= Solution
For a <stopping time> $T\le s$, <optional sampling> for $N$ gives $\mathbb E[N_s\mid\mathcal F_T]=N_T$, hence
$$
\mathbb E[M_TN_s]=\mathbb E[M_TN_T].
$$
This is justified by truncation and the <Cauchy-Schwarz inequality>, as in <a/iv/solution>. If $MN$ is a <martingale>, bounded optional sampling gives $\mathbb E[M_TN_T]=\mathbb E[M_0N_0]=0$, proving the required condition.
Conversely suppose the required condition holds. Every bounded <stopping time> $T$ lies below some deterministic $s$, so the displayed identity gives $\mathbb E[(MN)_T]=0$. The product is continuous and adapted, and its stopped values are integrable: the <Doob L2 maximal inequality> bounds the product of the two path suprema in $L^1$ by the <Cauchy-Schwarz inequality>. The <characterization of a martingale by bounded continuous-time stopped expectations> therefore makes $MN$ a <martingale>. Thus \b[orthogonality is equivalent to the backward-stopping <expectation> test]. In contrast to <weakly orthogonal continuous martingales>, this tests the product at random times, not only deterministic times.
Back to article page
