Local Time as an Occupation Density

Time Near a Level, Not Time At a Point

For a fixed level $a$, a Brownian path spends zero Lebesgue time at the single point $a$, almost surely. Yet the path may repeatedly cross that level, and a useful measure of those visits is obtained by looking at time spent in a narrow band around it. The band’s occupation time shrinks as its width shrinks; dividing by the width can leave a finite, informative limit.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. All time integrals below are ordinary Lebesgue integrals in time, not stochastic integrals. For $t\ge0$, $a\in\mathbb{R}$, and $\varepsilon>0$, define the normalized occupation of the open band around $a$ by

A_{\varepsilon}(t,a) =\frac{1}{2\varepsilon}\int_0^t \mathbf{1}_{\{|B_s-a|<\varepsilon\}}\,ds.

The strict inequality means that the band is $(a-\varepsilon,a+\varepsilon)$. Its width is $2\varepsilon$, which explains the denominator. This symmetric normalization is one convention for Brownian local time; the factor matters, so it is stated rather than assumed.

Definition: For fixed $t$ and $a$, Brownian local time $L_t^a$, in the symmetric occupation-density convention, is the $L^2(\mathbb{P})$ limit of $A_\varepsilon(t,a)$ as $\varepsilon\downarrow0$. Thus, for every fixed $t\ge0$ and $a\in\mathbb{R}$, $\mathbb{E}[(A_\varepsilon(t,a)-L_t^a)^2]\to0$. In particular, the definition concerns a limit of random variables, not the time spent at the single point $a$.

The standard local-time existence theorem guarantees this fixed-$(t,a)$ limit for Brownian motion. Its proof uses the construction of local time and occupation-density results developed in the standard treatment of those topics; it is not proved here. The fixed-pair statement is all that is needed for the definition. It does not, by itself, assert that the approximations converge simultaneously for every $t$ and $a$, or that a particular sample path has a continuous local-time function.

Why the Limit Is a Density

For a nonnegative ordinary density $g(x)$, the mass in $(a-\varepsilon,a+\varepsilon)$ is approximately $2\varepsilon g(a)$ when $g$ is continuous at $a$. Dividing that mass by $2\varepsilon$ recovers its density at $a$. The local-time approximation has the same form, with time spent in the band playing the role of mass. Accordingly, when the limiting density is finite, the interpretation is that time spent in a sufficiently narrow band is approximately its width times the local time.

\int_0^t\mathbf{1}_{\{|B_s-a|<\varepsilon\}}\,ds \ \approx\ 2\varepsilon L_t^a \qquad\text{for small }\varepsilon.

This is an interpretation of the defining approximation, not a claim that the time at the single level equals $L_t^a$. The occupation time on the left tends to zero as the band shrinks, while the normalized occupation can converge to a positive value. Local time therefore measures concentration of visits per unit of level width, rather than elapsed time at one point.

Proposition: For fixed $t\ge0$ and $a\in\mathbb{R}$, the expectation of the normalized band occupation has the limit $$ \lim_{\varepsilon\downarrow0}\mathbb{E}[A_\varepsilon(t,a)] =\int_0^t \frac{1}{\sqrt{2\pi s}} \exp\!\left(-\frac{a^2}{2s}\right)\,ds. $$ The integrand is defined for $s>0$; its value at $s=0$ does not affect the integral.

Proof. For $s>0$, $B_s$ has density $p_s(x)=(2\pi s)^{-1/2}\exp(-x^2/(2s))$. Tonelli’s theorem applies to the nonnegative indicator, so

\mathbb{E}[A_\varepsilon(t,a)] =\int_0^t \frac{1}{2\varepsilon} \mathbb{P}(|B_s-a|<\varepsilon)\,ds =\int_0^t\frac{1}{2\varepsilon} \int_{a-\varepsilon}^{a+\varepsilon}p_s(x)\,dx\,ds.

For each $s>0$, continuity of $p_s$ in $x$ implies that the inner average converges to $p_s(a)$. Also $p_s(x)\le(2\pi s)^{-1/2}$ for every $x$, so the average is bounded by $(2\pi s)^{-1/2}$. This bound is integrable on $(0,t]$, since $\int_0^t s^{-1/2}\,ds=2\sqrt{t}$. The dominated convergence theorem now gives the stated limit. When $t=0$, both sides are zero. $\square$

This calculation identifies the limiting mean of the approximations. It does not prove their $L^2$ convergence: convergence of expectations alone is not enough to establish convergence of random variables. The existence theorem is what justifies defining $L_t^a$ by the stronger limit above.

Worked Pathwise Models of the Density

The following deterministic calculations illustrate the normalization and the density idea. They are not claims that Brownian sample paths are differentiable or piecewise linear. Brownian local time is defined by the same normalized occupation principle, but its existence requires the Brownian local-time theorem.

Worked Example: One Linear Crossing

Let $x(s)=2s$ for $0\le s\le2$, and compute the normalized time in the band of radius $\varepsilon$ around level $a=2$, for $0<\varepsilon<2$.

The strict band condition is $|2s-2|<\varepsilon$. Dividing by $2$ gives $|s-1|<\varepsilon/2$, so the times in the band form the open interval $(1-\varepsilon/2,1+\varepsilon/2)$. The interval lies inside $[0,2]$ because $\varepsilon<2$, and its length is $\varepsilon$. Thus the normalized occupation is

\frac{1}{2\varepsilon} \int_0^2\mathbf{1}_{\{|2s-2|<\varepsilon\}}\,ds =\frac{\varepsilon}{2\varepsilon} =\frac12.

Check: the path crosses level $2$ once, at $s=1$, with slope $2$. The density contribution is the reciprocal slope $1/2$, matching the exact band calculation for every $0<\varepsilon<2$.

Worked Example: Two Crossings of a Triangular Path

On $0\le s\le2$, let $x(s)=2s$ for $0\le s\le1$, and $x(s)=2(2-s)$ for $1\le s\le2$. Find the normalized occupation around $a=1$, taking $0<\varepsilon<1$.

On the first segment, $|2s-1|<\varepsilon$ is equivalent to $(1-\varepsilon)/2<s<(1+\varepsilon)/2$, an interval of length $\varepsilon$. On the second segment, $|2(2-s)-1|<\varepsilon$ is equivalent to $(3-\varepsilon)/2<s<(3+\varepsilon)/2$, also an interval of length $\varepsilon$. Since $\varepsilon<1$, these intervals lie in their respective segments and are disjoint. The total time in the band is $2\varepsilon$, hence

\frac{1}{2\varepsilon} \int_0^2\mathbf{1}_{\{|x(s)-1|<\varepsilon\}}\,ds =\frac{2\varepsilon}{2\varepsilon}=1.

Check: there are two crossings, each with slope magnitude $2$. Each contributes $1/2$, and $1/2+1/2=1$, agreeing with the direct total-length calculation.

Worked Example: A Smooth Crossing with Nonconstant Speed

Let $x(s)=s^2$ on $0\le s\le2$, and take the level $a=1$. Compute the limit of the normalized band occupation.

For $0<\varepsilon<1$, the condition $|s^2-1|<\varepsilon$ is equivalent, since $s\ge0$, to $\sqrt{1-\varepsilon}<s<\sqrt{1+\varepsilon}$. The interval is inside $[0,2]$, and its length is $\sqrt{1+\varepsilon}-\sqrt{1-\varepsilon}$. Rationalizing this difference gives

\frac{\sqrt{1+\varepsilon}-\sqrt{1-\varepsilon}}{2\varepsilon} =\frac{2\varepsilon} {2\varepsilon\big(\sqrt{1+\varepsilon}+\sqrt{1-\varepsilon}\big)} =\frac{1}{\sqrt{1+\varepsilon}+\sqrt{1-\varepsilon}}.

As $\varepsilon\downarrow0$, the expression tends to $1/2$. Check: the level is crossed at $s=1$, where $x'(1)=2$, so the reciprocal speed is $1/2$, the same limit as the direct calculation.

What the Approximation Does—and Does Not—Say

The examples show why normalization is essential. An occupation time is measured in units of time; dividing by a level width produces a density with units of time per unit level. In a piecewise-smooth path with isolated crossings, the reciprocal speed at a crossing determines the contribution: slower motion spends more time in a fixed narrow band. Multiple crossings add their contributions, as in the triangular-path calculation.

Brownian motion has no ordinary derivative with which to perform this reciprocal-speed calculation. Its local time is instead obtained as a limit of the random band occupations. The deterministic examples explain the density interpretation but cannot substitute for the Brownian existence theorem.

A flat segment also warns against assuming that every path produces a finite density. If a path stays exactly at $a$ for a time interval of positive length, then the band occupation includes that entire interval for every $\varepsilon>0$; after division by $2\varepsilon$, the normalized occupation diverges as $\varepsilon\downarrow0$. Brownian local time is not defined by applying the deterministic crossing calculation to arbitrary continuous paths.

Common Mistake: Forgetting the Band Width

A common error is to observe that time in a shrinking band tends to zero and conclude that its local-time limit must be zero. The linear-crossing calculation gives a direct correction: for $x(s)=2s$ on $[0,2]$ at level $2$, the occupation time is $\varepsilon$, which tends to zero, but the normalized quantity is $\varepsilon/(2\varepsilon)=1/2$. It is the ratio to the band width—not the unnormalized occupation—that detects a finite density.

A different error is to divide by $\varepsilon$ while retaining the symmetric band $(a-\varepsilon,a+\varepsilon)$. That changes the normalization by a factor of two. The convention used here divides by $2\varepsilon$; any other convention must be stated explicitly and used consistently.

Check Your Understanding

Use the definition and the worked calculations to test the density interpretation.

For $x(s)=3s$ on $[0,2]$ at level $a=3$, compute the normalized occupation for $0<\varepsilon<3$.
For the triangular path in the second example, explain why the restriction $\varepsilon<1$ ensures that each band interval lies within its corresponding segment.
For $x(s)=s^2$ at level $1$, rationalize the band length and find its normalized limit as $\varepsilon\downarrow0$.
Why does convergence of $\mathbb{E}[A_\varepsilon(t,a)]$ alone not establish the $L^2$ limit used to define local time?
What happens to the normalized occupation if a path remains at the chosen level for a positive interval of time?