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Let $M_t $ be a martingale such that $\mathbb E(M_T)=0$.Let

$$M_T^n=(M_T \wedge n)\vee(-n)-\mathbb E((M_T \wedge n)\vee(-n))$$

and Let $$M_t^n=\mathbb E(M_T^n|\mathcal F_t)$$.Then how do I show that for all $\epsilon >0,$ $$\lim_{n \rightarrow \infty }P(\sup_{0\leq t \leq T}|M_t^n-M_T|\geq \epsilon)=0$$

I understand that $M_T^n$converges to $M_T$ pointwise as well as in $\mathcal L^1$ but how do I account for the supremum over [0,T]

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  • $\begingroup$ $T$ is a constant? Or a stopping time? $\endgroup$
    – Eric Auld
    Commented Aug 27, 2021 at 18:07
  • $\begingroup$ T is constant here $\endgroup$
    – abc
    Commented Aug 27, 2021 at 18:09

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I think you want to prove $$ \lim_{n \to \infty} \mathbb{P}(\sup_{0 \leq t \leq T} |M_t^n-M_t| \geq \varepsilon) \longrightarrow 0.$$ The idea is to use Doob's inequality. Note that $\left\{M_t^n=\mathbb{E}(M_T^n \mid \mathcal{F}_t), \mathcal{F}_t, t \geq 0 \right\}$ and $\left\{M_t=\mathbb{E}(M_T \mid \mathcal{F}_t), \mathcal{F}_t, t \geq 0 \right\}$ are both Doob's martingales and hence so is their difference. Hence, $$ \mathbb{P}(\sup_{0 \leq t \leq T} |M_t^n-M_t| \geq \varepsilon) \leq \varepsilon^{-1}\mathbb{E}|M_T-M_T^n|.$$ The last term goes to $0$ as $n \to \infty$ by DCT.

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