increment would not give the required bound.
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