The Joint Distribution of the Maximum and Endpoint

Why the Endpoint and Maximum Belong Together

Knowing where Brownian motion ends does not tell us how high it rose along the way. Conversely, knowing its highest value does not determine its endpoint. Joint questions retain both pieces of information: for example, what is the probability that the path rises above a barrier but ends below zero? Reflection turns such questions into calculations with a two-dimensional density.

Let $(B_s)_{s\ge0}$ be standard Brownian motion on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, with state space $\mathbb{R}$, time index $s\ge0$, and $B_0=0$. For a fixed $t>0$, define the running maximum $M_t=\max_{0\le s\le t}B_s$. Write $\phi_t(x)=\frac{1}{\sqrt{2\pi t}}\exp(-x^2/(2t))$ for the density of $B_t$, and let $\Phi(x)=\int_{-\infty}^{x}\phi_1(u)\,du$ be the standard normal distribution function.

Theorem (Joint density): For $t>0$, the pair $(M_t,B_t)$ has density $$ f_{M_t,B_t}(m,y)= \begin{cases} \displaystyle \frac{2(2m-y)}{t}\phi_t(2m-y),&m>\max(0,y),\\[4pt] 0,&\text{otherwise}. \end{cases} $$ The boundaries $m=0$ and $m=y$ carry no probability mass. Thus the density is supported on paths whose maximum is positive and at least as large as their endpoint.

The formula applies to standard Brownian motion started at zero and a fixed positive time. Its key input is the reflection principle: for a barrier $x>0$, reflecting the path after its first visit to $x$ maps paths that reach $x$ and end below $x$ to paths ending at the reflected endpoint $2x-B_t$. In density form, for $y<x$, $\mathbb{P}(M_t\ge x,\ B_t\in dy)=\phi_t(2x-y)\,dy$. This is the reflection principle from the earlier material; the derivation below uses it to identify the joint law.

Deriving the Density

Fix $m>0$. When $y<m$, the event $B_t\in dy$ splits into paths that stay at or below $m$ and paths that reach $m$. The endpoint density of the latter paths is given by reflection, so subtracting it from the unrestricted endpoint density gives $\mathbb{P}(M_t\le m,\ B_t\in dy)=[\phi_t(y)-\phi_t(2m-y)]\,dy$. When $y>m$, the event $M_t\le m$ is impossible because $B_t\le M_t$. Differentiating the density in this expression with respect to the upper bound $m$ gives the joint density on $m>\max(0,y)$.

Proof. We show that the expression in the theorem is the density of the whole pair, not just a formal derivative. Fix $0<a<b$ and an endpoint value $y<a$. The reflection principle gives the endpoint density on the event $M_t\le m$, for $m=a$ and $m=b$, as $\phi_t(y)-\phi_t(2m-y)$. Subtracting the expression at $a$ from the expression at $b$, the density for $a<M_t\le b$ and endpoint $y$ is $\phi_t(2a-y)-\phi_t(2b-y)$. On the other hand, integrating the proposed density over $a<m\le b$ gives $\int_a^b \frac{2(2m-y)}{t}\phi_t(2m-y)\,dm=\phi_t(2a-y)-\phi_t(2b-y)$, since $\frac{d}{dm}\phi_t(2m-y)=-\frac{2(2m-y)}{t}\phi_t(2m-y)$. Thus the proposed density gives the correct probabilities on such rectangles, which determine the distribution in the interior $m>\max(0,y)$.

To check that no probability is missing at the boundary or outside that interior, integrate the proposed density over $m>\max(0,y)$ for each fixed $y$. Set $u=2m-y$, so $dm=du/2$. If $y\ge0$, the lower limit for $u$ is $y$; if $y<0$, it is $-y$. In either case the integral is $\int_{|y|}^{\infty}\frac{u}{t}\phi_t(u)\,du=\phi_t(|y|)=\phi_t(y)$, using $\frac{d}{du}\phi_t(u)=-\frac{u}{t}\phi_t(u)$ and the symmetry $\phi_t(|y|)=\phi_t(y)$. Integrating this endpoint density over all $y$ gives $1$. The density therefore accounts for the full probability law, including showing that the excluded boundaries carry no mass. $\square$

The support condition has a direct pathwise meaning: $M_t\ge0$ because $B_0=0$, and $M_t\ge B_t$ because the terminal time is among the times over which the maximum is taken. The density is zero outside these constraints. Within the support, it depends on the reflected endpoint $2m-y$, rather than only on $m$ or only on $y$.

Worked Example: Evaluating the Density at a Point

Take $t=1$, $m=1$, and $y=0$. These values satisfy $1>\max(0,0)$, so the joint density is given by the theorem. Since $\phi_1(2)=\frac{1}{\sqrt{2\pi}}e^{-2}$, substitution gives

f_{M_1,B_1}(1,0) =\frac{2(2\cdot1-0)}{1}\phi_1(2) =4\phi_1(2) =\frac{4}{\sqrt{2\pi}}e^{-2}.

Check: the reflected endpoint in the formula is $2m-y=2$, and the coefficient is $2(2m-y)/t=4$, as used. This is a density value, not the probability of the single point $(M_1,B_1)=(1,0)$; a single point has probability zero under a density.

Marginal Checks

A joint density should reproduce each marginal law when the other coordinate is integrated out. These checks also show how the joint formula relates to the endpoint distribution and the law of the running maximum.

Corollary: The endpoint marginal is $\phi_t(y)$. The running maximum has density $2\phi_t(m)$ for $m>0$, and density zero for $m\le0$.

Proof. Integrating over the allowed $m$-values for fixed $y$ gives $\phi_t(y)$, as computed in the theorem’s proof. For the other marginal, fix $m>0$ and integrate over $y<m$. With $u=2m-y$, the integral becomes $\int_m^\infty \frac{2u}{t}\phi_t(u)\,du=2\phi_t(m)$. It is zero for $m<0$ because the maximum cannot be negative, and the theorem’s density assigns no mass to $m=0$. Both marginals therefore follow by integration of the joint density. $\square$

The maximum density integrates to one: $\int_0^\infty 2\phi_t(m)\,dm=1$, since $\phi_t$ is symmetric and integrates to one on the real line. In particular, for $a\ge0$, $\mathbb{P}(M_t>a)=2\int_a^\infty\phi_t(m)\,dm$. This is a useful consistency check, but joint events require the joint density: the maximum and endpoint are not independent.

Computing Joint Probabilities

For an event described by ranges of both variables, integrate the density over the corresponding region, respecting the support $m>\max(0,y)$. Reflection can make some of these integrals shorter: on a region where $y<m$, the density in $m$ is the negative derivative of $\phi_t(2m-y)$. The next examples illustrate both approaches.

Worked Example: The Maximum Exceeds One but the Endpoint Is Nonpositive

Find $\mathbb{P}(M_1>1,\ B_1\le0)$. For endpoints $y\le0$, the threshold $1$ is above $y$, so the reflection principle gives the density of paths that reach $1$ and end at $y$ as $\phi_1(2-y)$. Integrating over those endpoints,

\begin{aligned} \mathbb{P}(M_1>1,\ B_1\le0) &=\int_{-\infty}^{0}\phi_1(2-y)\,dy\\ &=\int_{2}^{\infty}\phi_1(u)\,du\\ &=1-\Phi(2)=\Phi(-2). \end{aligned}

The substitution is $u=2-y$, with $dy=-du$; as $y$ runs from $-\infty$ to $0$, $u$ runs from $\infty$ down to $2$. The last equality uses symmetry of the standard normal density. This probability is positive even though the endpoint is nonpositive: a path can rise above $1$ and later return below zero.

Check: the reflected endpoint $2-y$ is at least $2$ throughout the integration range. Thus the resulting normal tail is precisely $\Phi(-2)$, not the endpoint tail $\mathbb{P}(B_1\le0)=1/2$.

Worked Example: A Rectangle for the Maximum and Endpoint

Find $\mathbb{P}(1<M_1\le2,\ -1\le B_1\le0)$. Throughout this rectangle, $y\le0<1<m$, so it lies inside the density’s support. Integrating first with respect to $m$,

\begin{aligned} \mathbb{P}(1<M_1\le2,\ -1\le B_1\le0) &=\int_{-1}^{0}\int_1^2 2(2m-y)\phi_1(2m-y)\,dm\,dy\\ &=\int_{-1}^{0}\bigl[\phi_1(2-y)-\phi_1(4-y)\bigr]\,dy\\ &=\bigl[\Phi(3)-\Phi(2)\bigr]-\bigl[\Phi(5)-\Phi(4)\bigr]. \end{aligned}

For the inner integral, $\frac{d}{dm}\phi_1(2m-y)=-2(2m-y)\phi_1(2m-y)$, giving the displayed difference at $m=1$ and $m=2$. For the outer integrals, $u=2-y$ ranges from $3$ to $2$ as $y$ goes from $-1$ to $0$, giving $\Phi(3)-\Phi(2)$ after reversing limits. Likewise $u=4-y$ gives $\Phi(5)-\Phi(4)$.

Check: the probability is the difference between two normal tail intervals, $\int_2^3\phi_1(u)\,du$ and $\int_4^5\phi_1(u)\,du$. The first is larger because $\phi_1$ decreases on the positive half-line and the two intervals have the same length, so the computed probability is positive.

Worked Example: A Maximum Bound and a Positive Endpoint

Find $\mathbb{P}(M_1\le1,\ 0\le B_1\le\tfrac12)$. For every endpoint in this interval, $y\le1$, and the reflection formula for paths that stay at or below $1$ gives endpoint density $\phi_1(y)-\phi_1(2-y)$. Therefore

\begin{aligned} \mathbb{P}(M_1\le1,\ 0\le B_1\le\tfrac12) &=\int_0^{1/2}\bigl[\phi_1(y)-\phi_1(2-y)\bigr]\,dy\\ &=\Phi(\tfrac12)-\Phi(0)-\bigl[\Phi(2)-\Phi(\tfrac32)\bigr]\\ &=\Phi(\tfrac12)-\tfrac12-\Phi(2)+\Phi(\tfrac32). \end{aligned}

The first integral is the standard normal probability between $0$ and $\tfrac12$. In the second, the substitution $u=2-y$ changes the endpoints to $2$ and $\tfrac32$, so reversing the limits gives $\Phi(2)-\Phi(\tfrac32)$. The difference is nonnegative because it is the integral of the nonnegative endpoint density for paths that stay below $1$.

Check: at $y=\tfrac12$, the reflected endpoint is $2-\tfrac12=\tfrac32$, while at $y=0$ it is $2$. These are the limits in the subtracted normal interval.

A Common Independence Error

Common mistake. Multiplying the marginal probabilities as if $M_1$ and $B_1$ were independent gives the wrong answer for joint events. For example, the marginal law gives $\mathbb{P}(M_1>1)=2\Phi(-1)$, and symmetry gives $\mathbb{P}(B_1\le0)=\tfrac12$. The independence calculation would therefore be $2\Phi(-1)\cdot\tfrac12=\Phi(-1)$. But the reflection calculation above gives $\mathbb{P}(M_1>1,\ B_1\le0)=\Phi(-2)$. Since $-2<-1$ and $\Phi$ is strictly increasing, $\Phi(-2)<\Phi(-1)$. The product is not the joint probability. The correction is to use the joint density, or the reflected endpoint density when the event has a suitable form.

The same support constraint is an important check on any integral: a proposed region with $y>m$ is impossible, because the terminal value cannot exceed the running maximum. Once the region is restricted correctly, the reflected argument $2m-y$ and the factor $2(2m-y)/t$ must both be retained. Dropping the factor or integrating over endpoints that cannot occur changes the law.

Check Your Understanding

Use the support condition and the joint density to work through these questions.

For $t=2$, state the set of $(m,y)$ where the joint density is positive, and explain why each support restriction holds.
For $t=1$, express $\mathbb{P}(M_1>2,\ B_1\le0)$ as a standard normal tail, using reflection.
Integrate the joint density over $m>\max(0,y)$ and verify that the endpoint marginal is $\phi_t(y)$.
For $t=1$, set up the integral for $\mathbb{P}(1<M_1\le\tfrac32,\ -\tfrac12\le B_1\le0)$, checking that the whole rectangle is within the support.
Why does multiplying $\mathbb{P}(M_1>1)$ by $\mathbb{P}(B_1\le0)$ fail to give $\mathbb{P}(M_1>1,\ B_1\le0)$?