Boundary Local Time and the Reflection Regulator

Two Measures of Accumulated Boundary Contact

Reflected Brownian motion stays in the half-line by adding a nondecreasing correction to its Brownian input. That correction, called the regulator, grows only when the process is at zero. Local time also measures accumulated contact with a level, but its numerical value depends on the normalization used. With symmetric semimartingale local time, the regulator equals the local time. With a one-sided boundary occupation normalization, the corresponding quantity is twice the regulator.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix \(T>0\), and let \(x\ge0\). Assume the reflected process and its regulator satisfy the Skorokhod equation \(X_t=x+B_t+K_t\), where \(X_t\ge0\), \(K_0=0\), \(K\) is continuous and nondecreasing, and \(K\) can increase only when \(X=0\). The last condition means \(\int_0^T\mathbf{1}_{\{X_s>0\}}\,dK_s=0\), where the integral is pathwise with respect to the finite-variation process \(K\). These are the half-line reflection conditions established by the Skorokhod reflection map.

Definition: The symmetric local time at zero, denoted \(L_t^0(X)\), is normalized by the occupation approximation \(L_t^0(X)=\lim_{\varepsilon\downarrow0}\dfrac{1}{2\varepsilon}\int_0^t\mathbf{1}_{\{|X_s|<\varepsilon\}}\,d[X]_s\), in the standard local-time sense. Here \([X]\) is quadratic variation. Equivalently, this is the local time in the symmetric Tanaka formula \(|X_t|=|X_0|+\int_0^t\operatorname{sgn}(X_s)\,dX_s+L_t^0(X)\), with \(\operatorname{sgn}(0)=0\).

In this reflected setting, \([X]_t=t\): adding the continuous finite-variation process \(K\) does not change Brownian quadratic variation. Thus the occupation approximation here uses \(d[X]_s=ds\). Every Itô integral against \(B\) below has a bounded, progressively measurable integrand; on \([0,T]\) it belongs to \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^T H_s^2\,ds\le T\). Such integrals are square-integrable martingales: their conditional expectations satisfy \(\mathbb{E}[\int_0^t H_s\,dB_s\mid\mathcal{F}_u]=\int_0^u H_s\,dB_s\) for \(u\le t\), by the Itô integral construction and its martingale property.

The Regulator–Local-Time Identity

Theorem: For reflected Brownian motion satisfying the conditions above, almost surely for all \(t\in[0,T]\), \(K_t=L_t^0(X)\), when \(L^0(X)\) is symmetric local time with the stated normalization.

Proof. We apply the symmetric Tanaka formula to \(X\). Since \(X_t\ge0\), \(|X_t|=X_t\), and \(\operatorname{sgn}(X_s)=\mathbf{1}_{\{X_s>0\}}\). In the integral against \(dX=dB+dK\), the \(dK\) contribution vanishes: \(dK\) is supported on \(\{X=0\}\), where the sign is zero. It remains to identify the Brownian integral over \(\{X>0\}\).

The occupation-density formula implies \(\int_0^T\mathbf{1}_{\{X_s=0\}}\,d[X]_s=0\), because the singleton \(\{0\}\) has zero Lebesgue measure in the level variable. Since \(d[X]_s=ds\), this says \(\int_0^T\mathbf{1}_{\{X_s=0\}}\,ds=0\) almost surely. The bounded progressively measurable integrand \(\mathbf{1}_{\{X_s=0\}}\) is therefore in \(\mathcal{L}^2\). By the Itô isometry, \(\int_0^t\mathbf{1}_{\{X_s=0\}}\,dB_s=0\) in \(L^2\), for each \(t\). Consequently \(\int_0^t\mathbf{1}_{\{X_s>0\}}\,dB_s=B_t\). Substitution in Tanaka’s formula gives \(X_t=x+B_t+L_t^0(X)\). Comparing with the Skorokhod equation \(X_t=x+B_t+K_t\) yields \(K_t=L_t^0(X)\). The continuous versions of both processes make the equality simultaneous for all \(t\in[0,T]\). \(\square\)

The proof uses both boundary conditions, not just nonnegativity. The support condition removes the regulator from the Tanaka integral, while the zero Lebesgue measure of the boundary set lets the Brownian integral over \(\{X>0\}\) recover the full Brownian motion. Omitting either step can lead to an incorrect factor.

Where the Factor of Two Enters

Corollary: For reflected Brownian motion, the one-sided boundary occupation approximation \(A_\varepsilon(t)=\dfrac{1}{\varepsilon}\int_0^t\mathbf{1}_{\{0\le X_s<\varepsilon\}}\,d[X]_s\) converges to \(2K_t\) in the local-time occupation sense. Hence the one-sided boundary local time defined by this approximation is \(2K_t\), not \(K_t\).

Proof. Since \(X\ge0\), the indicator \(\mathbf{1}_{\{|X_s|<\varepsilon\}}\) in the symmetric occupation approximation agrees with \(\mathbf{1}_{\{0\le X_s<\varepsilon\}}\). The possible endpoint \(X_s=0\) contributes no \(d[X]\)-mass, because \(\int_0^t\mathbf{1}_{\{X_s=0\}}\,d[X]_s=0\). Therefore \(A_\varepsilon(t)=2\left(\dfrac{1}{2\varepsilon}\int_0^t\mathbf{1}_{\{|X_s|<\varepsilon\}}\,d[X]_s\right)\). The term in parentheses converges to \(L_t^0(X)=K_t\); multiplying by two proves the claim. \(\square\)

Worked Example: Converting Between the Two Normalizations

Suppose \(K_t=3\) at a fixed time, and the symmetric occupation approximation is close to its limit. What are the corresponding symmetric and one-sided local-time values?

By the theorem, the symmetric local time is \(L_t^0(X)=K_t=3\). The one-sided approximation uses \(1/\varepsilon\) rather than \(1/(2\varepsilon)\), so its limit is \(2L_t^0(X)=2\cdot3=6\).

Check: dividing the one-sided value by two recovers the regulator: \(6/2=3=K_t\). The factor comes from the denominator in the normalization, not from an additional push at the boundary.

Worked Checks for Brownian Inputs

Worked Example: Expected Regulator at Time Two

Take \(x=0\) and \(T=2\). The Skorokhod formula gives \(K_T=\max(0,-\inf_{0\le s\le T}B_s)\). Compute \(\mathbb{E}[K_2]\), and then identify the expected symmetric local time.

By the Brownian reflection principle, for \(y>0\), \(\mathbb{P}(K_2>y)=2\mathbb{P}(B_2>y)=2\bigl(1-\Phi(y/\sqrt{2})\bigr)\). The tail-integral formula for a nonnegative random variable gives \(\mathbb{E}[K_2]=\int_0^\infty\mathbb{P}(K_2>y)\,dy\). Changing variables \(u=y/\sqrt{2}\) yields \(\mathbb{E}[K_2]=2\sqrt{2}\int_0^\infty(1-\Phi(u))\,du\). For a standard normal variable \(Z\), \(\int_0^\infty(1-\Phi(u))\,du=\mathbb{E}[Z^+]=1/\sqrt{2\pi}\), so \(\mathbb{E}[K_2]=2/\sqrt{\pi}\). By the theorem, \(\mathbb{E}[L_2^0(X)]=2/\sqrt{\pi}\) as well.

Check: the one-sided boundary occupation normalization has expected limit \(2\mathbb{E}[K_2]=4/\sqrt{\pi}\). Halving it gives \(2/\sqrt{\pi}\), the symmetric local-time expectation just computed.

Worked Example: A Diffusion Coefficient of Three

Now let the input be \(3B_t\) with \(x=0\), and let \(K_t\) be the Skorokhod regulator in \(X_t=3B_t+K_t\). Find the expected regulator at \(T=2\), and compare it with symmetric local time.

The regulator is \(K_2=\max(0,-3\inf_{0\le s\le2}B_s)\). Since \(3>0\), this is three times the regulator for input \(B\). The preceding tail calculation gives \(\mathbb{E}[K_2]=3(2/\sqrt{\pi})=6/\sqrt{\pi}\). Here \([X]_t=9t\), so the occupation approximation uses \(d[X]_s=9\,ds\); the regulator–local-time theorem still gives \(\mathbb{E}[L_2^0(X)]=6/\sqrt{\pi}\).

Check: the one-sided boundary normalization gives \(2\mathbb{E}[K_2]=12/\sqrt{\pi}\). The diffusion coefficient changes the accumulated contact, but it does not change the factor relating the two normalizations.

Worked Example: Recovering the Regulator by Tanaka’s Formula

Use Tanaka’s formula to identify the regulator for any reflected process \(X_t=x+B_t+K_t\) satisfying the stated assumptions, without first computing the running minimum of \(B\).

For \(f(z)=|z|\), the derivatives away from zero are \(f'(z)=\operatorname{sgn}(z)\); the symmetric Tanaka correction is \(L_t^0(X)\). Because \(X\ge0\), \(f(X_t)=X_t\), and the stochastic integral is \(\int_0^t\mathbf{1}_{\{X_s>0\}}\,dB_s=B_t\), as shown in the theorem proof. The finite-variation contribution is \(\int_0^t\operatorname{sgn}(X_s)\,dK_s=0\), because \(dK\) is supported on \(X=0\). Thus Tanaka gives \(X_t=x+B_t+L_t^0(X)\). Subtracting \(x+B_t\) from the Skorokhod equation gives \(K_t=L_t^0(X)\).

Check: both decompositions have the same initial value \(x\) and the same Brownian term \(B_t\). Their remaining continuous terms must therefore agree, which recovers the regulator exactly.

A Common Normalization Mistake

A common error is to use the one-sided occupation integral but label its limit “symmetric local time.” For example, if \(A_\varepsilon(t)=\varepsilon^{-1}\int_0^t\mathbf{1}_{\{0\le X_s<\varepsilon\}}\,d[X]_s\), the symmetric local-time approximation is \(A_\varepsilon(t)/2\), not \(A_\varepsilon(t)\). Since \(A_\varepsilon(t)\) tends to \(2K_t\), calling that limit the symmetric local time would incorrectly give \(K_t=L_t^0(X)/2\). The correction is to check the denominator in the occupation approximation before applying Tanaka’s formula: with the symmetric denominator \(2\varepsilon\), the limit is \(K_t=L_t^0(X)\).

This distinction matters whenever boundary local time is used as a coefficient in a reflected equation or as a measure of boundary occupation. A statement such as “the regulator is local time” is incomplete until the normalization is specified. With the symmetric Tanaka normalization used here, they are equal; with the one-sided normalization above, the boundary local time is twice the regulator.

Check Your Understanding

Keep the local-time normalization explicit in each answer.

Which support condition on \(dK\) makes its contribution to the symmetric Tanaka integral vanish?
Why does \(\int_0^t\mathbf{1}_{\{X_s=0\}}\,dB_s=0\) hold for reflected Brownian motion?
If the symmetric local time is \(5\), what is the limiting value of the one-sided boundary occupation approximation defined above?
For input \(2B_t\) and \(T=1\), use Brownian scaling and the regulator–local-time identity to find \(\mathbb{E}[K_1]\) and \(\mathbb{E}[L_1^0(X)]\).
Explain why changing the occupation denominator from \(2\varepsilon\) to \(\varepsilon\) doubles the limiting boundary local time.