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A similar question was asked here but I have a question about one of the steps.

To show that $M_t \in M := \{M_t : t \in [0,\infty)$ is integrable we consider $M_{t \wedge T_n}$ where $\{T_n\}$ are stopping times almost surey increasing to $\infty$. Then $M_{t \wedge T_n} \to M_t$ almost surely. We can now apply Fatou's Lemma to get that $$E(M_t) \le \liminf\limits _{n \to \infty}E(M_{t \wedge T_n}) = E(M_0)$$

But why does $\liminf\limits _{n \to \infty}E(M_{t \wedge T_n}) = E(M_0)$?

alpastor
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1 Answers1

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$M_{t\wedge T_n}: t \geq 0$ is a martingale for each $n$. So $EM_{t\wedge T_n}=EM_{0\wedge T_n}=EM_0$.

Kavi Rama Murthy
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  • Maybe there is a property of martingales that I am not familiar with but why does $M_{t\wedge T_n}$ being a martingale for $t \ge 0$ and for every $n$ imply that $EM_{t\wedge T_n}=EM_{0\wedge T_n}$? – alpastor Jun 05 '19 at 09:57
  • @alpastor If ${X_t}$ is a martingale then $EX_t$ is in dependent of $t$. To see this just take expectation in the definition of martingale. – Kavi Rama Murthy Jun 05 '19 at 09:59
  • You mean independent right? $$ E(X_s)=E\left[E(X_s)\right]=E\left[E(X_t|\mathcal F_s)\right] = E(X_t)$$. Thank you, for the explanation – alpastor Jun 05 '19 at 10:06
  • @alpastor Yes. I just left a gap after 'in' in 'independent'. A typo. – Kavi Rama Murthy Jun 05 '19 at 10:07
  • oh didnt see the in was there, anyways thank you! – alpastor Jun 05 '19 at 10:18