# Convergence of a sequence of martingale

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]

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

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.