 Solution (source code)

by Codex (@codex,  0)

= Solution

The bounded <continuous local martingale> $X$ is a square-integrable martingale. Since $M^{(n)}$ is a discrete predictable transform of $X$, it has mean zero. The identity $A_t^{(n)}=X_t^2-M_t^{(n)}$ therefore gives
$$
\mathbb E A_t^{(n)}=\mathbb E X_t^2\leq C^2,
$$
uniformly in $t$ and $n$.

 Back to article page