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

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

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  • $\begingroup$ 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 $EM_{t\wedge T_n}=EM_{0\wedge T_n}$? $\endgroup$
    – alpastor
    Commented Jun 5, 2019 at 9:57
  • $\begingroup$ @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. $\endgroup$
    – Kavi Rama Murthy
    Commented Jun 5, 2019 at 9:59
  • $\begingroup$ 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 $\endgroup$
    – alpastor
    Commented Jun 5, 2019 at 10:06
  • $\begingroup$ @alpastor Yes. I just left a gap after 'in' in 'independent'. A typo. $\endgroup$
    – Kavi Rama Murthy
    Commented Jun 5, 2019 at 10:07
  • $\begingroup$ oh didnt see the in was there, anyways thank you! $\endgroup$
    – alpastor
    Commented Jun 5, 2019 at 10:18

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