Let $W$ be a standard one dimensional Brownian motion and let $A$ be the process defined by : $$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_s}\right)\mathrm{d}s$$

and for $t\geq 0$, we define the following stopping time: $$\sigma_t := \inf \{s\geq 0: \ A_s > t\}$$

Notice that a.s: $$\forall \ t\geq 0: \quad \sigma_t \leq t$$

Now, define the martingale $M$ as a time changed Brownian motion, i.e. $M := \left(W_{A_t}\right)_{t\geq 0}$. Can we prove that $M$ goes to infinity in probability ? Thanks.

  • $\begingroup$ actually, are you sure this is the right martingale you have in mind? Shouldn't it be $M_{t}=B_{A_{t}}$ equivalently $M_{\sigma_{t}}=B_{t}$? $\endgroup$ Commented Dec 25, 2023 at 23:29
  • $\begingroup$ Yes, you are right, it is actually $M_t:= B_{A_t}$. I have edited the original post. Can we prove that M is a true martingale? $\endgroup$
    – Greyearl
    Commented Dec 25, 2023 at 23:34
  • $\begingroup$ every Itô integral has zero expectation and so the increment disappears. $\endgroup$ Commented Dec 25, 2023 at 23:37
  • $\begingroup$ That’s true, it might require a localization argument as the intregrand is not bounded, but it can be proven. Thank you ! $\endgroup$
    – Greyearl
    Commented Dec 25, 2023 at 23:40
  • $\begingroup$ This question is surprinzingly close to the conjecture I made in my partial answer at mathoverflow.net/questions/460751/another-curious-martingale/… However, I think that there is a typo here. The martingale that I considered is $(X_{\sigma_t})_{t \ge 0}$, and not $(W_{A_t})$. It do not know whether this last process is a martingale. $\endgroup$ Commented Dec 27, 2023 at 21:07

1 Answer 1


Here we study


First, as mentioned Martingale Convergence here

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.

By Itô isometry we have

$$E[M^{2}_{t}]=\int^t 1+Ee^{W_{s}}ds=\int^t 1+e^{\frac{s}{2}}ds \to +\infty,$$

which also shows that $<M>_{\infty}=\infty$. In general for continuous martingales we have that

"$X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite" iff $<X>_{\infty}<\infty.$

The left-right direction that is relevant here is as follows: we use the stopping time $\tau_{n}:=\inf\{t\geq 0: |X_{t}|\geq n\}$ and the bound $$E[<X>_{t\wedge \tau_{n}}]=E[X^{2}_{t\wedge \tau_{n}}]\leq n^{2}$$

to get that the event $\{\sup|X_t|<\infty\}=\bigcup_{n}\{\sup|X_t|<n\}$ implies that the finiteness $<X>_{\infty}<\infty$. (For the other direction we can use instead $\tau_{n}:=\inf\{t\geq 0: <X>_{t}\geq n\}$.)

So we can't have the first case in the above theorem, because it would give a contradiction and so we have the second case.

For the "in probability" part, we will use a reverse tail inequality: when $\mathbb {E} [X]=0,\,\mathbb {E} [X^{2}]=1$ then

$$\Pr(X\geq 0)\geq {\frac {2{\sqrt {3}}-3}{\mathbb {E} [X^{4}]}},$$

from Berger's "The Fourth Moment Method. By symmetry of Brownian motion we write

$$P[M_{t}\leq 0]=P[0\leq \tilde{M}_{t}:=\int_{0}^{t}\sqrt{1+e^{-W_{s}}}dW_{s}].$$

We still have $E[\tilde{M}_{t}]=0$ and we will need the fourth moment bound

$$E\left(\int f dW\right)^{4}\leq c_{2}\left(\int Ef^{2} ds\right)^{2},$$

from Corollary 4 in "Notes on the Itô Calculus" by Steven P. Lalley. Therefore, for $X_{t}:=\tilde{M}_{t}\frac{1}{\sqrt{E[\tilde{M}_{t}^{2}]}}$ and using the above bound we get

$$P[\tilde{M}_{t}\geq 0]\geq {\frac {2{\sqrt {3}}-3}{\mathbb {E} [X^{4}]}}\geq \frac {2{\sqrt {3}}-3}{c_{2}}>0$$

for all $t>0$.

  • $\begingroup$ Thanks for you answer. Now that we have convergence to infinity in the $L^2$ sense, does that imply convergence in probability even though the limit is infinite? $\endgroup$
    – Greyearl
    Commented Dec 25, 2023 at 23:28
  • $\begingroup$ it does not converge to infinity. See the theorem I cited. The infinite L2 shows that we don't have finite limit and thus we have the second case. $\endgroup$ Commented Dec 25, 2023 at 23:30
  • 1
    $\begingroup$ sorry I didn’t read it well, thanks. $\endgroup$
    – Greyearl
    Commented Dec 25, 2023 at 23:31
  • $\begingroup$ If I have understood well the theorem, it rules out almost sure convergence of continuous martingales to infinity. Do you think that convergence to infinity in the probability sense might still hold? $\endgroup$
    – Greyearl
    Commented Dec 25, 2023 at 23:45
  • 2
    $\begingroup$ "the limsup/liminf convergence is almost surely which is stronger than probability". What do you mean by this? Why does the almost sure limsup/liminf "convergence" to $\pm\infty$ prevent $X_t$ from converging to $\infty$ in probability? $\endgroup$ Commented Dec 26, 2023 at 7:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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