The Joint Law of the Brownian Maximum and Endpoint

Why the Endpoint and Maximum Must Be Considered Together

The terminal value $B_t$ describes where a Brownian path ends; the running maximum records whether it has crossed a barrier along the way. These are different kinds of information. For example, a path may end below a level $a$ after having crossed $a$, or it may end below $a$ without ever reaching it. The reflection principle distinguishes these possibilities and gives the joint law of the maximum and endpoint.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_s)_{s\ge0},\mathbb{P})$ carrying a standard Brownian motion $(B_s)_{s\ge0}$. Thus $B_0=0$, its paths are continuous, and its increments over disjoint time intervals are independent, with $B_v-B_u\sim N(0,v-u)$ for $u<v$. Fix $t>0$, and define the running maximum $M_t=\sup_{0\le s\le t}B_s$. In particular, $M_t\ge0$. Write $p_t(y)=\dfrac{1}{\sqrt{2\pi t}}\exp\!\left(-\dfrac{y^2}{2t}\right)$ for the density of $B_t$, and write $\Phi$ for the standard normal distribution function.

Definition: The joint distribution function of $(M_t,B_t)$ is $F_t(m,x)=\mathbb{P}(M_t\le m,\ B_t\le x)$. A joint density, when it exists, is a function $f_t(m,x)$ whose integral over any Borel set gives the probability that $(M_t,B_t)$ belongs to that set.

The Reflection Identity

For a barrier $a>0$, let $\tau_a=\inf\{s\ge0:B_s=a\}$, with $\tau_a=\infty$ if the path never reaches $a$. The key idea is to reflect the part of the path after its first hit of $a$. An endpoint $y<a$ then becomes the endpoint $2a-y>a$. The strong Markov property and the symmetry of Brownian increments ensure that this reflection preserves probability.

Proposition: If $a>0$ and $C$ is a Borel subset of $(-\infty,a)$, then $$ \mathbb{P}(M_t\ge a,\ B_t\in C)=\mathbb{P}(B_t\in 2a-C), $$ where $2a-C=\{2a-y:y\in C\}$. In particular, for $b<a$, $$ \mathbb{P}(M_t\ge a,\ B_t\le b)=\mathbb{P}(B_t\ge 2a-b). $$

Proof. On the event $\{\tau_a\le t\}$, define a reflected path by keeping $B_s$ unchanged up to $\tau_a$, and setting its value at $s\ge\tau_a$ to $2a-B_s$. The strong Markov property at the bounded stopping time $\tau_a\wedge t$ says that, after a hit before $t$, the increments from the hitting time form a Brownian motion independent of the past. Brownian motion and its negative have the same law, so reflection preserves the law of the path after the hit. If the original endpoint is $y$, the reflected endpoint is $2a-y$. For $y\in C\subset(-\infty,a)$, the original path has hit $a$, and its reflected endpoint lies in $2a-C$. Conversely, an endpoint in $2a-C\subset(a,\infty)$ forces the continuous path to have hit $a$; reflecting again recovers an endpoint in $C$. Thus the reflection gives a probability-preserving correspondence between the two events. Taking $C=(-\infty,b]$ gives the stated special case; endpoint equalities have probability zero because $B_t$ has a density. $\square$

The proposition is stronger than the marginal maximum formula: it keeps the endpoint restriction in the event. Taking $C=(-\infty,a)$, or using the usual reflection argument for endpoints above $a$, gives $\mathbb{P}(M_t\ge a)=2\mathbb{P}(B_t\ge a)$ for $a>0$. Consequently $M_t$ has no atom at any positive $a$, and $\mathbb{P}(M_t=0)=0$ by letting $a$ decrease to zero.

The Joint Distribution Function and Density

Theorem: For $t>0$ and $m\ge0$, $$ F_t(m,x)= \begin{cases} \Phi\!\left(\dfrac{x}{\sqrt{t}}\right)+ \Phi\!\left(\dfrac{2m-x}{\sqrt{t}}\right)-1,&x<m,\\[6pt] 2\Phi\!\left(\dfrac{m}{\sqrt{t}}\right)-1,&x\ge m. \end{cases} $$ For $m<0$, $F_t(m,x)=0$. The joint density is $$ f_t(m,x)=\frac{2(2m-x)}{t}\,p_t(2m-x), \qquad m>0,\quad x<m, $$ and is zero outside this region. There is no probability mass on its boundary.

Proof. First take $m>0$ and $x<m$. Since $B_t\le x$ implies $B_t<m$, the reflection identity with $a=m$ gives $\mathbb{P}(M_t\ge m,\ B_t\le x)=\mathbb{P}(B_t\ge2m-x)$. Subtract this from $\mathbb{P}(B_t\le x)$, using the absence of an atom at $M_t=m$: $$ F_t(m,x)=\Phi\!\left(\frac{x}{\sqrt{t}}\right) -\left[1-\Phi\!\left(\frac{2m-x}{\sqrt{t}}\right)\right]. $$ This is the first formula. If $x\ge m$, the condition $B_t\le x$ is automatic whenever $M_t\le m$, since $B_t\le M_t$. Hence $F_t(m,x)=\mathbb{P}(M_t\le m)=2\Phi(m/\sqrt{t})-1$, by the marginal reflection formula. For $m=0$ the same expression is zero, because $M_t\ge0$ and $\mathbb{P}(M_t=0)=0$. For $m<0$, the event is empty.

On the open region $m>0,\ x<m$, differentiate the first expression first with respect to $x$, then with respect to $m$. The derivative of $\Phi(z)$ is $p_1(z)$, and $p_t(z)=p_1(z/\sqrt{t})/\sqrt{t}$. Therefore $$ \frac{\partial F_t}{\partial x} =p_t(x)-p_t(2m-x), \qquad \frac{\partial^2 F_t}{\partial m\,\partial x} =\frac{2(2m-x)}{t}\,p_t(2m-x). $$ This is the stated density. To check that it accounts for all probability, fix $x$. If $x\ge0$, integrate the density over $m\in[x,\infty)$; with $u=2m-x$, the integral is $\int_x^\infty (u/t)p_t(u)\,du=p_t(x)$. If $x<0$, integrate over $m\in[0,\infty)$; the same substitution gives $\int_{-x}^\infty (u/t)p_t(u)\,du=p_t(-x)=p_t(x)$. These equalities follow from $p_t'(u)=-(u/t)p_t(u)$. Thus integrating over the maximum coordinate recovers the endpoint density.

There is no mass on $m=0$, because $\mathbb{P}(M_t=0)=0$, nor on $x=m$. For the latter, the process $(B_t-B_{t-s})_{0\le s\le t}$ has the law of Brownian motion: its increments over disjoint reversed intervals are independent centered normal variables with the corresponding interval lengths. Moreover $M_t-B_t=\sup_{0\le s\le t}(B_{t-s}-B_t)$, which has the law of a Brownian maximum by symmetry. That maximum is zero with probability zero. The displayed density therefore gives the whole joint law, not merely its part away from the boundary. $\square$

Using the Joint Law

Worked Example: A High Maximum with a Low Endpoint

For $t=1$, calculate $\mathbb{P}(M_1\ge1,\ B_1\le0)$. The conditions $b=0<a=1$ meet the reflection identity’s hypothesis, so

\mathbb{P}(M_1\ge1,\ B_1\le0) =\mathbb{P}(B_1\ge2) =1-\Phi(2).

Check: reflection maps paths that hit $1$ and end at or below $0$ to paths ending at or above $2$. The result is at most $\mathbb{P}(B_1\ge2)$, in fact equal to it; it is also at most $\mathbb{P}(B_1\le0)=1/2$, as the expression requires.

Worked Example: A Barrier Crossing and an Endpoint Interval

For $t=1$, find the probability that $M_1\ge1$ and $-\tfrac12\le B_1\le\tfrac12$. The whole endpoint interval lies below the barrier, so reflection maps it to $[2-\tfrac12,\,2-(-\tfrac12)]=[\tfrac32,\tfrac52]$. Hence

\begin{aligned} \mathbb{P}\left(M_1\ge1,\ -\frac12\le B_1\le\frac12\right) &=\mathbb{P}\left(\frac32\le B_1\le\frac52\right)\\ &=\Phi\!\left(\frac52\right)-\Phi\!\left(\frac32\right). \end{aligned}

Check: the reflected interval has the same length as the original and lies entirely above $1$, so every endpoint in it implies a barrier crossing. Its probability is nonnegative because $\tfrac52>\tfrac32$ and $\Phi$ is increasing.

Worked Example: Staying Below a Barrier and Ending in an Interval

For $t=1$, calculate $\mathbb{P}(M_1\le1,\ 0\le B_1\le\tfrac12)$. For endpoint values $y<1$, the density of paths that stay below $1$ and end at $y$ is $p_1(y)-p_1(2-y)$, by subtracting the reflected paths that crossed the barrier. Integrating over the requested interval gives

\begin{aligned} \mathbb{P}\left(M_1\le1,\ 0\le B_1\le\frac12\right) &=\int_0^{1/2}\bigl[p_1(y)-p_1(2-y)\bigr]\,dy\\ &=\Phi\!\left(\frac12\right)-\Phi(0) -\left[\Phi(2)-\Phi\!\left(\frac32\right)\right]. \end{aligned}

Check: the first difference is the probability $0\le B_1\le\tfrac12$; the bracket is the probability $\tfrac32\le B_1\le2$, the reflected interval removed by the barrier condition. The answer is nonnegative: on $0\le y\le\tfrac12$, $y^2<(2-y)^2$, so $p_1(y)>p_1(2-y)$ throughout the interval.

Interpretation and a Common Pitfall

The support condition $x\le m$ is unavoidable: a path cannot end above its own maximum. Within that support, the density depends on both coordinates through $2m-x$. This dependence is the quantitative reason the endpoint and maximum are not independent. In particular, a terminal value below a barrier does not tell us whether that barrier was crossed earlier; the reflected term measures exactly the paths that did cross.

Common mistake: subtracting the wrong reflected tail. For $m>0$ and $x<m$, an incorrect calculation is $\mathbb{P}(M_t\le m,B_t\le x)=\Phi(x/\sqrt t)-\Phi((2m-x)/\sqrt t)$. This subtracts the upper-tail complement incorrectly and can even produce a negative value. The reflected paths have probability $1-\Phi((2m-x)/\sqrt t)$, so the correction is $\Phi(x/\sqrt t)-[1-\Phi((2m-x)/\sqrt t)]$. For instance, at $t=1,m=1,x=0$, the incorrect expression is $\tfrac12-\Phi(2)<0$, impossible for a probability; the correct value is $\Phi(0)+\Phi(2)-1>0$.

For a barrier event with endpoint values entirely below the barrier, reflection often gives the shortest calculation. For a stay-below-barrier event, subtract the reflected endpoint density and integrate. The conditions matter: the simple endpoint reflection identity above requires the endpoint set to lie below the barrier. When an endpoint interval straddles the barrier, split it at the barrier and treat the part below separately; endpoint values above the barrier already force a crossing.

Check Your Understanding

Use the joint law and reflection identity, keeping track of which endpoint sets lie below the barrier.

For $t>0$ and $0<b<a$, express $\mathbb{P}(M_t\ge a,\ B_t\le b)$ using $\Phi$.
For $m>0$, explain why $F_t(m,x)$ is constant in $x$ when $x\ge m$.
Integrate the joint density over $m\in[\max(0,x),\infty)$ and verify that the result is $p_t(x)$.
For $t=1$, write the reflected-interval expression for $\mathbb{P}(M_1\ge2,\ -1\le B_1\le0)$.
Why does the formula for $F_t(m,x)$ use a different expression when $x\ge m$ than when $x<m$?