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Assume $M$ is local continuous martingale started from $0$. How would one go about showing that if it is a martingale bounded in $L^2$, then $E[\langle M\rangle_{\infty}]\lt \infty $?

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This is not true. The condition $\mathbb{E}[\langle M,M \rangle_\infty] < \infty$ implies $M$ is a true martingale, but you can have continuous local martingales that are bounded in $L^2$ but not true martingales. For example, if $B_t$ is a Brownian motion in $\mathbb{R}^3$ started from $(1,0,0)$ then $M_t := \frac{1}{|B_t|}$ is a continuous local martingale bounded in $L^2$, but is not a true martingale. We can look at $M_t - 1$ instead to satisfy $M_0 = 0$.

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  • $\begingroup$ That's confusing. Did I misunderstand the question entirely? It's actually an iff, so in particular what I wrote must hold. It's an exercise in several sources, for example the following: web.ma.utexas.edu/users/gordanz/notes/semimartingales.pdf $\endgroup$
    – Dalamar
    May 5, 2021 at 0:38
  • $\begingroup$ The iff in that exercise (19.7) is that $M \in \mathcal M_0^{2,c}$ iff $\mathbb{E}[\langle M,M \rangle_\infty] < \infty$, i.e. we need to assume $M$ is a true martingale rather than just a local martingale to conclude $\mathbb{E}[\langle M,M \rangle_\infty] < \infty$. $\endgroup$ May 5, 2021 at 0:43
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Under your hypotheses the process $K_t:=M_t^2-\langle M\rangle_t$ is a continuous local martingale. Let $(T_n)$ be an increasing sequence of stopping times with $\lim_nT_n=\infty$ (a.s.) and each stopped process $K^{T_n}$ a martingale. By Doob's inequality, for each $t>0$ and each $n$, $$ \Bbb E[\langle M\rangle_{T_n\wedge t}]=\Bbb E[M^2_{T_n\wedge t}]\le\Bbb E[\sup_{0\le s\le t}M^2_s]\le 4\Bbb E[M^2_t]\le C<\infty, $$ for some constant $C$. By monotonicity, $\Bbb E[\langle M\rangle_\infty]\le C<\infty$.

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