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I need some help with the following:

$1$. I have a $M$ continuous martingale, how can I show that if $t_n \underset{n}{\rightarrow}t$ then: $$ M_{t_n}\overset{L^1}{\underset{n}{\rightarrow}}M_t $$ assuming that $\mathbb{E}\left[\underset{t\in[0,T]}{\sup}|M_t|\right]<+\infty$

$2$. Given that $M$ is square integrable how to show that $t \mapsto \mathbb{E}\left[M_t^2\right]$ is continuous.

thanks

Davide Giraudo
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MJBTF
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  • Have you learned about uniform integrability yet? Once you do you'll be able to see that the hypothesis $\Bbb E\left[\sup_{t\in[0,T]}|M_t|\right]<\infty$ for 1. is overkill. The convergent sequence ${t_n}$ is bounded above by $T$. Because $M_s=\Bbb E[M_T\mid\mathcal F_s]$ for each $s\in[0,T]$, the collection ${M_s: 0\le s\le T}$ is uniformly integrable. This plus the pointwise convergence of $M_{t_n}$ to $M_t$ implies the $L^1$ convergence. – John Dawkins Dec 12 '20 at 17:33

1 Answers1

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  1. $M_t$ is continuous, $M_{t_n}(\omega) \to M(t)(\omega)$, hence we have almost sure convergence.

Further, $|M_{t_n}| \le \sup_{t \in [0,1]} M_t = \eta$. As $\mathbb{E} \eta < \infty$ we have $M_{t_n} \to M_t$ in $L_1$ by dominated convergence theorem.

  1. In general case it's not true - so I think you meant that $M$ is countinious in the second problem also.

As in previous case, it's sufficient to prove that $$\sup_{t \in [0,1]} M_{t}^2 \in L_1. \quad (*)$$

Previous version of proof of (*) (it's not correct, the correct version is above) It follows from Jensen's inequality for conditional expectations. Indeed, we have $\mathbb{E} (M_T | \mathcal{F}_t) = M_t$. Hence

$$ \eta = \mathbb{E} ( M_T^2 | \mathcal{F}_t) \ge \Bigl(\mathbb{E} ( M_T | \mathcal{F}_t) \Bigr)^2 = M_t^2 \Longrightarrow \sup_{t \in [0,T]} M_{t}^2 \le \eta$$ and $\mathbb{E} \eta = \mathbb{E} M_T^2< \infty$, because our martingale is in $L_2$.

New version of proof of (*). $X_t$ is martingale and hence $X_t^2$ is submartingale and $X_t^2 \ge 0$. Hence by Doob's maximal inequality (see https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality or https://fabricebaudoin.wordpress.com/2012/04/10/lecture-11-doobs-martingale-maximal-inequalities/) we get

$$\mathbf{E}\left[\sup_{0 \leq s \leq T} |X_s|^p\right] \leq \left(\frac{p}{p-1}\right)^p\mathbf{E}[|X_T|^p]$$ and if $p=2$ we have $\mathbf{E}\left[\sup_{0 \leq s \leq T} |X_s|^2\right] < \infty$.

Botnakov N.
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