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Exercise 5.3.1 in Durrett's "Probability Theory and Examples" states

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

Here are my thoughts: It looks like we need to use theorem 5.2.8 (Martingale convergence theorem), which states that $X_n$ converges almost surely if $\sup EX_n^+<\infty$. I am trying to understand why we can just do the following (or at least I'm guessing we can't):

Suppose $X_n > 0$ (edit: on some set with positive measure), then $0< \sup X_n < \infty$, and therefore $EX_n^+ \le \sup X_n < \infty$. So we're done.

Otherwise, if $X_n \le 0$ a.e., then $\sup X_n^+ = 0$ a.e. Once again, we're done.

Am I missing something? Since I didn't use all the information, I probably am.

Thanks for the help.

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  • $\begingroup$ Actually I think you can, but I also think the point of the problem was to get you to try to use Theorem 5.3.1 and deduce that $P(D) = 0$. $\endgroup$
    – gogurt
    Commented Dec 16, 2015 at 17:47

1 Answer 1

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You've stepped off a cliff in deducing that the expectation $E(X^+_n)$ is bounded above by the random variable $\sup_n X_n$.

Suggestion: Employ the argument used by Durrett in the proof of his Theorem 5.3.1. Fix a positive real $K$, define the stopping time $T=T_K$ to be the first time $n$ that $X_n$ is larger than $K$, and observe that the stopped process satisfies $$ X_{n\wedge T}\le K+\sup_m\xi_m^+, $$ so that $$ E(X_{n\wedge T})\le K+E(\sup_m\xi_m^+)<\infty,\qquad\forall n. $$ Now apply the submartingale convergence theorem to the stopped process. This yields a.s. convergence of $X_n$ on the event $\{T_K=\infty\}=\{\sup_mX_m\le K\}$. Finally, vary $K$.

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  • $\begingroup$ I'm probably 6 years late, but I cannot understand the part where K is varied ? $\endgroup$
    – user447237
    Commented Jan 9, 2022 at 13:47
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    $\begingroup$ By this I simply meant use the fact that $\cup_{K=1}^\infty\{\sup_mX_m\le K\}=\{\sup_m X_m<\infty\}$ has probability 1. You've shown that $(X_n)$ converges a.s on each piece $\{\sup_mX_m\le K\}$, $K=1,2,\ldots$, so it converges on their union. $\endgroup$ Commented Jan 9, 2022 at 17:03
  • $\begingroup$ oh !! got it. thanks a lot. $\endgroup$
    – user447237
    Commented Jan 9, 2022 at 17:38
  • $\begingroup$ The convergence theorem requires $\sup E(X_n^+)< \infty$. Could you explain how to get $\sup E(X_{n \wedge T}^+) < \infty$? Could it has both infinity positive part and negative part? $\endgroup$
    – MichaelS
    Commented Aug 3, 2023 at 10:07
  • $\begingroup$ I could just as well have noted (and should have) that $X_{n\wedge T}^+\le K+\sup_m\xi_m^+$. $\endgroup$ Commented Aug 4, 2023 at 17:29

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