The Skorokhod Reflection Map on the Half-Line

The Deterministic Reflection Problem

Suppose a continuous path is allowed to move freely unless it tries to cross below zero. A nondecreasing correction can push it upward, but the correction should not act unnecessarily when the path is already positive. The half-line Skorokhod reflection map makes this rule precise: it takes a continuous input path and returns both the smallest correction needed to keep the path nonnegative and the resulting reflected path.

Fix a time horizon $T>0$. The stochastic setting in which this construction may be used is a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. The reflection construction below is deterministic: it applies separately to each continuous input path $x:[0,T]\to\mathbb{R}$ with $x(0)\ge0$. No stochastic integral is used here. For a random continuous input, the same formulas can be applied path by path wherever the input path is continuous and starts nonnegative.

Definition: A solution of the half-line reflection problem for a continuous input $x$ with $x(0)\ge0$ is a pair of continuous functions $(y,L)$ on $[0,T]$ such that $L(0)=0$, $L$ is nondecreasing, $y(t)=x(t)+L(t)\ge0$ for every $t$, and $L$ can increase only when $y=0$. Precisely, the last condition is $\int_0^T \mathbf{1}_{\{y(s)>0\}}\,dL(s)=0$, where $dL$ denotes the Lebesgue–Stieltjes measure induced by the nondecreasing function $L$.

The restriction $x(0)\ge0$ is necessary with this convention: $L(0)=0$ forces $y(0)=x(0)$. The support condition says that the correction does not increase on any region where the reflected path is strictly above the boundary. The integral in this condition is an ordinary Lebesgue–Stieltjes integral, not an Itô integral.

The Running-Minimum Formula

Theorem: For every continuous $x:[0,T]\to\mathbb{R}$ with $x(0)\ge0$, the unique solution of the half-line reflection problem is $$ L(t)=\max\left(0,-\min_{0\le s\le t}x(s)\right),\qquad y(t)=x(t)+L(t),\qquad 0\le t\le T. $$ Moreover, $L(t)$ is the least possible value at time $t$ among all nondecreasing corrections $A$ with $A(0)=0$ and $x+A\ge0$.

Proof. Define $m(t)=\min_{0\le s\le t}x(s)$ and $L(t)=\max(0,-m(t))$. Since $x$ is continuous on a compact interval, the minimum exists. The running minimum $m$ is nonincreasing, so $L$ is nondecreasing. Also $m(0)=x(0)\ge0$, hence $L(0)=0$. The function $m$, and therefore $L$, is continuous: uniform continuity of $x$ implies that extending the interval $[0,t]$ by a small amount can lower its minimum by at most the maximum change in $x$ over that small amount. For every $t$, $x(t)\ge m(t)$. If $m(t)\ge0$, then $L(t)=0$ and $y(t)=x(t)\ge0$. If $m(t)<0$, then $L(t)=-m(t)$, so $y(t)=x(t)-m(t)\ge0$.

To check the support condition, consider a time $t$ with $y(t)>0$. If $L(t)=0$, then $x(t)>0$; continuity gives a neighborhood of $t$ on which $x$ remains positive, so the running negative minimum stays zero there. If $L(t)>0$, then $x(t)>m(t)$, because $y(t)=x(t)-m(t)>0$. Continuity gives a neighborhood of $t$ on which $x$ stays strictly above $m(t)$. In that neighborhood no new running minimum below $m(t)$ is reached, so $L$ is constant there. Thus $L$ is locally constant on the open set $\{t:y(t)>0\}$, which gives $\int_0^T\mathbf{1}_{\{y(s)>0\}}\,dL(s)=0$.

For minimality, let $A$ be any nondecreasing correction with $A(0)=0$ and $x(s)+A(s)\ge0$ for every $s$. For $s\le t$, this inequality gives $A(s)\ge -x(s)$, and monotonicity gives $A(t)\ge A(s)$. Also $A(t)\ge0$. Taking the largest of these lower bounds over $s\le t$ yields $A(t)\ge\max(0,-\min_{0\le s\le t}x(s))=L(t)$.

It remains to prove uniqueness for solutions satisfying the support condition. Let $(y_1,L_1)$ and $(y_2,L_2)$ be two such solutions. Minimality gives $L_1\le L_2$ and $L_2\le L_1$, since each is an admissible nondecreasing correction for the same input. Hence $L_1=L_2$, and $y_1=x+L_1=x+L_2=y_2$. This proves existence, minimality, and uniqueness. $\square$

Computing the Regulator and Reflected Path

Worked Example: A Single Downward Excursion

On $[0,3]$, let $$ x(t)= \begin{cases} 1-2t,&0\le t\le1,\\ t-2,&1\le t\le3. \end{cases} $$ Compute the regulator $L$ and reflected path $y$.

On $[0,1]$, $x$ decreases from $1$ to $-1$. It first reaches zero at $1-2t=0$, or $t=\tfrac12$. The running minimum is $x(t)$ from $\tfrac12$ through $1$, and it is $-1$ at $t=1$. On $[1,3]$, the input increases from $-1$ to $1$, so the running minimum remains $-1$. Therefore $$ L(t)= \begin{cases} 0,&0\le t\le\tfrac12,\\ 2t-1,&\tfrac12\le t\le1,\\ 1,&1\le t\le3. \end{cases} $$ Adding $L$ to $x$ gives $$ y(t)= \begin{cases} 1-2t,&0\le t\le\tfrac12,\\ 0,&\tfrac12\le t\le1,\\ t-1,&1\le t\le3. \end{cases} $$

Check: At $t=\tfrac12$, $x=1-2(\tfrac12)=0$ and $L=0$, so $y=0$. At $t=1$, $x=-1$ and $L=1$, so $y=0$. At $t=3$, $x=1$ and $L=1$, so $y=2$. The regulator increases only on $[\tfrac12,1]$, where $y=0$, as required.

Worked Example: Two Boundary Contacts

On $[0,4]$, define $$ x(t)= \begin{cases} t,&0\le t\le1,\\ 3-2t,&1\le t\le2,\\ t-3,&2\le t\le3,\\ t-3,&3\le t\le4. \end{cases} $$ Find the running-minimum regulator and verify the resulting path.

The input is nonnegative on $[0,1]$. On $[1,2]$, it decreases from $1$ to $-1$, crossing zero when $3-2t=0$, at $t=\tfrac32$. It attains its minimum $-1$ at $t=2$. On $[2,4]$, it increases from $-1$ to $1$, so the running minimum stays at $-1$. Thus $$ L(t)= \begin{cases} 0,&0\le t\le\tfrac32,\\ 2t-3,&\tfrac32\le t\le2,\\ 1,&2\le t\le4. \end{cases} $$ Consequently, $$ y(t)= \begin{cases} t,&0\le t\le1,\\ 3-2t,&1\le t\le\tfrac32,\\ 0,&\tfrac32\le t\le2,\\ t-2,&2\le t\le4. \end{cases} $$

Check: At $t=\tfrac32$, the input is $3-2(\tfrac32)=0$, while $L=0$. At $t=2$, the input is $-1$, while $L=1$. At $t=3$, the input is $0$, while $L=1$, so $y(3)=1$. The regulator increases only from $\tfrac32$ to $2$, exactly while the reflected path is zero; after $2$, the path is positive and the regulator is constant.

Stability Under Changes in the Input

Proposition: If $x,z:[0,T]\to\mathbb{R}$ are continuous and start nonnegative, and $\Phi(x)=x+L_x$, $\Phi(z)=z+L_z$ are their reflected paths, then $$ \sup_{0\le t\le T}|L_x(t)-L_z(t)|\le \sup_{0\le t\le T}|x(t)-z(t)|,\qquad \sup_{0\le t\le T}|\Phi(x)(t)-\Phi(z)(t)|\le2\sup_{0\le t\le T}|x(t)-z(t)|. $$

Proof. Put $\delta=\sup_{0\le s\le T}|x(s)-z(s)|$. For each fixed $t$, the inequalities $x(s)\ge z(s)-\delta$ and $z(s)\ge x(s)-\delta$ hold for all $s\le t$. Taking minima over the same interval $[0,t]$ gives $\min_{s\le t}x(s)\ge\min_{s\le t}z(s)-\delta$ and $\min_{s\le t}z(s)\ge\min_{s\le t}x(s)-\delta$. Therefore the two running minima differ by at most $\delta$. The map $u\mapsto\max(0,-u)$ changes by at most the change in $u$, so the running-minimum formula gives $|L_x(t)-L_z(t)|\le\delta$. Taking the supremum proves the first bound. For every $t$, the triangle inequality gives $|\Phi(x)(t)-\Phi(z)(t)|=|x(t)-z(t)+L_x(t)-L_z(t)|\le |x(t)-z(t)|+|L_x(t)-L_z(t)|\le2\delta$. Taking the supremum proves the second bound. $\square$

Worked Example: Different Regulators, Equal Reflected Paths

On $[0,3]$, let $x(t)=-t$ for $0\le t\le1$, $x(t)=t-2$ for $1\le t\le2$, and $x(t)=0$ for $2\le t\le3$. Set $z(t)=x(t)+\tfrac14$. Compute both regulators and paths, and compare them.

The minimum of $x$ reaches $-1$ at $t=1$, so $L_x(t)=t$ on $[0,1]$ and $L_x(t)=1$ on $[1,3]$. The minimum of $z$ reaches $-\tfrac34$ at $t=1$, so $L_z(t)=0$ on $[0,\tfrac14]$, $L_z(t)=t-\tfrac14$ on $[\tfrac14,1]$, and $L_z(t)=\tfrac34$ on $[1,3]$. Adding the respective regulators gives $$ \Phi(x)(t)= \begin{cases} 0,&0\le t\le1,\\ t-1,&1\le t\le2,\\ 1,&2\le t\le3, \end{cases} \qquad \Phi(z)(t)= \begin{cases} \tfrac14-t,&0\le t\le\tfrac14,\\ 0,&\tfrac14\le t\le1,\\ t-1,&1\le t\le2,\\ 1,&2\le t\le3. \end{cases} $$

The inputs are uniformly $\tfrac14$ apart. The regulators differ by at most $\tfrac14$, and they differ by exactly $\tfrac14$ throughout $[1,3]$. Yet the reflected paths coincide on $[\tfrac14,3]$. For instance, at $t=3$, $x(3)=0$, $z(3)=\tfrac14$, $L_x(3)=1$, and $L_z(3)=\tfrac34$, so $\Phi(x)(3)=1$ and $\Phi(z)(3)=1$.

Check: On $[0,\tfrac14]$, the path difference is $\tfrac14-t$, whose largest value is $\tfrac14$ at $t=0$. From $\tfrac14$ to $1$, both reflected paths are zero; from $1$ to $2$, both equal $t-1$; from $2$ to $3$, both equal $1$. Thus the uniform distance between reflected paths is $\tfrac14$, attained at $t=0$, not on the interval where the regulators differ.

Common Mistakes and Why the Map Matters

Using the current input instead of its running minimum. In the first example, at $t=3$ the current input is $x(3)=1$. The incorrect pointwise calculation $\max(0,-x(3))=0$ would miss the accumulated correction. The correct regulator is $L(3)=\max(0,-\min_{0\le s\le3}x(s))=1$, because the input previously reached $-1$ at $t=1$. The running minimum records past boundary pressure that a formula based only on the current value discards.

Confusing regulator differences with reflected-path differences. The regulator is added to the input, so a change in regulator can be offset by a change in input. In the last example, at $t=3$ the regulators differ by $\tfrac14$, but the inputs differ by $\tfrac14$ in the opposite direction; both reflected paths equal $1$. The stability proposition bounds the reflected-path difference by twice the input difference, but does not assert that the difference equals the regulator difference.

The key point is pathwise: the reflected value at time $t$ depends on the current input and on the lowest input value seen up to $t$. This makes the construction useful for turning a continuous path into a nonnegative one while enforcing a precise boundary rule. Its formula also makes approximation and comparison straightforward: control the input uniformly, and the regulator and reflected path are controlled uniformly as well.

Check Your Understanding

Use the running-minimum formula and the boundary-support condition to justify each answer.

For an input with $x(0)\ge0$, explain why a correction $L(0)=0$ cannot produce a nonnegative reflected path if $x(0)<0$.
Let $x(t)=2-t$ on $[0,3]$. Compute $L(t)$ and $y(t)$, specifying the time at which the regulator first becomes positive.
For two inputs at uniform distance $\delta$, derive the bound $\sup_t|L_x(t)-L_z(t)|\le\delta$ from their running minima.
In the third worked example, compute $L_x(2)$, $L_z(2)$, $\Phi(x)(2)$, and $\Phi(z)(2)$.
Give a reason the formula $\max(0,-x(t))$ at a single time $t$ need not equal the correct regulator $L(t)$.