# Tag Info

• 23.9k
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### Let $M$ be a continuous martingale, $r >0$, and $\tau := \inf \{t \ge 0 : \langle M \rangle_t \ge r \}$. Then $M_\tau$ is square-integrable

Yes, your proof is correct. One could also use Fatou's lemma for a slightly more direct proof and a tighter bound: After showing $\mathbb{E}[|M_{t \wedge \tau}|^2] \le r$, we have \begin{align*} \...
• 10.9k
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### Brownian motion: how is DCT for conditional expectation applied on this stopped martingale?

By Vitali's convergence theorem, it suffices to show that $( M_{\tau (s_2) \wedge \tau_n}, n \in \mathbb N)$ is uniformly integrable. By OST, $( M_{t \wedge \tau (s_2) \wedge \tau_n}, t \ge 0)$ is a ...
• 13.7k
1 vote
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### Proof of Doobs decomposition theorem

$\mathbb{E}[Y_{n+1}|\mathcal{F}_n]$ is $\mathcal{F}_n$-measurable and $\mathcal{F}_n \subseteq \mathcal{F}_t$ for $0 \leq n \leq t$. Hence, $\mathbb{E}[Y_{n+1}|\mathcal{F}_n]$ is $\mathcal{F}_t$-...
• 21.2k
1 vote
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1 vote
Accepted

• 73

### Is M defined as below a martingale?

Which Bessel function do you have in mind? (There are several, and your notation $B_1$ is non-standard.) A more appropriate condition for testing whether you have a martingale might be Kazamaki's: If ...
• 23.9k
1 vote
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### Application of Cesaro lemma

You just need the particular case of arithmetic mean of Cesàro's lemma, which yields that $\frac{1}{n}\sum_{k=0}^{n-1}M_k$ converges almost surely as $n\to+\infty$ to the same limit as that of $M_n$. ...
• 4,989
Accepted

### Given a local martingale $M$ the running supremum $N_t=\sup_{0\leq s\leq t}|M_s|$ is locally integrable

Well, $M$ may (and often will) have a jump at $S_n$, so it is quite possible that $|M_{S_n}|>n$. So actually $$N_{S_n} \le |M_{S_n}| \vee \sup_{s<S_n} |M_s| \le |M_{S_n}| \vee n.$$
• 24.5k