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For a Brownian motion $B_t$ started from $0$ (it is also a martingale), denote by $T_a=\inf\{t\ge 0: B_t=a\}$ the stopping time. Consider the stopped process $B_{t\land T_a}$ (this is still a martingale), then $|B_{t\land T_a}|$ is bounded by $a$. (Recall that $T_a<\infty$ a.s.) Why can we say $B_{t\land T_a}$ is uniformly integrable?

At first I thought this is because that $E\left[\lvert B_{t\land T_a}\rvert^2\right]\le C$ for some constants $C$, then $B_{t\land T_a}$ is uniformly integrable.

However, I just saw this question:Almost sure bounded imply finite expectation?. I believe the answer means the finite bounded cannot imply the expectation. Thus, we cannot get $E[|B_{t\land T_a}|^2]\le C$ from $|B_{t\land T_a}|\le a$.

Davide Giraudo
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Hermi
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  • Yes you can: The integral preserves $\le$, so if $\lvert B_{t\land T_a}\rvert\le a$, then $$\mathsf E(\lvert B_{t\land T_a}\rvert^2)\le\mathsf E(a^2)= a^2.$$ This shows that $B_{t\land T_a}$ is uniformly $L^2$, which is stronger than uniformly $L^1$ since the $L^1$ norm is $\le$ the $L^2$-norm on probability spaces. – Maximilian Janisch Feb 02 '22 at 00:47
  • @MaximilianJanisch Does it contradict with the question: https://math.stackexchange.com/questions/1653877/almost-sure-bounded-imply-finite-expectation?noredirect=1&lq=1? – Hermi Feb 02 '22 at 00:49
  • No, to the contrary. That answer you linked states in the middle $$"|X|1 \equiv \mathbb{E}[|X|] \leq |X|{\infty} < \infty"$$ so $L^1$-norm is $\le$ the $L^\infty$ norm on probability spaces, which also follows from what I said. – Maximilian Janisch Feb 02 '22 at 00:51
  • @MaximilianJanisch But we also have $E|B_{t\land T_a}|\le Ea=a$, right? – Hermi Feb 02 '22 at 00:53
  • This is true for $a\ge 0$, yes. Actually if $a<0$ the whole argument doesn't work because then $B_{t\land T_a}$ is not a.s. bounded anymore :( – Maximilian Janisch Feb 02 '22 at 00:54
  • @MaximilianJanisch Ok, I see. Can we prove it? I mean if a r.v. $|X|\le a$ for $a>0$, then $E |X|\le |a|$? – Hermi Feb 02 '22 at 00:54
  • It's just $$\int f\le \int g$$ for $f\le g$ and you can find a proof of this in any textbook which constructs the Lebesgue integral (e.g. Achim Klenke, Wahrscheinlichkeitstheorie. 3. Auflage. Springer-Verlag Berlin/Heidelberg (2013).) – Maximilian Janisch Feb 02 '22 at 00:55
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    @MaximilianJanisch I see. Thanks! – Hermi Feb 02 '22 at 00:59

1 Answers1

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Uniform boundedness of $\{B_{t\land T_a}\}_{t \geq 0}$ implies that it is uniformly integrable. Indeed, by noting that for an $t\geq 0$, $$E(B_{t \land T_a } 1(|B_{t\land T_a}| \geq 2a )) = E(0) = 0$$ we immediately see that the family $\{B_{t\land T_a}\}_{t \geq 0}$ is uniformly integrable, by appealing to the definition of uniform integrability.

Jose Avilez
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  • Yes, I think this result is for martingale. If for general random variables. I think that "uniform boundness+L^1" implies the uniformly integrable. – Hermi Feb 02 '22 at 03:32