Reflected Brownian Motion from a Skorokhod Equation

Reflection by a Running Maximum

Brownian motion can move below zero, but a reflected process is constrained to remain on the nonnegative half-line. The Skorokhod construction enforces this constraint by adding the smallest continuous, nondecreasing correction that keeps the path nonnegative. The correction is called the regulator: it remains constant while the reflected path is positive and increases only when that path is at the boundary.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, adapted to the filtration. Brownian paths are continuous almost surely. Fix an initial value $x\ge0$. The construction below is pathwise: first apply it to any continuous input path $y$ with $y(0)=x$, then take $y(t)=x+B_t$. The deterministic result is what makes the stochastic construction work.

Definition: For a continuous path $y:[0,T]\to\mathbb{R}$ with $y(0)\ge0$, define its regulator and reflected path by $$ L_t=\sup_{0\le s\le t}(-y(s))^+,\qquad X_t=y(t)+L_t,\qquad 0\le t\le T, $$ where $a^+=\max(a,0)$. A solution of the Skorokhod equation for $y$ is a pair of continuous paths $(X,L)$ such that $X=y+L\ge0$, $L_0=0$, $L$ is nondecreasing, and $\int_0^T\mathbf{1}_{\{X_s>0\}}\,dL_s=0$. The integral here is with respect to the Lebesgue–Stieltjes measure of the nondecreasing path $L$.

For Brownian input $y(t)=x+B_t$, this gives $$ X_t=x+B_t+L_t,\qquad L_t=\sup_{0\le s\le t}(-x-B_s)^+. $$ The equation is often written $dX_t=dB_t+dL_t$. The $dL_t$ term is a pathwise finite-variation correction, not an Ito integral and not an ordinary drift coefficient of the form $\mu(t,X_t)\,dt$. No stochastic integral is needed for this construction.

The Skorokhod Construction and Its Properties

Theorem: For each continuous $y$ with $y(0)\ge0$, the paths $X=y+L$ and $L_t=\sup_{0\le s\le t}(-y(s))^+$ solve the Skorokhod equation. This solution is unique. Moreover, if $y,z$ are two such inputs, then $$ \sup_{0\le t\le T}|L^y_t-L^z_t|\le\sup_{0\le t\le T}|y(t)-z(t)|,\qquad \sup_{0\le t\le T}|X^y_t-X^z_t|\le2\sup_{0\le t\le T}|y(t)-z(t)|. $$

Proof. Set $m(t)=\sup_{0\le s\le t}(-y(s))^+$. It is nonnegative, nondecreasing, and continuous: the running maximum of a continuous function is continuous. Also $m(0)=0$, since $y(0)\ge0$. For each $t$, $m(t)\ge(-y(t))^+$, so $y(t)+m(t)\ge0$. This proves nonnegativity of $X$.

It remains to verify the boundary-support condition. On any open interval on which $X(t)=y(t)+m(t)>0$, the function $m$ must be constant. Indeed, if $m$ increased between two times in that interval, a new running maximum of $(-y)^+$ would be attained at some time $s$ in the interval. Since the new maximum is positive, $-y(s)=m(s)$, which gives $X(s)=0$, a contradiction. Thus $m$ is constant on each connected component of the open set $\{t:X(t)>0\}$. Its Lebesgue–Stieltjes measure therefore assigns zero mass to that open set, proving $\int_0^T\mathbf{1}_{\{X_s>0\}}\,dm(s)=0$. This establishes existence.

For uniqueness, let $(\widetilde X,\widetilde L)$ be any solution. Since $\widetilde X(s)=y(s)+\widetilde L(s)\ge0$, we have $\widetilde L(s)\ge(-y(s))^+$. Monotonicity then gives $\widetilde L(t)\ge m(t)$ for every $t$. To prove the reverse inequality, suppose $\widetilde L(t)>m(t)$. Because $\widetilde L$ is continuous and starts at zero, it accumulates positive Stieltjes measure at times $s\le t$ when $\widetilde L(s)>m(t)$. But at any such time where that measure can increase, the support condition requires $\widetilde X(s)=0$; hence $\widetilde L(s)=-y(s)\le m(t)$, a contradiction. More explicitly, the Stieltjes measure of $\{s\le t:\widetilde L(s)>m(t)\}$ is $\widetilde L(t)-m(t)>0$, since $\widetilde L$ is continuous. The support condition says that this measure can only be carried by $\{\widetilde X=0\}$, where $\widetilde L(s)=-y(s)\le m(t)$. This is impossible. Therefore $\widetilde L(t)=m(t)$, and then $\widetilde X=y+m$; uniqueness follows.

Finally, for two inputs $y,z$, the map $a\mapsto a^+$ is 1-Lipschitz, and taking suprema changes values by at most the largest pointwise difference. Consequently $\sup_t|L^y_t-L^z_t|\le\sup_t|y(t)-z(t)|$. Since $X^y-X^z=(y-z)+(L^y-L^z)$, the triangle inequality gives the stated bound for $X$. $\square$

Reflected Brownian Motion as a Strong Solution

Apply the deterministic construction to $y(t)=x+B_t$. The process $L_t=\sup_{0\le s\le t}(-x-B_s)^+$ is adapted: it depends only on the Brownian path up to time $t$, so it is $\mathcal{F}_t$-measurable. It is continuous and nondecreasing, and $X_t=x+B_t+L_t$ is continuous and nonnegative. Thus the construction uses the given Brownian motion on the given probability space; it establishes a strong solution. The pathwise uniqueness in the theorem says that any two solutions driven by this same Brownian path and with the same initial value agree almost surely at every time.

A weak solution, in contrast, is formulated on a probability space that need only carry a Brownian motion and a process satisfying the equation; the Brownian motion need not be fixed in advance. The same deterministic uniqueness argument applies on any such space. Since the constructed pair is a measurable function of the input path, Brownian motion’s fixed law then determines the law of the reflected process. Thus this construction also gives uniqueness in law. These statements concern the reflection equation and do not identify $L$ with a particular normalization of Brownian local time.

Worked Example: A Path That First Stays at the Boundary

Take the continuous piecewise-linear input $y$ with values $y(0)=0$, $y(1)=-2$, $y(2)=-1$, and $y(3)=1$, linear between the listed times. Compute $L$ and $X$ on each segment.

On $0\le t\le1$, $y(t)=-2t$, so $(-y(t))^+=2t$. This is increasing, hence $L_t=2t$ and $X_t=-2t+2t=0$. On $1\le t\le2$, $y(t)=-2+(t-1)=t-3$. Its minimum so far is still $-2$, so $L_t=2$ and $X_t=t-1$. On $2\le t\le3$, $y(t)=-1+2(t-2)=2t-5$. This rises from $-1$ to $1$, so the running negative minimum remains $-2$, $L_t=2$, and $X_t=2t-3$.

Check: At the knots, the construction gives $(L_0,L_1,L_2,L_3)=(0,2,2,2)$ and $(X_0,X_1,X_2,X_3)=(0,0,1,3)$. In particular $X=y+L$ at each knot. The regulator increases on $[0,1]$, where $X=0$, and is constant thereafter, as required.

Worked Example: A Positive Initial Value

Let $x=1$, and choose a piecewise-linear Brownian-input path $B$ with values $B(0)=0$, $B(1)=-\tfrac12$, $B(2)=-2$, and $B(3)=-\tfrac32$. The input to the reflection map is $y=1+B$, with knot values $1,\tfrac12,-1,-\tfrac12$.

On $[0,1]$, $y(t)=1-\tfrac12t$, which stays positive, so $L_t=0$ and $X_t=1-\tfrac12t$. On $[1,2]$, write $u=t-1$. Then $y(t)=\tfrac12-\tfrac32u$, which reaches zero when $\tfrac12-\tfrac32u=0$, or $u=\tfrac13$, that is, $t=\tfrac43$. Before that time, $L=0$ and $X=y$. After that time, the running negative minimum of $y$ is $y(t)$, so $L_t=-y(t)$ and $X_t=0$. In particular $L_2=1$.

On $[2,3]$, $y(t)=-1+\tfrac12(t-2)$ increases from $-1$ to $-\tfrac12$. The running minimum stays at $-1$, so $L_t=1$ and $X_t=\tfrac12(t-2)$.

Check: At $t=2$, $y=-1$, $L=1$, and $X=0$; at $t=3$, $y=-\tfrac12$, $L=1$, and $X=\tfrac12$. The regulator’s increase begins exactly when $y$ first reaches zero and stops once the input starts recovering from its minimum.

Worked Example: A Later, Deeper Minimum

Let $y$ be piecewise linear through $(0,0),(1,-1),(2,-\tfrac25),(3,-\tfrac32),(4,-\tfrac7{10})$. Find the running regulator and reflected values at the knots.

At time $0$, the running negative maximum is $L_0=0$. At time $1$, the input has reached $-1$, so $L_1=1$ and $X_1=-1+1=0$. At time $2$, the input is $-\tfrac25$, which is above the earlier minimum $-1$; thus $L_2=1$ and $X_2=-\tfrac25+1=\tfrac35$. At time $3$, the input reaches the new minimum $-\tfrac32$, so $L_3=\tfrac32$ and $X_3=0$. At time $4$, the input is $-\tfrac7{10}$, above that minimum; hence $L_4=\tfrac32$ and $X_4=-\tfrac7{10}+\tfrac32=\tfrac45$.

Check: The regulator values $0,1,1,\tfrac32,\tfrac32$ are nondecreasing. At each knot, $X=y+L$ gives $0,0,\tfrac35,0,\tfrac45$, all nonnegative. It increases only when the path establishes a new negative minimum, at which the reflected path is zero.

What the Construction Says—and a Common Mistake

For Brownian input, almost every continuous Brownian path is transformed by the same running-maximum rule. The regulator records the depth of the lowest input value seen so far below zero; it does not continually push the process upward while it is away from the boundary. This pathwise description is useful both for reasoning about reflection and for constructing it without solving a separate stochastic integral equation.

A common error is to use the current negative part $L_t=(-y(t))^+$ instead of the running maximum. In the third example, at time $1$ this gives $L_1=1$, but at time $2$ it gives $L_2=\tfrac25$. That proposed correction decreases from $1$ to $\tfrac25$, violating the requirement that a regulator be nondecreasing. It also forgets that the earlier minimum was $-1$: the correct values at time $2$ are $L_2=1$ and $X_2=\tfrac35$, not $L_2=\tfrac25$ and $X_2=0$. The running supremum is essential because it retains the deepest deficit encountered so far.

The regulator is defined here by the Skorokhod equation and its boundary-support condition. Its relation to boundary local time requires an additional normalization argument; it is not part of this pathwise construction.

Check Your Understanding

Use the running-maximum formula and the boundary-support condition to check your reasoning.

For the input $y(t)=\tfrac12-t$ on $0\le t\le1$, find $L_t$ and $X_t$ before and after the input first reaches zero.
Explain why $L_t=(-y(t))^+$ need not be nondecreasing for a continuous input path.
For any solution $(X,L)$, show from $X_s\ge0$ and monotonicity of $L$ that $L_t\ge\sup_{0\le s\le t}(-y(s))^+$.
State the boundary-support condition using the Lebesgue–Stieltjes measure of $L$.
Why does the pathwise construction establish a strong solution when the input is $x+B_t$?

Reflected Brownian Motion from a Skorokhod Equation

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