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$\newcommand{\P}{\mathbb{P}}$I would like to prove that stopping time:

$\sigma=\inf\{t\geqslant0: |B_t|=\frac{1}{2}(1+\sqrt{1+t})\}$

is integrable.

This is part of the answer to the question here.

I've tried to estimate ${\P}(\sigma\geqslant t)$ in the following way: $$\begin{array}{rcl}\P(\sigma\geqslant t) & = & \P(\forall s<t |B_s|\leqslant \frac{1}{2}(1+\sqrt{1+s} ) \\ & = & \P(\forall s<t |B_1|\sqrt{s}\leqslant \frac{1}{2}(1+\sqrt{1+s}))\\ &= &\P\left(\forall s<t |B_1|\leqslant \frac{1}{2}\left(\frac{1}{s}+\sqrt{1+\frac{1}{s}}\right)\right) \end{array}$$

I'm not sure if it is correct (I get something which doesn't converge).

Thank you in advance.

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Cofibration
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  • Your reasoning is not correct. For each fixed $s$ the random variable $B_s$ equals in distribution $\sqrt{s} B_1$, but the process $(B_s){s \leq t}$ does not equal (in distribution) $(\sqrt{s} B_1){s \leq t}$. For this reason you can't simply use the scaling property to get the 2nd "=". – saz Apr 02 '17 at 12:21
  • Thank you, it was stupid mistake. So I don't know how to estimate this probability. – Cofibration Apr 02 '17 at 12:24
  • You could simply ask @zhoraster for a hint how to prove the integrability (i.e. write a comment to the answer which you linked in your question). – saz Apr 02 '17 at 12:32
  • This can be estimate in the same way as in original question. If $\sigma$ is bigger than t, then this would mean that the Brownian motion $W$ is below the bound of $1/2(1+sqrt(1+t))$ you just need to change the bounderies in $G_t$ – quallenjäger Apr 02 '17 at 13:15
  • I'm not really sure how to choose $G_t$.If I take $G_t={f: \forall s\in[0,t] f(s)\in (-\frac{1}{2}(1+\sqrt{1+s}, \frac{1}{2}(1+\sqrt{1+s})}$ I will get nothing interesting becasue $\frac{1}{2} (G_t-G_t)=G_t$. Could you explain your idea? – Cofibration Apr 02 '17 at 14:39

1 Answers1

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First of all, note that $\sigma$ is almost surely finite; this is e.g. a consequence of the law of the iterated logarithm.

It follows from the optional stopping theorem that $M_t := B_{t \wedge \sigma}^2 - t \wedge \sigma$ is a martingale; in particular

$$\mathbb{E}(M_t) = \mathbb{E}(M_0)=0,$$

i.e.

$$\mathbb{E}(B_{t \wedge \sigma}^2) = \mathbb{E}(t \wedge \sigma). \tag{1}$$

By the very definition of $\sigma$, we have

$$\mathbb{E}(B_{t \wedge \sigma}^2) \leq \frac{1}{4} \mathbb{E}\bigg[ (1+\sqrt{1+t \wedge \sigma})^2 \bigg]$$

and therefore

$$\mathbb{E}(t \wedge \sigma) \leq \frac{1}{4} \mathbb{E}\bigg[ (1+\sqrt{1+t \wedge \sigma})^2 \bigg] \leq \frac{1}{2} \mathbb{E}(2+t \wedge \sigma).$$

Hence,

$$\frac{1}{2} \mathbb{E}(t \wedge \sigma) \leq 1.$$

Applying Fatou's lemma we conclude

$$\mathbb{E}(\sigma) \leq \liminf_{n \to \infty} \mathbb{E}(n \wedge \sigma) \leq 2.$$

saz
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