Find the distribution of maximum of $B_t-t$

Question

Let $B_t$ be a standard Brownian motion. It is easy to show that $\sup B_t-t<\infty$ a.s. . The question is, can we determinate the distribution of $\sup_{t\in [0,\infty)}B_t-t$?

Answer 1

The process $M$ defined by $M_t = \exp(2B_t-2t)$ is a non-negative continuous martingale starting at $1$ and converging to $0$. Given $a \ge 0$, the values of $M$ until $\tau_a := \inf\{t \ge a : B_t-t \ge a\}$ belong to $[0,\exp(2a)]$. Hence optional topping theorem applies at time $T_a$ so $$1 = \mathbb{E}[M_0] = \mathbb{E}[M_{T_a}] = \exp(2a)\mathbb{P}[T_a < +\infty].$$ As a result, $$P[\sup_{t\in [0,\infty)}B_t-t \ge a] = P[T_a < +\infty] = \exp(-2a).$$

Answer 2

Here is a more general result:

Note that $$P(\max_{s\in[0,t]}(W_s-b s)<a)=P(\tau>t), \tag{A}\label{A}$$ where $W_\cdot$ is the standard Wiener process, $a\in[0,\infty)$, $b\in\mathbb R$, $$\tau:=\tau_{a,b}:=\inf\{s>0\colon X_s=0\},$$ $$X_t:=a+bt+W_t.$$

Proposition 1: For any real $t>0$ and any real $x$, \begin{equation*} P(X_t\in dx,\tau>t)=f_t(x-a-bt)(1-e^{-2ax/t})\,1(x>0)\,dx, \end{equation*} where $f_t$ is the pdf of $W_t$.

Integrating here in $x$ and using \eqref{A}, we get $$P(\max_{s\in[0,t]}(W_s-b s)<a)=P(\tau>t) \\ =\frac{1}{2} \left(\text{erf}\left(\frac{a+b t}{\sqrt{2} \sqrt{t}}\right)-e^{-2 a b} \text{erfc}\left(\frac{a-b t}{\sqrt{2} \sqrt{t}}\right)+1\right). \tag{B}\label{B}$$ Letting here $t\to\infty$, we get $$P(\max_{s\in[0,\infty)}(W_s-b s)<a)=1-e^{-2 a b_+}, \tag{C}\label{C}$$ where $b_+:=\max(0,b)$.

It remains to provide the

Proof of Proposition 1: The process $X_\cdot$ is continuous. So, $X_t>0$ on the event $\{\tau>t\}$. So, without loss of generality $x>0$.

For $s\in[0,t]$, let $B_s:=W_s-\frac st\,W_t$. Then $(B_s)_{s\in[0,t]}$ is (a Brownian bridge and it is) independent of $W_t$. So, \begin{equation} \begin{aligned} &P(X_t\in dx,\tau>t) \\ &=P(X_t\in dx,\ X_s>0\ \forall s\in[0,t]) \\ &=P(X_t\in dx,\ W_s>-a-bs\ \forall s\in[0,t]) \\ &=P(W_t\in dx-a-bt,\ B_s>-a-bs-\tfrac st\,(x-a-bt)\ \forall s\in[0,t]) \\ &=P(W_t\in dx-a-bt,\ B_s>-a-\tfrac st\,(x-a)\ \forall s\in[0,t]) \\ &=P(W_t\in dx-a-bt)\,P(B_s>-a-\tfrac st\,(x-a)\ \forall s\in[0,t]) \\ &=\frac{P(W_t\in dx-a-bt)}{P(W_t\in dx-a)}\,P(W_t\in dx-a)P(B_s>-a-\tfrac st\,(x-a)\ \forall s\in[0,t]) \\ &=\frac{P(W_t\in dx-a-bt)}{P(W_t\in dx-a)}\,P(W_t\in dx-a,\ B_s>-a-\tfrac st\,(x-a)\ \forall s\in[0,t]) \\ &=\frac{P(W_t\in dx-a-bt)}{P(W_t\in dx-a)}\,P(W_t\in dx-a,\ W_s>-a\ \forall s\in[0,t]) \\ &=\frac{P(W_t\in dx-a-bt)}{P(W_t\in dx-a)}\,P(W_t\in a-dx,\ W_s<a\ \forall s\in[0,t]). \end{aligned} \tag{0}\label{eq:} \end{equation} The first three equalities in \eqref{eq:} follow by the definitions of $\tau$, $X_\cdot$, and $B_\cdot$; the 5th and 7th equalities there follow by the independence of $(B_s)_{s\in[0,t]}$ from $W_t$; the penultimate equality in \eqref{eq:} follows again by the definition of $B_\cdot$; and the last equality in \eqref{eq:} follows by symmetry. Next, \begin{equation} \frac{P(W_t\in dx-a-bt)}{P(W_t\in dx-a)}=\frac{f_t(x-a-bt)}{f_t(x-a)}. \tag{1}\label{1} \end{equation} Further, letting $M_t:=\max_{s\in[0,t]}W_s$ and using the reflection principle for the Brownian motion, we get \begin{equation} \begin{aligned} & P(W_t\in a-dx,\ W_s<a\ \forall s\in[0,t]) \\ &=P(W_t\in a-dx,\ M_t<a) \\ &=P(W_t\in a-dx)-P(W_t\in a-dx,\ M_t\ge a) \\ &=P(W_t\in a-dx)-P(W_t\in a+dx) \\ &=f_t(a-x)\,dx-f_t(a+x)\,dx =f_t(x-a)(1-e^{-2ax/t})\,dx. \end{aligned} \tag{2}\label{2} \end{equation} Collecting \eqref{eq:}, \eqref{1}, and \eqref{2}, we complete the proof. $\quad\Box$