# Show this statement holds for a right continuous martingale.

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?

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

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 $$1$$.
• Hi, could you tell me how $(1) \implies (2)$ via rightcontinuity if $\mathbb{E}(\sup_s M_s^2) = 0$?
• 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 $\mathbb{P}(M_t = 0, \, \forall 0 \leq t\leq T) = 1$. May 24, 2020 at 0:11