2
$\begingroup$

Exercise 5.3.1 on page 205 of [Probability: Theory and Examples, 4th edition, Durrett, 2010] reads:

Exercise 5.3.1. Let $X_n$, $n \ge 0$, be a submartingale with $\sup X_n < \infty$. Let $\xi_n = X_n − X_{n−1}$ and suppose $\mathbb{E}(\sup \xi_n^+) < \infty$. Show that $X_n$ converges a.s.

In [Probability: Theory and Examples, 5th edition, Durrett, 2019], the same question appears as Exercise 4.2.4. on page 223.

I would like to cite this result, but I hesitate to cite an exercise. Could anyone kindly provide a reference for this conclusion, or for a variant similar to the following?

Theorem. Let $X_n$ be a submartingale with $\mathbb{E}(\sup (X_n-X_{n-1})^+) < \infty$ (or more strongly, $\sup|X_n-X_{n-1}|\le c <\infty$, which is adequate for my application). Then $$ \mathbb{P}\left(\{X_n \text{ converges}\}\bigcup \{\sup X_n = \infty\}\right) = 1 $$

I would be surprised if this kind of result has never been mentioned in any other monographs/textbooks, although I have found nothing after searching.

Any comments or criticism will be appreciated. Thank you. (I am looking for a reference. Nevertheless, if you would like to post a proof, it is also very welcome.)

$\endgroup$

2 Answers 2

1
$\begingroup$

You can essentially follow the same proof as Theorem $5.3.1$, which appears right above this exercise. Briefly, let $0<K<\infty$, and let $N:=\inf\{n:X_n \ge K\}$. Then $\{X_{n\wedge N}\}$ is a submartingale, and $X_{n\wedge N}^+ \le K + \xi_N^+$. It follows by the martingale convergence theorem that $\lim_{n\to\infty}X_{n\wedge N}$ exists a.s., and in particular that $\lim_{n\to\infty}X_n$ exists on $\{N=\infty\}$. Letting $K\to\infty$, and using the fact that $\mathbb P(\limsup_{n\to\infty}X_n<\infty) = 1$, we see that $\lim_{n\to\infty}X_n$ exists a.s.

In terms of citation, you could probably get away with either citing the exercise, or Theorem $5.3.1$ itself. You might even do both, i.e. "by a straightforward generalization of [Durrett, Theorem $5.3.1$] (see [Durrett, Exercise $5.3.1$]), one has..." But you could also just prove it directly, since it wouldn't take more than a couple of lines.

$\endgroup$
0
$\begingroup$

If you are okay with the stronger statement(uniform bounded increments), then you can cite Theorem 4.3.1 on Durrett's PTE(5th ed.) directly, just several pages after the exercise. It works for submartingales as well, as suggested by @Jason, you just need to change $D$ to be $\{\limsup X_n=+\infty\}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .