Normalization Conventions for Brownian Local Time

Why Brownian Local Time Has More Than One Normalization

Brownian local time measures how much time a path spends near a level. The phrase “near a level” can mean a symmetric band or a one-sided interval, and the limits use different interval lengths. A second choice is whether to divide the symmetric-band occupation time by its full width or by its half-width. These choices alter the numerical local-time field and therefore alter the constants in Tanaka’s formula and the occupation-density formula.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\), with \(B_0=0\). For an adapted, progressively measurable process \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\), the Ito integral \(\int_0^T H_s\,dB_s\) is defined in \(L^2(\mathbb{P})\). The occupation-density result from the earlier material says that Brownian motion has a local-time field \(L_t^a\) such that, for nonnegative Borel \(f\),

\int_0^t f(B_s)\,ds=\int_{\mathbb{R}}f(a)L_t^a\,da.

We use a version for which \(a\mapsto L_t^a\) is continuous. This convention fixes the normalization: \(L_t^a\) is the density of time occupation with respect to Lebesgue measure in the level variable.

Definition: For a fixed level \(a\), the symmetric occupation limit is \(L_t^a=\lim_{\varepsilon\downarrow0}(2\varepsilon)^{-1}\int_0^t\mathbf{1}_{\{|B_s-a|<\varepsilon\}}\,ds\). The right-sided and left-sided occupation limits are respectively \(L_t^{a,+}=\lim_{\varepsilon\downarrow0}\varepsilon^{-1}\int_0^t\mathbf{1}_{\{a\le B_s<a+\varepsilon\}}\,ds\) and \(L_t^{a,-}=\lim_{\varepsilon\downarrow0}\varepsilon^{-1}\int_0^t\mathbf{1}_{\{a-\varepsilon\le B_s<a\}}\,ds\). When these limits agree, their common value is the standard local time \(L_t^a\).

The symmetric band has width \(2\varepsilon\), while either one-sided interval has width \(\varepsilon\). Thus the displayed denominators are all the interval widths. In particular, replacing the symmetric denominator \(2\varepsilon\) by \(\varepsilon\) does not preserve the same field: it doubles it.

Why the Three Occupation Limits Agree

Proposition: Suppose the occupation-density formula holds and \(a\mapsto L_t^a\) is continuous at a fixed level \(a\). Then the symmetric, right-sided, and left-sided occupation limits exist and all equal \(L_t^a\).

Proof. Apply the occupation-density formula to the indicators of the three intervals. The symmetric quotient is \((2\varepsilon)^{-1}\int_{a-\varepsilon}^{a+\varepsilon}L_t^x\,dx\), the right-sided quotient is \(\varepsilon^{-1}\int_a^{a+\varepsilon}L_t^x\,dx\), and the left-sided quotient is \(\varepsilon^{-1}\int_{a-\varepsilon}^{a}L_t^x\,dx\). For any \(\delta>0\), continuity at \(a\) gives a \(\rho>0\) such that \(|L_t^x-L_t^a|<\delta\) whenever \(|x-a|<\rho\). If \(0<\varepsilon<\rho\), the difference between each average and \(L_t^a\) has absolute value at most the average of \(|L_t^x-L_t^a|\), hence is at most \(\delta\). Each quotient therefore tends to \(L_t^a\). \(\square\)

This argument also identifies precisely what is being averaged. If instead we define the rescaled symmetric-band field

A_t^a=\lim_{\varepsilon\downarrow0}\frac{1}{\varepsilon} \int_0^t\mathbf{1}_{\{|B_s-a|<\varepsilon\}}\,ds,

then the same occupation-density calculation gives \(A_t^a=2L_t^a\), not \(L_t^a\). The field itself changes with the normalization.

Worked Example: Comparing the Three Averages

Suppose, at a fixed \(t\) and \(a\), the occupation density is \(L_t^x=3+2(x-a)\) for \(|x-a|\le\varepsilon\). Compute the three occupation averages and the rescaled symmetric-band average.

For the right-sided average, substitute \(u=x-a\):

\frac{1}{\varepsilon}\int_a^{a+\varepsilon}(3+2(x-a))\,dx =\frac{1}{\varepsilon}\int_0^\varepsilon(3+2u)\,du =\frac{3\varepsilon+\varepsilon^2}{\varepsilon} =3+\varepsilon.

The left-sided average is \(\varepsilon^{-1}\int_{-\varepsilon}^0(3+2u)\,du =\varepsilon^{-1}(3\varepsilon-\varepsilon^2)=3-\varepsilon\). For the symmetric average, the odd term integrates to zero:

\frac{1}{2\varepsilon}\int_{-\varepsilon}^{\varepsilon}(3+2u)\,du =\frac{6\varepsilon}{2\varepsilon}=3.

As \(\varepsilon\downarrow0\), all three averages tend to \(3=L_t^a\). The rescaled symmetric-band average divides by \(\varepsilon\), so it is \(6\), tending to \(6=2L_t^a\). Check: the full symmetric band integrates to \(6\varepsilon\); dividing by its width \(2\varepsilon\) gives \(3\), and dividing by \(\varepsilon\) gives \(6\).

Tanaka Formulas Under Each Convention

With the standard normalization \(L_t^a\), the Tanaka formulas are

|B_t-a|=|B_0-a|+\int_0^t\operatorname{sgn}(B_s-a)\,dB_s+L_t^a,

(B_t-a)^+=(B_0-a)^++\int_0^t\mathbf{1}_{\{B_s>a\}}\,dB_s+\frac12L_t^a.

Here \(x^+=\max(x,0)\), and we set \(\operatorname{sgn}(0)=0\). In both formulas the Ito integrand is adapted and progressively measurable, and its absolute value is bounded by \(1\); therefore its squared time integral has expectation at most \(t\), as required for membership in \(\mathcal{L}^2\). The corresponding left-part formula is \((a-B_t)^+=(a-B_0)^+-\int_0^t\mathbf{1}_{\{B_s<a\}}\,dB_s+\tfrac12L_t^a\). The values assigned to indicators at equality do not affect the time integrals: the occupation-density formula implies \(\int_0^t\mathbf{1}_{\{B_s=a\}}\,ds=0\).

These are statements of the standard Tanaka formula; its derivation is treated in the subsequent material on Tanaka’s formula. The normalization changes in these identities follow directly by substitution, as the next result records.

Proposition: If \(A_t^a=2L_t^a\) is used as the local-time field, then the absolute-value correction is \(\tfrac12A_t^a\), and the positive-part and negative-part corrections are each \(\tfrac14A_t^a\).

Proof. The standard absolute-value formula has correction \(L_t^a\). Since \(L_t^a=A_t^a/2\), this correction is \(A_t^a/2\). The standard positive-part formula has correction \(L_t^a/2\), which becomes \(A_t^a/4\). The negative-part correction is also \(L_t^a/2\), so it too becomes \(A_t^a/4\). These substitutions establish each stated coefficient. \(\square\)

Worked Example: A Tanaka Correction at a Fixed Level

At a particular time and level, suppose the standard local-time value is \(L_t^a=3\). Compute the correction in each formula using both conventions.

Under the standard convention, the absolute-value correction is \(3\), and each one-sided positive- or negative-part correction is \(3/2\). The rescaled field is \(A_t^a=2L_t^a=6\). Under that convention the absolute-value correction is \(A_t^a/2=6/2=3\), and each one-sided correction is \(A_t^a/4=6/4=3/2\). Check: both conventions give identical corrections in the path identities; only the numerical value assigned to the local-time field and its coefficient change.

Occupation-Density Formulas and Their Constants

The standard occupation-density formula uses \(L_t^a\) as the density. For any nonnegative Borel function \(f\), or any Borel \(f\) for which the integrals are absolutely integrable, it reads

\int_0^t f(B_s)\,ds=\int_{\mathbb{R}}f(a)L_t^a\,da.

Replacing \(L_t^a\) by \(A_t^a=2L_t^a\) gives the equivalent formula

\int_0^t f(B_s)\,ds=\frac12\int_{\mathbb{R}}f(a)A_t^a\,da.

The factor \(1/2\) is essential. In particular, integrating the density over all levels gives \(\int_{\mathbb{R}}L_t^a\,da=t\), whereas the rescaled field satisfies \(\int_{\mathbb{R}}A_t^a\,da=2t\).

Worked Example: Checking the Occupation-Density Constant

Take \(f(a)=1\) for every \(a\). Compute the occupation integral and the corresponding level integrals under both conventions.

The left side is \(\int_0^t1\,ds=t\). Under the standard convention the occupation-density formula gives \(\int_{\mathbb{R}}L_t^a\,da=t\). Since \(A_t^a=2L_t^a\), its level integral is \(\int_{\mathbb{R}}A_t^a\,da=2t\); the rescaled formula therefore gives \(\tfrac12(2t)=t\). At \(t=1\), these values are respectively \(1\), \(2\), and \(\tfrac12\cdot2=1\). Check: both formulas recover the same elapsed time, and omitting the factor \(1/2\) from the rescaled formula would incorrectly give \(2t\).

Expected Local Time as an Independent Check

Worked Example: Expected Local Time at the Origin

For \(t=1\), compute \(\mathbb{E}[L_1^0]\) and the expected absolute-value correction in Tanaka’s formula.

Since \(B_0=0\), the standard absolute-value Tanaka formula gives \(|B_1|=\int_0^1\operatorname{sgn}(B_s)\,dB_s+L_1^0\). The integrand is progressively measurable and bounded by \(1\), so it is in \(\mathcal{L}^2\). Its Ito integral is a square-integrable martingale starting at zero; in particular, its expectation at time \(1\) is zero. Taking expectations therefore gives \(\mathbb{E}[L_1^0]=\mathbb{E}|B_1|\).

Because \(B_1\) has density \((2\pi)^{-1/2}e^{-x^2/2}\), symmetry and direct integration yield

\mathbb{E}|B_1| =2\int_0^\infty \frac{x}{\sqrt{2\pi}}e^{-x^2/2}\,dx =\frac{2}{\sqrt{2\pi}}\left[-e^{-x^2/2}\right]_{0}^{\infty} =\sqrt{\frac{2}{\pi}}.

Thus the standard local-time correction has expectation \(\sqrt{2/\pi}\), while the rescaled field has \(\mathbb{E}[A_1^0]=2\sqrt{2/\pi}\) and its absolute-value correction has expectation \(\tfrac12\mathbb{E}[A_1^0]=\sqrt{2/\pi}\). Check: taking expectations in Tanaka’s identity again gives \(\mathbb{E}|B_1|=\sqrt{2/\pi}\), in agreement with the direct Gaussian calculation.

Common Factor-of-Two Pitfall

A common error is to use the symmetric band \(|B_s-a|<\varepsilon\), divide its occupation time by \(\varepsilon\), and then treat the result as the standard field \(L_t^a\). That quotient tends to \(A_t^a=2L_t^a\), not \(L_t^a\). For example, using \(A_t^a\) as though it were \(L_t^a\) in the absolute-value Tanaka formula would insert a correction \(A_t^a\) instead of \(A_t^a/2\). At the value \(L_t^a=3\), this incorrect calculation gives correction \(6\), whereas the correct correction is \(A_t^a/2=6/2=3\).

The reliable procedure is to record the occupation-limit denominator first, state which field that limit defines, and then transform every formula coefficient using that field’s relation to the standard density. The symmetric limit with denominator \(2\varepsilon\) and either one-sided limit with denominator \(\varepsilon\) give \(L_t^a\); the symmetric limit with denominator \(\varepsilon\) gives \(2L_t^a\). Tanaka and occupation-density formulas must use the same choice consistently.

Check Your Understanding

For each question, specify the normalization being used before comparing constants.

Why does the right-sided occupation average converge to \(L_t^a\) when \(a\mapsto L_t^a\) is continuous at \(a\)?
If \(A_t^a\) is defined by dividing the symmetric-band occupation time by \(\varepsilon\), express \(A_t^a\) in terms of \(L_t^a\).
Write the positive-part Tanaka correction using \(A_t^a\), and compare it with the correction using \(L_t^a\).
What multiplicative constant must appear in the occupation-density formula when its density is \(A_t^a\)?
For \(L_t^a=4\), calculate the rescaled field and the absolute-value Tanaka correction under both conventions.