Show that a process is Brownian motion

For a Brownian motion $(B_t)_{t \geq 0}$ consider $$X_t=\int_0^t\sin\left(\frac{1}{B_s}\right)\;ds+B_t.$$ I want to show that $(X_t)_t$ is a Brownian motion.

Clearly, one can check the definition of BM and invoke the fact that $\int_{t_i}^{t_{i+1}}\sin(B_s^{-1})\;ds$ is measurable wrt. $\sigma(\{B_s,t_i\leq s\leq t_{i+1}\})$ together with the Markov property. However, this might be a bit tedious.

I was wondering whether I can show that $(X_t)_t$ is a martingale and therefore conclude with Levy's characterization.

• You want to show that $(X_t)_t$ is a BM with respect to some probability measure or with respect to the same probability measure as for your original BM? – saz Feb 13 '18 at 14:53
• W.r.t. some probability measure would be simply Girsanov's theorem. – julian Feb 13 '18 at 15:05

Let $(B_t)_{t \geq 0}$ be a Brownian motion on a probabiliy space $(\Omega,\mathcal{A},\mathbb{P})$. The process

$$X_t := \int_0^t \sin \left( \frac{1}{B_s} \right) \,ds + B_t$$

is a Brownian motion with respect to some probability measure $\mathbb{Q}$ on $(\Omega,\mathcal{A})$ (this is a direct consequence of Girsaonov's theorem); as far as I can see, it is, however, not a Brownian motion on the original probability space.

To show that $(X_t)_{t \geq 0}$ is not a Brownian motion on $(\Omega,\mathcal{A},\mathbb{P})$ it clearly suffices to show that $(X_t)_{t \geq 0}$ is not a $\mathbb{P}$-martingale. Since $(B_t)_{t \geq 0}$ is a martingale, we know that $(X_t)_{t \geq 0}$ is a martingale iff

$$M_t := \int_0^t \sin \left( \frac{1}{B_s} \right) \, ds$$

is a martingale. Denote by $(\mathcal{F}_t)_{t \geq 0}$ the canonical filtration of $(B_t)_{t \geq 0}$. If we fix $s \leq t$, then it follows from the independence and stationarity of the increments of $(B_t)_{t \geq 0}$ that

\begin{align*} \mathbb{E}(M_t \mid \mathcal{F}_s) &= \int_0^s \sin(B_r^{-1}) \, dr + \int_s^t \mathbb{E}(\sin(B_r^{-1}) \mid \mathcal{F}_s) \, dr \\ &= M_s+ f(t-s,B_s) \end{align*}

where

$$f(u,x) := \int_0^{u} \mathbb{E} \left( \sin \left[ \frac{1}{x+ B_r} \right] \right) \, dr.$$

Thus, $(M_t)_{t \geq 0}$ is a martingale iff

$$f(t-s,B_s) = 0 \quad \text{\mathbb{P}-almost surely for all s \leq t}. \tag{1}$$

Clearly, we can choose $\delta_1,\delta_2>0$ such that

$$\forall \left| x- \frac{\pi}{2} \right| \leq \delta_1, |y| \leq \delta_2: \quad \frac{1}{x+y} \in \left[ \frac{\pi}{4}, \frac{3\pi}{4} \right].$$

Moreover, the continuity of the sample paths of Brownian motion implies that there exists $u>0$ such that

$$\mathbb{P}(|B_r| \leq \delta_2) \geq \frac{3}{4} \quad \text{for all r \in [0,u].}$$

This implies

$$\mathbb{E}\left( 1_{|B_r| \leq \delta_2} \sin \left[ \frac{1}{x+B_r} \right] \right) \, dr \geq \sin \left( \frac{\pi}{4} \right) \frac{3}{4}$$

for all $|x-\pi/2|\leq \delta_1$ and $r \in [0,u]$. On the other hand,

$$\left| \mathbb{E}\left( 1_{|B_r| > \delta_2} \sin \left[ \frac{1}{x+B_r} \right] \right) \right| \leq \mathbb{P}(|B_r|>\delta_2) = 1- \mathbb{P}(|B_r| \leq \delta_2)\leq \frac{1}{4}$$

for all $r \in [0,u]$. Hence,

$$f(u,x) \geq \int_0^u \left[ \sin \left( \frac{\pi}{4} \right) \frac{3}{4} - \frac{1}{4} \right] \, dr > \frac{u}{4} >0$$

for any $|x-\pi/2| \leq \delta_1$. Since $\mathbb{P}(|B_s-\pi/2| \leq \delta_1)>0$ for any $s>0$, this implies that $(1)$ does not hold true for $t :=s+u$.

• Why is $\int_s^t E[\sin(B_r^{-1})|\mathcal{F}_s]\;dr=\int_0^{t-s}E[\sin((B_s+B_r)^{-1})]\;dr$? – julian Feb 13 '18 at 16:37
• @julian I'm not claiming that this holds, please check carefully what I wrote. The idea is to write $$\frac{1}{B_r} = \frac{1}{(B_r-B_s)+B_s}$$ and use that $(B_r-B_s)_{r \geq s}$ is independent from $\mathcal{F}_s$. – saz Feb 13 '18 at 16:40
• @saz: I'm not sure I understand why this argument is necessary. Since $|\sin(1/x)| \leq 1$ it is clear that $|M_t-M_s| \leq \int_s^t|\sin(1/B_u)|du \leq |t-s|$ for all $s,t$ (and thus $M$ has Lipchitz paths almost surely). Therefore $M$ is a continuous process whose quadratic variation is zero (because clearly any Lipchitz function has zero quadratic variation). Hence $M$ cannot be a martingale. What is wrong with this argument? – Shalop Feb 19 '18 at 6:53
• @Shalop Nothing wrong about it, as far as I can see. – saz Feb 19 '18 at 8:13