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Let $T > 0$ and consider a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ on which there is a filtration

$$\{\mathcal{F}_t: 0 \leq t \leq T\}$$

Let $M:=\{M_t: 0 \leq t \le T\}$ be a martingale w.r.t. this filtration such that $M$ is right continuous.

If $$\mathbb{E}\left(\sup_{0 \leq s \leq T} M_s^2\right)< \infty$$

then a proof I'm reading claims the following:

(1) $\sup_{0 \leq s \leq T} M_s = 0 $ almost surely

(2) Consequently, because the martingale is right continuous, $M_s = 0$ for every $0 \leq s \leq T$ with probability one.

Why are $(1)$ and $(2)$ true?

Thanks in advance!

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  • $\begingroup$ Do you mean $\mathbb{E}\left(\sup_{0\leq s \leq T}M_s^2 \right) = 0$? $\endgroup$
    – Michh
    May 21, 2020 at 20:41
  • $\begingroup$ Nope, I don't. Note that in (1) there are no squares. $\endgroup$
    – user745578
    May 21, 2020 at 20:41

1 Answer 1

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This is not true: take $(B_s)_{0 \leq s \leq T}$ a Brownian motion. By Doob's maximal inequality, $$\mathbb{E} \left[\sup_{0\leq s\leq T}B_s^2\right] \leq 4 \mathbb{E}\left[B_T^2\right]<\infty.$$ But $B$ is certainly not the zero function with probability $1$.

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  • $\begingroup$ Hi, could you tell me how $(1) \implies (2)$ via rightcontinuity if $\mathbb{E}(\sup_s M_s^2) = 0$? $\endgroup$
    – user745578
    May 22, 2020 at 21:29
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    $\begingroup$ I am not sure how (2) follows from (1) but here's another way: since $0 \leq \mathbb{E}[M_t^2] \leq \mathbb{E}[\sup_s M_s^2] = 0$, we have that $\forall 0\leq t \leq T$, $\mathbb{P}(M_t =0) = 1$. Now use the right continuity of $M$ to deduce that $\mathbb{P}(M_t = 0, \, \forall 0 \leq t\leq T) = 1$. $\endgroup$
    – Michh
    May 24, 2020 at 0:11

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