Brownian running maximum ($M_t$)

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

Stop the Exponential martingale for Brownian motion $e^{\theta B_s-{\theta }^{2}s/2}$ when $B$ first reaches $\delta$ or at $t$. On the hitting event its value is at least $e^{\theta \delta -{\theta }^{2}t/2}$, so that event has probability at most $e^{-\theta \delta +{\theta }^{2}t/2}$. Minimize over $\theta >0$ and apply the same bound to $-B$. The union bound gives the displayed inequality for $t>0$; at zero the probability is zero.

For Brownian motion, write the later maximum as $B_b+Z+H$, where $Z=B_c-B_b$ has a nondegenerate normal distribution and $H={\sup }_{0\le u\le d-c}(B_{c+u}-B_c)$. The independent increments make $Z$ independent of the earlier history and of $H$. Conditional on those two objects, equality of the two maxima prescribes one value of $Z$, an event of probability zero. This argument uses the positive gap $c-b$ and does not assume independence of the two absolute maxima.

For $a>0$ and $t>0$, the distribution function is $\Phi ((a-\nu t)/\sqrt{t})-e^{2a\nu }\Phi ((-a-\nu t)/\sqrt{t})$. The Brownian reflection principle gives the zero-drift terminal density below the barrier as ${\varphi }_t(y)-{\varphi }_t(2a-y)$ for $y<a$. Weight it by $e^{\nu y-{\nu }^{2}t/2}$ using the Girsanov theorem and complete squares in the two Gaussian integrals.

For $W_t^{\nu }=B_t+\nu t$, $M_t^{\nu }={\sup }_{s\le t}{W}_s^{\nu }$, $a>0$ and $x\le a$, the joint distribution function is $\Phi ((x-\nu t)/\sqrt{t})-e^{2a\nu }\Phi ((x-2a-\nu t)/\sqrt{t})$. The Brownian reflection principle gives killed endpoint density ${\varphi }_t(y)-{\varphi }_t(2a-y)$ at zero drift. Multiplication by $e^{\nu y-{\nu }^{2}t/2}$ and completion of squares gives the formula.

Let $f$ be positive, continuous and nondecreasing, with ${\int }_0^{1}f(t)dt/t<\infty$. The small-value probability for the Brownian running maximum is at most a constant times the value. Brownian scaling and dyadic integral comparison make the exceptional probabilities $\mathbb{P}(S_{2^{-n-1}}<2^{-n/2}f(2^{-n}))$ summable. The Borel-Cantelli first lemma and monotonicity interpolate between dyadic times. Apply this argument to $mf$ for every positive integer $m$ on a countable intersection of probability-one events to obtain the infinite lower limit.

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.