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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.

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  • $\begingroup$ 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

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Here we study

$$M_{t}=B_{A_{t}}\stackrel{d}{=}\int_{0}^{t}\sqrt{1+e^{W_{s}}}dW_{s}.$$

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 $t>0$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ "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

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