Brownian reflection principle

Reflecting a Brownian path after its first hit of a level $b>0$ preserves Wiener measure and maps an endpoint $x\le b$ to $2b-x\ge b$.

Reflecting a Brownian motion $B$ after a stopping time $T$ by ${\hat{B}}_t=2B_T-B_t$ for $t>T$, and leaving it unchanged for $t\le T$, preserves its path probability distribution. For finite $T$, the Strong Markov property and symmetry of the future Brownian motion prove this. When $T=\infty$, the path is left unchanged.

For $M_t={\sup }_{0\le s\le t}{B}_s$, the reflection principle givesso $M_t$ and $|B_t|$ have the same distribution.

For standard one-dimensional Brownian motion and a fixed $h>0$, the Brownian reflection principle gives the survival probability $2\Phi (h/\sqrt{t})-1$. The standard normal distribution function has derivative $1/\sqrt{2\pi }$ at zero, giving the displayed asymptotic. Thus the persistence exponent is one half, with leading constant $h\sqrt{2/\pi }$.

Stopping Brownian motion at its first crossing below $-b$, the nonnegative local martingale $(W_{t\wedge \tau }+b)/b$ starts at one and ends at zero. The maximal identity for a continuous nonnegative local martingale tending to zero gives tail $b/(b+x)$ for the stopped maximum, hence density $b/(b+x{)}^2$ for $x>0$. There is no atom at zero.

Let $M_t={\sup }_{0\le s\le t}{B}_s$ for a standard Brownian motion. For $m\ge 0$, the Brownian reflection principle givesOn $b<m$, the pair $(B_t,M_t)$ therefore has joint probability density

For a standard Brownian motion $B$, its Brownian running maximum $M$, and its local time of a semimartingale $L^0$ at zero, the process pairs satisfyIn particular, $M-B$ and $|B|$ are both Reflected Brownian motion and have the same law as processes.

Brownian motion attains its maximum on $[0,1]$ at a unique time $M^{\ast }$ almost surely. This time has the arcsine distribution

Let $\tau$ be the first attainment of the maximum of standard Brownian motion on $[0,T]$, with $T>0$. Time reversal shows $\tau <T$ with probability one. Then $Y_t\le 0$ for $0\le t\le T-\tau$, an initial interval of positive length. Standard Brownian motion has probability zero of this property by the Brownian reflection principle, so $Y$ is not Brownian. Moreover $\tau$ is not a stopping time for the natural Brownian filtration: for $0<t<T$, $\mathbb{P}(\tau \le t\mid {\mathcal{F}}_t)=2\Phi ((M_t-B_t)/\sqrt{T-t})-1$, which is strictly between zero and one on $\{B_t<0\}$. The Strong Markov property is therefore inapplicable at this random time.