1
$\begingroup$

I have seen the following versions of the stopping time Theorems in Class and I am confused about some of the conditions.

Theorem (Stopping time thm for Martingales): Let $(X_n)_{n \in \mathbb{N}}$ be a martingale and $T$ be a stopping time, then the stopped process $(X_{n \wedge T})_{n \in \mathbb{N}}$ is also a martingale. Furtheremore if one of the conditions:

1) $T$ is bounded
2) $T < \infty$ a.s. and $|X_n| \leq Y \in L^1$ (dominated by $L^1$ RV)
3) $T \in L^1$ and the increments $(X_{n+1}-X_n)_{n \in \mathbb{N}}$ are bounded

4) (Wikipedia) $|X_{n \wedge T}| \leq c$ for some constant $c$
is satisfied then, $$\mathbb{E}(X_T)=\mathbb{E}(X_0) $$

My question will only concern 2 and 4 (from Wikipedia). I believe 1 is a rather useless criteria because for most practical stopping times it's rather hard to judge apriori whether they are bounded or not.

Theorem (Stopping time thm for UI Martingales): Let $(M_n)_{n \in \mathbb{N}}$ be an UI martingale and $T$ be a stopping time, then the stopped process $(M_{n \wedge T})_{n \in \mathbb{N}}$ is also an UI martingale and have $$\mathbb{E}(M_T)=\mathbb{E}(M_0) $$


My Questions: In the theorem for UI martingales there are no conditions on the stopping time whatsoever, in particular the stopping time does not have to be finite a.s. (in which case the terminal process $M_T$ is indeed a.s. well defined)

It makes sense to me, that there is no condition on the stopping time in the UI version of the thm because UI martingales are closed and therefore always convergence a.s. with limit in $L^1$, i.e. $M_\infty$ exists and is well defined. The same logic can of course not be transferred to general martingales where we apriori don't know if $X_\infty$ exists.

However in the Thm for Martingales 2) I don't see why we would require $T< \infty$ a.s. because the condition $|X_n| \leq Y \in L^1$ readily implies that $(X_n)_{n \in \mathbb{N}}$ is UI right? (Because a martingale that is bounded in $L^1$ converges a.s. to a finite limit and the dominance ensures that our convergence transfers to $L^1$ convergence, and thus the martingale is UI)

Same goes for condition 4), wouldn't it be enough to require that $|X_{n \wedge T} | \leq Y \in L^1$ which ensures that the stopped process is again UI?


TL;DR: In 2) is $T < \infty$ a.s. necessary?
In 4) can the constant $c$ be replaced by a RV $Y \in L^1$?

$\endgroup$

1 Answer 1

2
$\begingroup$

What you say is essentially correct: In (2) the condition $T < \infty$ a.s. is not necessary. In (4) the constant $c$ can be replaced by a RV $Y \in L^1$.

Here's the story (I guess). The writers of these theorems are trying to be concise and precise, not trying to give the most general version (explaining (4)). And in (2) if they did not assume $T < \infty$ a.s. then they would need to add a sentence explaining that the not-yet-defined $M_\infty$ was the limit guaranteed by an earlier Theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.