Distributions of the Brownian Maximum and Minimum

From a Path Extremum to a Distribution

The endpoint $B_t$ records where Brownian motion finishes, but many questions concern whether it reaches a level at any time before $t$. The running maximum and minimum capture those pathwise events. The reflection principle turns a crossing event into an endpoint probability, allowing the full one-dimensional distributions of these extrema to be calculated from the normal distribution.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, with $B_0=0$. Fix $t\ge0$, and define

M_t=\sup_{0\le s\le t}B_s,\qquad m_t=\inf_{0\le s\le t}B_s.

Continuity of Brownian paths ensures these extrema are finite and attained on $[0,t]$. For $t>0$, $M_t\ge0$ and $m_t\le0$, since the path starts at zero. Let $\Phi(x)=\mathbb{P}(Z\le x)$ for a standard normal random variable $Z$. We use the reflection principle in the form that, for $a>0$, paths that hit $a$ and finish below $a$ have the same probability as paths that finish above $a$.

Distribution of the Running Maximum

Theorem: For $t>0$ and $a>0$, $$ \mathbb{P}(M_t\ge a)=2\mathbb{P}(B_t\ge a)=2\left(1-\Phi\left(\frac{a}{\sqrt{t}}\right)\right). $$ Consequently, the cumulative distribution function $F_{M_t}(x)=\mathbb{P}(M_t\le x)$ is $0$ for $x\le0$, and $2\Phi(x/\sqrt{t})-1$ for $x>0$.

Proof. Fix $a>0$. Split the event $\{M_t\ge a\}$ according to whether the endpoint is at least $a$ or below $a$. If $B_t\ge a$, continuity and $B_0=0<a$ imply the path has hit $a$, so $\{B_t\ge a\}\subseteq\{M_t\ge a\}$. By the reflection principle, the probability of paths that hit $a$ and end below $a$ equals $\mathbb{P}(B_t>a)$. The endpoint $B_t$ has a normal density, so $\mathbb{P}(B_t=a)=0$, and therefore $\mathbb{P}(M_t\ge a)=\mathbb{P}(B_t\ge a)+\mathbb{P}(B_t>a)=2\mathbb{P}(B_t\ge a)$. Since $B_t/\sqrt{t}$ is standard normal, this is the stated tail probability. For $x>0$, the same formula with strict crossing follows by taking levels decreasing to $x$: $\mathbb{P}(M_t>x)=\lim_{n\to\infty}\mathbb{P}(M_t\ge x+1/n)=2\mathbb{P}(B_t>x)$. Thus $\mathbb{P}(M_t=x)=0$, and $F_{M_t}(x)=1-2(1-\Phi(x/\sqrt{t}))=2\Phi(x/\sqrt{t})-1$. At zero, $\mathbb{P}(M_t>0)=\lim_{n\to\infty}\mathbb{P}(M_t\ge1/n)=1$, so there is no atom at zero. For $x<0$, $M_t\ge0$ makes the CDF zero; at $x=0$ it is also zero. $\square$

The formula distinguishes the chance of ever crossing a positive level from the chance of ending above it. In fact, the crossing probability is twice the endpoint tail. The factor of two accounts for paths that cross the level and then finish below it, paired by reflection with paths that finish above it.

Worked Example: Maximum on a Four-Unit Interval

For $t=4$, find the probability that the running maximum is at most $2$, the probability it exceeds $2$, and the density at $2$.

First standardize the level: $2/\sqrt{4}=1$. The CDF formula gives $\mathbb{P}(M_4\le2)=2\Phi(1)-1$. The strict crossing probability is $\mathbb{P}(M_4>2)=2(1-\Phi(1))$. These add to $2\Phi(1)-1+2-2\Phi(1)=1$, as required.

For $x>0$, differentiate the CDF using $\Phi'(z)=(2\pi)^{-1/2}e^{-z^2/2}$. The density is $f_{M_t}(x)=\sqrt{2/(\pi t)}e^{-x^2/(2t)}$. At $t=4,x=2$, this is $\sqrt{2/(4\pi)}e^{-4/8}=e^{-1/2}/\sqrt{2\pi}$.

Check: integrating this density from $0$ to $2$, with the substitution $z=x/2$, gives $\int_0^1\sqrt{2/\pi}e^{-z^2/2}\,dz=2\Phi(1)-1$, the CDF value above. There is no probability mass below zero or at zero.

Distribution of the Running Minimum

The process $(-B_s)_{s\ge0}$ is also a standard Brownian motion: its increments are independent centered normal variables with the same variances as those of $B$. Its running maximum over $[0,t]$ is $-m_t$. Thus the minimum is the negative of a random variable with the maximum distribution. This sign symmetry determines both its CDF and its density.

Proposition: For $t>0$, the CDF of $m_t$ is $$ F_{m_t}(x)= \begin{cases} 2\Phi(x/\sqrt{t}),&x<0,\\ 1,&x\ge0. \end{cases} $$ Its density is $f_{m_t}(x)=\sqrt{2/(\pi t)}e^{-x^2/(2t)}$ for $x<0$, and zero for $x>0$. There is no atom at zero.

Proof. Since $-m_t$ has the same distribution as $M_t$, for $x<0$ we have $\mathbb{P}(m_t\le x)=\mathbb{P}(-m_t\ge -x)=\mathbb{P}(M_t\ge -x)$. The maximum tail formula gives $2(1-\Phi((-x)/\sqrt{t}))$. The standard normal density is symmetric, so $1-\Phi(-x/\sqrt{t})=\Phi(x/\sqrt{t})$, yielding $2\Phi(x/\sqrt{t})$. For $x\ge0$, $m_t\le0$ almost surely, so the CDF is one. Differentiation on $x<0$ gives the stated density; the CDF approaches zero as $x$ increases to zero and equals one at zero, but its left limit at zero is $2\Phi(0)=1$. Hence there is no jump, and no atom at zero. $\square$

Although the maximum and minimum densities have the same formula on their respective supports, their supports lie on opposite sides of zero. The equality of the formulas reflects sign symmetry, not equality of the two random variables on the same path.

Worked Example: A Negative Minimum Level

For $t=9$, calculate the probability that the minimum is at most $-3$, the probability it is greater than $-3$, and the density at $-3$.

Here $(-3)/\sqrt{9}=-1$. The minimum CDF gives $\mathbb{P}(m_9\le-3)=2\Phi(-1)=2(1-\Phi(1))$, where the last equality uses normal symmetry. Its complement is $\mathbb{P}(m_9>-3)=1-2\Phi(-1)=2\Phi(1)-1$.

The density at $x=-3$ is $\sqrt{2/(9\pi)}e^{-(-3)^2/(2\cdot9)}=\sqrt{2/(9\pi)}e^{-1/2}$.

Check: by symmetry, $-m_9$ has the distribution of $M_9$. Therefore $\mathbb{P}(m_9\le-3)=\mathbb{P}(-m_9\ge3)=\mathbb{P}(M_9\ge3)=2(1-\Phi(3/\sqrt9))=2(1-\Phi(1))$, agreeing with the direct calculation.

Densities and Quantiles

For $t>0$, differentiating the maximum CDF on its positive support gives $f_{M_t}(x)=\sqrt{2/(\pi t)}e^{-x^2/(2t)}$ for $x>0$. The density integrates to one: substituting $z=x/\sqrt{t}$, its total mass is $\int_0^\infty\sqrt{2/\pi}e^{-z^2/2}\,dz=1$. The minimum density has the same expression on $x<0$ and also integrates to one by reflection. Neither law has an atom at zero.

For a continuous distribution function $F$, its $p$-quantile for $0<p<1$ is the unique $q$ satisfying $F(q)=p$, when $F$ is strictly increasing on its support. The maximum and minimum CDFs are strictly increasing on their respective supports, so solving their formulas gives

Quantile formulas: For $t>0$ and $0<p<1$, $$ q_p(M_t)=\sqrt{t}\,\Phi^{-1}\left(\frac{p+1}{2}\right),\qquad q_p(m_t)=\sqrt{t}\,\Phi^{-1}\left(\frac{p}{2}\right). $$

Proof. For the maximum, its $p$-quantile is positive and solves $2\Phi(q/\sqrt{t})-1=p$. Rearranging gives $\Phi(q/\sqrt{t})=(p+1)/2$; applying the increasing inverse function $\Phi^{-1}$ yields the first formula. For the minimum, the quantile is negative and solves $2\Phi(q/\sqrt{t})=p$. Dividing by two and applying $\Phi^{-1}$ gives the second formula. Both solutions lie on the required supports: $(p+1)/2$ belongs to $(1/2,1)$, while $p/2$ belongs to $(0,1/2)$. Strict increase ensures uniqueness. $\square$

Worked Example: Median Maximum and Lower-Quartile Minimum

On the interval $[0,4]$, find the median of $M_4$ and the 25th percentile of $m_4$, and verify both answers.

For the maximum, $t=4$ and $p=1/2$, so $q_{1/2}(M_4)=2\Phi^{-1}((1/2+1)/2)=2\Phi^{-1}(3/4)$. Substitution into the CDF gives $F_{M_4}(2\Phi^{-1}(3/4))=2\Phi(\Phi^{-1}(3/4))-1=2(3/4)-1=1/2$.

For the minimum, $p=1/4$, so $q_{1/4}(m_4)=2\Phi^{-1}((1/4)/2)=2\Phi^{-1}(1/8)$. Its CDF at that value is $F_{m_4}(2\Phi^{-1}(1/8))=2\Phi(\Phi^{-1}(1/8))=2(1/8)=1/4$.

Check: $\Phi^{-1}(1/8)=-\Phi^{-1}(7/8)$ by normal symmetry. Hence the minimum's 25th percentile is the negative of the maximum's 75th percentile, as sign reversal requires.

Degenerate Time and a Common Mistake

When $t=0$, the interval contains only the starting time, so $M_0=m_0=0$ with probability one. The formulas involving $1/\sqrt{t}$ and densities apply only for $t>0$; at zero the laws are point masses at zero. For positive time, all positive levels have zero probability of being an exact maximum level, and all negative levels have zero probability of being an exact minimum level. Consequently, replacing a strict crossing event by a non-strict one does not change its probability at a fixed nonzero threshold.

Common mistake: treating the maximum like the endpoint. A tempting but incorrect calculation is $\mathbb{P}(M_1\le1)=\Phi(1)$, as if the event were just $B_1\le1$. The endpoint event permits paths that crossed $1$ and returned below it. The correct calculation is $\mathbb{P}(M_1\le1)=1-2(1-\Phi(1))=2\Phi(1)-1$. Using $\Phi(1)\approx0.8413$, the incorrect value is $0.8413$, while the correct value is approximately $0.6827$; the difference is the probability of paths that cross the level but finish below it. The reflection principle is essential precisely because a path event cannot generally be replaced by an endpoint event.

These distributions are useful whenever a calculation depends only on the highest or lowest level reached by Brownian motion over a fixed horizon. They provide crossing probabilities and quantiles without requiring the joint law of the extremum and endpoint. The maximum and minimum considered separately are fully described here; their joint behavior with each other or with $B_t$ requires additional information.

Check Your Understanding

Use the CDF, symmetry, and quantile formulas above. Assume $t>0$ unless a question specifies otherwise.

For $a>0$, express $\mathbb{P}(M_t>a)$ in terms of $\Phi$, and explain why it equals $\mathbb{P}(M_t\ge a)$.
Write the CDF of $m_t$ at a negative value $x$, then use normal symmetry to express it using $\Phi(-x/\sqrt{t})$.
Find the $p$-quantile of $M_t$ and verify it by substitution into its CDF.
What are the laws of $M_0$ and $m_0$, and why do the positive-time density formulas not apply at $t=0$?
Explain why $\mathbb{P}(B_t\le a)$ is not, in general, the same as $\mathbb{P}(M_t\le a)$ for $a>0$.