## Lecture 13. Some basic properties of the Brownian motion

In this Lecture we present some basic properties of the Brownian motion paths.

Proposition. Let $(B_t)_{t\ge 0}$ be a standard Brownian motion.

$\mathbb{P} \left( \inf_{ t\ge 0} B_t = - \infty , \sup_{t \ge 0} B_t =+\infty \right)=1.$

Proof

Since the process $(-B_t)_{t\ge 0}$ is also a Brownian motion, in order to prove that

$\mathbb{P} \left( \inf_{ t\ge 0} B_t = - \infty , \sup_{t \ge 0} B_t=+\infty \right)=1,$

we just have to check that

$\mathbb{P} \left( \sup_{t \ge 0} B_t =+\infty \right)=1.$

Let $N\in \mathbb{N}$. From the scaling property of Brownian motion, we have

$\mathbb{P} \left( c \sup_{t \ge 0} B_t \le N\right)=\mathbb{P}\left( \sup_{t \ge 0} B_t \le N\right), c>0.$

Therefore we have

$\mathbb{P} \left( \sup_{t \ge 0} B_t \le N\right)=\mathbb{P}\left( \sup_{t \ge 0} B_t = 0 \right).$

Now, we may observe that

$\mathbb{P} \left( \sup_{t \ge 0} B_t = 0 \right) \le \mathbb{P}\left(B_1 \le 0, \sup_{t \ge 1} B_t =0\right)=\mathbb{P} \left(B_1 \le 0, \sup_{t\ge 0} (B_{t+1}-B_1) =-B_1\right).$

Since the process $(B_{t+1}-B_1)_{t\ge 0}$ is a Brownian motion independent of $B_1$, we have for $c >0$,

$\mathbb{P} \left(B_1 \le 0, \sup_{t \ge 0}\left (B_{t+1}-B_1 \right) =-B_1\right)=\mathbb{P} \left(B_1 \le 0, c \sup_{t \ge 0} (B_{t+1}-B_1) =-B_1\right).$

Therefore we get

$\mathbb{P} \left(B_1 \le 0, \sup_{t \ge 0} (B_{t+1}-B_1) =-B_1 \right) =\mathbb{P} \left(B_1 \le 0, \sup_{t \ge 0} (B_{t+1}-B_1) =0\right)$
$=\mathbb{P} \left(B_1 \le 0\right) \mathbb{P} \left(\sup_{t \ge 0}(B_{t+1}-B_1) =0\right)$
$=\frac{1}{2} \mathbb{P}\left( \sup_{t \ge 0} B_t = 0 \right)$

Thus,

$\mathbb{P} \left( \sup_{t \ge 0} B_t = 0 \right) \le \frac{1}{2} \mathbb{P} \left( \sup_{t \ge 0} B_t = 0 \right)$

and we can deduce that

$\mathbb{P} \left( \sup_{t \ge 0} B_t = 0 \right)=0,$

and

$\mathbb{P} \left( \sup_{t \ge 0} B_t \le N\right)=0.$

Since this holds for every $N$, it implies that

$\mathbb{P} \left( \sup_{t \ge 0} B_t =+\infty \right)=1.$

$\square$

By using this proposition we deduce the following proposition whose proof is let as an exercise to the reader.

Proposition (Recurrence property of Brownian motion)
Let $(B_t)_{t\ge 0}$ be a Brownian motion. For every $t \ge 0$ and $x \in \mathbb{R}$,

$\mathbb{P} ( \exists s \ge t, B_s=x)=1.$

Martingale theory provides powerful tools to study Brownian motion. We list in the Proposition below some martingales that are naturally associated with the Brownian motion and that will play an important role in the sequel.

Proposition. Let $(B_t)_{t\ge 0}$ be a standard Brownian motion. The following processes are martingales (with respect to their natural filtration):

1. $(B_t)_{t\ge 0}$;
2. $(B_t^2-t)_{t\ge 0}$;
3. $\left( e^{\lambda B_t -\frac{\lambda^2}{2}t} \right)_{ t \ge 0}$, $\lambda \in \mathbb{C}$.

Proof

1. First, we note that for $t \ge 0$, $\mathbb{E}(\mid B_t\mid)<+\infty$ because $B_t$ is a Gaussian random variable. Now for $t \ge s$, $\mathbb{E}(B_t-B_s\mid \mathcal{F}_s)=\mathbb{E}(B_t-B_s )=0$, therefore $\mathbb{E}(B_t\mid\mathcal{F}_s)=B_s.$
2. For $t \ge 0$, $\mathbb{E}( B_t^2)=t<+\infty$ and for $t \ge s$, $\mathbb{E}((B_t-B_s)^2 \mid \mathcal{F}_s)=\mathbb{E}((B_t-B_s)^2 )=t-s$, therefore $\mathbb{E}(B_t^2-t\mid \mathcal{F}_s)=B_s^2-s.$
3. For $t \ge 0$, $\mathbb{E}\left( \left| e^{\lambda B_t -\frac{\lambda^2}{2}t} \right|\right) < +\infty$, because $B_t$ is a Gaussian random variable. Then we have for $t \ge s$, $\mathbb{E}(e^{\lambda (B_t-B_s)} \mid \mathcal{F}_s)=\mathbb{E}(e^{\lambda (B_t-B_s)})=e^{\frac{\lambda^2}{2} (t-s)}$, and therefore $\mathbb{E} \left(e^{\lambda B_t -\frac{\lambda^2}{2}t} \mid \mathcal{F}_s\right)=e^{\lambda B_s -\frac{\lambda^2}{2}s}.$

$\square$

The previous martingales may be used to explicitly compute the distribution of some functionals associated to Brownian motion.

Proposition. Let $(B_t)_{t\ge 0}$ be a standard Brownian motion. We denote for $a>0$,

$T_a=\inf \{ t \ge 0,\text{ } B_t=a \}.$

For every $\lambda >0$, we have

$\mathbb{E} \left( e^{-\lambda T_a}\right)=e^{-a \sqrt{2\lambda}}.$

Therefore, the distribution of $T_a$ is given by the density function

$\mathbb{P} (T_a \in dt)=\frac{a}{(2\pi t)^{3/2}}e^{-\frac{a^2}{2t}} dt,\quad t>0.$

Proof

Let $\alpha >0$. For $N \ge 1$, we denote by $T_N$ the almost surely bounded stopping time:

$T_N=T_a \wedge N.$

Applying Doob’s stopping theorem to the martingale $\left( e^{\alpha B_t -\frac{\alpha^2}{2} t} \right)_{t \ge 0}$ yields:

$\mathbb{E} \left( e^{\alpha B_{T_a \wedge N} -\frac{\alpha^2}{2} (T_a \wedge N)} \right)=1.$

But for $N \ge 1$,

$e^{\alpha B_{T_a \wedge N} -\frac{\alpha^2}{2} (T_a \wedge N)} \le e^{\alpha a}.$

Therefore from Lebesgue dominated convergence theorem, by letting $n \rightarrow +\infty$, we obtain

$\mathbb{E} \left( e^{\alpha B_{T_a } -\frac{\alpha^2}{2} T_a } \right)=1.$

Since by continuity of the Brownian paths we have,

$B_{T_a } =a,$

we conclude,

$\mathbb{E} \left(e^{ -\frac{\alpha^2}{2} T_a} \right)=e^{-\alpha a}.$

The formula for the density function of $T_a$ is obtained by inverting the previous Laplace transform $\square$

This entry was posted in Stochastic Calculus lectures. Bookmark the permalink.