Does there exist an almost surely continuous martingale $X$ with $X_t \to +\infty$ almost surely?
Remark: Note that such a martingale exists in discrete time, or equivalently in continuous time if the a.s. continuity requirement is dropped.
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2 Answers
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No. see Martingale Convergence
Theorem 4 Let X be a continuous martingale. Then, almost surely, one of the following is satisfied
- $X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite.
- $\limsup_{t\rightarrow\infty}X_t=\infty$ and $\liminf_{t\rightarrow\infty}X_t=-\infty$. In this case, the process hits every value in $\mathbb R$ at arbitrarily large times.
answered Dec 17, 2023 at 7:52
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$\begingroup$ Wow, very interesting theorem. The link is not working however, could you rectify that? $\endgroup$ Commented Dec 17, 2023 at 7:55
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$\begingroup$ I suppose the proof is via the Dambin-Dubins-Schwartz theorem, the two possibilities correspond to the time change being bounded or unbounded respectively. (Edit: it seems the post provides an alternate proof.) $\endgroup$ Commented Dec 17, 2023 at 8:00
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$\begingroup$ The result that you linked follows immediately from Lemma 5 in those notes. However, I cannot make any sense of the 5-line proof of that Lemma 5, which in particular has this: "$n-1_{\{\tau_n > 0\}}X^{\tau_n}$ is a nonnegative supermartingale". Here $\tau_n:=\inf\left\{t\in{\mathbb R}_+\colon X_t\ge n\right\}$. First here, $X^{\tau_n}$ is undefined when $\tau_n=\infty$. Second, assuming that $\tau_n<\infty$ a.s., we have $n-1_{\{\tau_n > 0\}}X^{\tau_n}=n\,1(X_0\ge n)$, which is not a supermartingale. $\endgroup$ Commented Dec 20, 2023 at 20:18
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1$\begingroup$ @NateRiver : Thank you for your response. I see now that I misunderstood the notation, in at least two places. $\endgroup$ Commented Dec 21, 2023 at 0:21
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I think Iosif's Fatou lemma argument can be fixed, as follows.
Assume without loss of generality that $X_0 = 0$.
Suppose to the contrary that $X_t \to +\infty$ a.s. Then it must be that $\inf_{t \ge 0} X_t > -\infty$ a.s. Choose a number $b < 0$ sufficiently negative that the event $A_b = \{ \inf_{t \ge 0} X_t > b\}$ has positive probability. Let $T_b = \inf\{ t : X_t \le b\}$ be the corresponding stopping time, and let $Y_t = X_{t \wedge T_b}$. By optional stopping, $Y_t$ is again a martingale, so $E[Y_t] = E[Y_0] = 0$ for all $t$. Note also that on the event $A_b$, we have $Y_t = X_t$ for all $t$ and so $Y_t \to +\infty$ a.s. on $A_b$.
On the other hand, $Y_t \ge b$ (by continuity!) so we can apply Fatou's lemma to $Y_t$, concluding that $E[\liminf Y_t] \le E[Y_0] = 0$. But $\liminf Y_t = +\infty$ a.s. on $A_b$, where $P(A_b) > 0$, so this is absurd.
Note that we used the continuity in a subtle way: if $X_t$ had not been continuous, we could still define the stopped martingale $Y_t$, but we would not be able to say $Y_t \ge b$ (because $X_t$ could cross $b$ for the first time by jump). In such a case, $Y_t$ could end up being unbounded below, and the argument fails.
answered Dec 17, 2023 at 16:45