The Maximum of Absolute Brownian Motion

Brownian Motion Between Two Barriers

A one-sided barrier can be handled by reflecting Brownian paths at their first passage. For two barriers, a path can hit either side, and correcting for paths that encounter both requires more than adding two one-sided probabilities. A useful approach is to calculate the probability of staying inside the interval directly. It leads to two convergent series: an image series that is effective when the barriers are far away relative to the time, and an eigenfunction series that makes the distribution especially explicit.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ carrying standard Brownian motion $(B_t)_{t\ge0}$, with state space $S=\mathbb{R}$, time index $t\ge0$, and $B_0=0$. Thus the paths are continuous and the increments over disjoint time intervals are independent centered normal random variables with variance equal to the interval length. Use the natural filtration $\mathcal{F}_t=\sigma(B_s:0\le s\le t)$. For $t>0$, define the maximum absolute value up to time $t$ by $M_t^{\ast}=\sup_{0\le s\le t}|B_s|$. For $a>0$, the event $M_t^{\ast}<a$ means the path stays strictly between $-a$ and $a$ throughout $[0,t]$.

Definition: Let $g_t(z)=(2\pi t)^{-1/2}\exp(-z^2/(2t))$, the density of a centered normal random variable with variance $t$. Write $\Phi(x)=\int_{-\infty}^{x}g_1(z)\,dz$ for the standard normal distribution function and $Q(x)=1-\Phi(x)$ for its upper tail.

The Image Series

The method of images represents the density of paths that have not yet reached either boundary as a signed sum of ordinary Gaussian densities. In this formula, the positive terms are images of the starting point, while the negative terms cancel paths that have crossed a boundary. The formula is valid for $a>0$, $t>0$, and $-a<y<a$.

Standard heat-kernel result: The transition density at time $t$, for Brownian motion started at $x\in(-a,a)$ and stopped when it exits $(-a,a)$, is the Dirichlet heat kernel on that interval. Its method-of-images representation is $$ K_a(t,x,y)=\sum_{k\in\mathbb{Z}}\left[g_t(y-x+4ka)-g_t(y+x+(4k+2)a)\right],\qquad -a<y<a. $$ The Dirichlet heat-kernel identification with the stopped Brownian transition density is a standard result; its general proof is not given here. The series converges absolutely for $t>0$, and its terms can be integrated over $(-a,a)$.

For clarity, the image formula can be checked directly as a heat-kernel construction. Each summand solves the heat equation in the spatial variable, and the Gaussian tails make the series and its derivatives converge on compact subsets for $t>0$. At $y=a$, the positive and negative terms cancel after reindexing; the same cancellation holds at $y=-a$. As $t$ decreases to zero, only the direct Gaussian term contributes to the initial mass inside the interval; the other image points lie outside it. These properties give the Dirichlet heat kernel, whose identification with stopped Brownian motion is the standard result stated above.

Starting at $x=0$, integrate the kernel over the interval. The event of remaining in the interval through time $t$ is exactly the event that the stopped process has not exited by that time. In the integral, shifting the variable in each Gaussian term tiles the real line by intervals of length $2a$, with alternating signs. Specifically, the interval corresponding to index $k$ is $((2k-1)a,(2k+1)a)$, and its sign is $(-1)^k$. Hence

\begin{aligned} \mathbb{P}(M_t^{\ast}<a) &=\sum_{k\in\mathbb{Z}}(-1)^k\int_{(2k-1)a}^{(2k+1)a}g_t(z)\,dz\\ &=1-4\sum_{k=0}^{\infty}(-1)^k Q\!\left(\frac{(2k+1)a}{\sqrt{t}}\right). \end{aligned}

To see the second equality, the central interval contributes $1-2Q(a/\sqrt{t})$. The two intervals at distance $2a$ from the origin contribute $2\bigl(Q(a/\sqrt{t})-Q(3a/\sqrt{t})\bigr)$ with a negative sign. The next pair contributes $2\bigl(Q(3a/\sqrt{t})-Q(5a/\sqrt{t})\bigr)$ with a positive sign. Continuing this grouping gives the displayed tail series. Since $Q(x)$ decreases to zero as $x$ increases, the terms decrease to zero and the alternating series converges. In particular, truncating after the term with argument $(2m+1)a/\sqrt{t}$ gives an error in the survival probability of at most $4Q((2m+3)a/\sqrt{t})$.

The Eigenfunction Series and the Maximum Distribution

The same probability has a series with exponential terms. It is useful when the required probability is small, since its terms decay rapidly with the square of the index.

Theorem: For every $a>0$ and $t>0$, $$ \mathbb{P}(M_t^{\ast}<a)=\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1} \exp\!\left(-\frac{(2k+1)^2\pi^2t}{8a^2}\right). $$ The series is absolutely convergent. Its partial sums also have alternating-series error bounds.

Proof. The Dirichlet heat kernel on $(-a,a)$ also has the eigenfunction expansion $$ K_a(t,x,y)=\frac{1}{a}\sum_{n=1}^{\infty} \exp\!\left(-\frac{n^2\pi^2t}{8a^2}\right) \sin\!\left(\frac{n\pi(x+a)}{2a}\right) \sin\!\left(\frac{n\pi(y+a)}{2a}\right). $$ This is the standard Fourier expansion of the Dirichlet heat kernel. At $x=0$, the first sine factor is $\sin(n\pi/2)$, which vanishes for even $n$. For odd $n=2k+1$, it equals $(-1)^k$. Also, $$ \int_{-a}^{a}\sin\!\left(\frac{n\pi(y+a)}{2a}\right)\,dy =\frac{2a}{n\pi}\bigl(1-\cos(n\pi)\bigr) =\frac{4a}{n\pi} $$ when $n$ is odd. Integrating the kernel over $y\in(-a,a)$ therefore gives the stated series. For absolute convergence, put $c=\pi^2t/(8a^2)>0$. The absolute value of its $k$-th term is bounded by $e^{-c(2k+1)}/(2k+1)$, and the latter terms are bounded by a convergent geometric sequence. Finally, the positive term magnitudes $e^{-c(2k+1)^2}/(2k+1)$ decrease to zero, so the alternating-series remainder after any term is no greater than the first omitted term, multiplied by $4/\pi$. $\square$

For a fixed $t>0$, the formula gives the distribution of $M_t^{\ast}$ as the barrier $a$ varies. It also shows that this distribution has no atoms at positive thresholds.

Proof. Fix $t>0$. On any compact range $0<a_0\le a\le a_1$, the absolute values of the eigenfunction-series terms are bounded by $$ \frac{4}{\pi(2k+1)} \exp\!\left(-\frac{(2k+1)^2\pi^2t}{8a_1^2}\right). $$ The sum of these bounds is finite, so the series converges uniformly on that range. Each term is continuous in $a$; uniform convergence therefore makes $f(a)=\mathbb{P}(M_t^{\ast}<a)$ continuous for $a>0$. If $b_n\downarrow a$, the events $\{M_t^{\ast}<b_n\}$ decrease to $\{M_t^{\ast}\le a\}$. Continuity of probability for decreasing events gives $\mathbb{P}(M_t^{\ast}\le a)=\lim_{n\to\infty}f(b_n)=f(a)$. Thus $\mathbb{P}(M_t^{\ast}=a)=0$. $\square$

Worked Examples

Worked Example: A Barrier of One at Time One

Set $a=1$ and $t=1$. Substituting these values into the eigenfunction series gives $$ \mathbb{P}(M_1^{\ast}<1) =\frac{4}{\pi}\left(e^{-\pi^2/8}-\frac{1}{3}e^{-9\pi^2/8} +\frac{1}{5}e^{-25\pi^2/8}-\cdots\right). $$

The first two terms give the approximation $\frac{4}{\pi}(e^{-\pi^2/8}-\frac{1}{3}e^{-9\pi^2/8})$. The omitted terms start with a positive term. Their magnitudes decrease, since the exponential decreases as $2k+1$ increases and the denominator $2k+1$ increases. Therefore the true probability is at least this two-term partial sum and exceeds it by at most the first omitted term, $\frac{4}{5\pi}e^{-25\pi^2/8}$.

Check: The first omitted term has index $k=2$, so $2k+1=5$, and its exponent is $-5^2\pi^2/8=-25\pi^2/8$. This is exactly the stated error bound.

Worked Example: A Barrier of Two at Time One

Set $a=2$ and $t=1$. In the image series, $a/\sqrt{t}=2$, and the successive tail arguments are $2,6,10,\ldots$. Thus $$ \mathbb{P}(M_1^{\ast}\ge2) =4\left(Q(2)-Q(6)+Q(10)-\cdots\right). $$

Keeping the first two terms gives $4(Q(2)-Q(6))$. Since $Q(2)>Q(6)>Q(10)>\cdots>0$, the alternating-series bound shows that this partial sum is a lower bound on the exit probability, with error at most $4Q(10)$. Equivalently, the survival probability lies between $1-4(Q(2)-Q(6)) - 4Q(10)$ and $1-4(Q(2)-Q(6))$.

Check: The first omitted image-series term has index $k=2$, so its tail argument is $(2k+1)a/\sqrt{t}=5\cdot2=10$. Its magnitude is $4Q(10)$, as claimed.

Worked Example: A Barrier of Three at Time Two

Set $a=3$ and $t=2$. The exponent coefficient is $\pi^2t/(8a^2)=2\pi^2/(8\cdot9)=\pi^2/36$. The series begins $$ \mathbb{P}(M_2^{\ast}<3) =\frac{4}{\pi}\left(e^{-\pi^2/36} -\frac{1}{3}e^{-9\pi^2/36} +\frac{1}{5}e^{-25\pi^2/36}-\cdots\right). $$

The one-term partial sum is $\frac{4}{\pi}e^{-\pi^2/36}$. Because the next term is negative and the term magnitudes decrease, the probability is no greater than this one-term sum and is no less than that sum minus the magnitude of the second term. Consequently, $$ \frac{4}{\pi}e^{-\pi^2/36}-\frac{4}{3\pi}e^{-\pi^2/4} \le \mathbb{P}(M_2^{\ast}<3) \le \frac{4}{\pi}e^{-\pi^2/36}. $$

Check: The second exponent is $-9\pi^2/36=-\pi^2/4$, and its coefficient is $(4/\pi)(1/3)=4/(3\pi)$. These are precisely the subtracted error-bound terms.

Why the Two-Sided Calculation Matters

A common mistake is to use the one-sided reflection principle twice and conclude that $\mathbb{P}(M_t^{\ast}\ge a)=4Q(a/\sqrt{t})$. Each one-sided crossing event has probability $2Q(a/\sqrt{t})$, but adding the two probabilities counts paths that hit both barriers twice. The image series gives the correction explicitly: $$ \mathbb{P}(M_t^{\ast}\ge a) =4\left[Q\!\left(\frac{a}{\sqrt{t}}\right) -Q\!\left(\frac{3a}{\sqrt{t}}\right) +Q\!\left(\frac{5a}{\sqrt{t}}\right)-\cdots\right]. $$

For example, at $a=2,t=1$, adding the two one-sided crossing probabilities gives $4Q(2)$. The correct expression is $4(Q(2)-Q(6)+Q(10)-\cdots)$, not $4Q(2)$. The correction terms account for the overlap between upper and lower barrier crossings. The image series is particularly convenient for seeing this correction, while the eigenfunction series is convenient for controlling errors at longer times or narrower barriers. In either representation, check that $a>0$ and $t>0$; at $t=0$, the process starts at zero and the maximum is zero, so the positive-time formulas are not needed.

Check Your Understanding

Use the series and its stated conditions to answer the following questions.

For $a=1,t=2$, what is the coefficient of $\pi^2$ in the exponent of the first eigenfunction-series term?
In the image series, what is the first omitted tail argument if the terms with arguments $a/\sqrt{t}$ and $3a/\sqrt{t}$ have been retained?
Why does adding the upper- and lower-barrier crossing probabilities overcount some paths?
Explain why the eigenfunction-series probability is continuous in $a>0$.
For $a=3,t=2$, write the magnitude of the second term of the eigenfunction series before simplifying its exponent.