Constructing Local Time by Smoothing

From a Smooth Formula to Local Time

The absolute-value function has a corner at zero, so the ordinary twice-differentiable version of Itô’s formula cannot be applied to it directly. Smoothing removes the corner. The second derivative of the smoothed function is concentrated near the corner, and its Itô correction records how much Brownian motion spends near a specified level. To obtain local time, it is essential to prove convergence of both the smoothed correction and the stochastic integral.

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 a finite horizon $T$, every stochastic integral below is an Itô integral. Its integrand is progressively measurable and belongs to $\mathcal{L}^2$, the class of progressively measurable $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Processes formed by composing the continuous adapted process $B$ with a Borel function are progressively measurable; the integrands used below are bounded, so they belong to $\mathcal{L}^2$.

For $\varepsilon>0$, smooth the absolute value by setting $f_\varepsilon(x)=\sqrt{x^2+\varepsilon^2}$. Its first two derivatives and the corresponding Itô correction are

f_\varepsilon'(x)=\frac{x}{\sqrt{x^2+\varepsilon^2}},\qquad f_\varepsilon''(x)=\frac{\varepsilon^2}{(x^2+\varepsilon^2)^{3/2}},\qquad \frac12 f_\varepsilon''(x)=\rho_\varepsilon(x) =\frac{\varepsilon^2}{2(x^2+\varepsilon^2)^{3/2}}.

The functions $f_\varepsilon'$ are bounded in absolute value by $1$. Also $0\le f_\varepsilon(x)-|x|\le\varepsilon$, since $|x|\le\sqrt{x^2+\varepsilon^2}\le |x|+\varepsilon$. The correction kernel is nonnegative and has total mass one:

\int_{-\infty}^{\infty}\rho_\varepsilon(x)\,dx =\frac12\left[\frac{x}{\sqrt{x^2+\varepsilon^2}}\right]_{-\infty}^{\infty} =\frac12(1-(-1))=1.

Thus $\rho_\varepsilon$ is a normalized approximation to a point mass at zero. In particular, $\rho_\varepsilon(B_s-a)$ weights times when $B_s$ is near the level $a$. The construction below proves the limit needed to give that interpretation a precise local-time meaning.

Worked Example: Checking the Kernel’s Scale and Normalization

Evaluate the correction kernel at the centre and at one smoothing width from the centre, then check its total mass.

At $x=0$, direct substitution gives $\rho_\varepsilon(0)=\varepsilon^2/(2\varepsilon^3)=1/(2\varepsilon)$. At $x=\varepsilon$, the denominator is $2(2\varepsilon^2)^{3/2}=2^{5/2}\varepsilon^3$, so $\rho_\varepsilon(\varepsilon)=1/(2^{5/2}\varepsilon)$. The value at one smoothing width is smaller than the central value by the factor $2^{-3/2}$.

The mass check uses the antiderivative $\tfrac12 x/\sqrt{x^2+\varepsilon^2}$. Its limits are $\tfrac12$ as $x\to\infty$ and $-\tfrac12$ as $x\to-\infty$, so the integral is $1$. Check: the central height is proportional to $1/\varepsilon$, while the kernel’s width is proportional to $\varepsilon$; the exact integral confirms that this concentration does not change the normalization.

The Smoothed Itô Identity

Fix a level $a\in\mathbb{R}$. Applying Itô’s formula to the twice continuously differentiable function $x\mapsto f_\varepsilon(x-a)$ gives, for $0\le t\le T$,

f_\varepsilon(B_t-a) =f_\varepsilon(B_0-a) +\int_0^t f_\varepsilon'(B_s-a)\,dB_s +\frac12\int_0^t f_\varepsilon''(B_s-a)\,ds.

Here the first derivative is $f_\varepsilon'(B_s-a)=(B_s-a)/\sqrt{(B_s-a)^2+\varepsilon^2}$, and the second-derivative correction is $\tfrac12 f_\varepsilon''(B_s-a)=\rho_\varepsilon(B_s-a)$. Define the smoothed local-time process by

A_t^{a,\varepsilon} =\int_0^t \rho_\varepsilon(B_s-a)\,ds =f_\varepsilon(B_t-a)-f_\varepsilon(B_0-a) -\int_0^t f_\varepsilon'(B_s-a)\,dB_s.

This process is continuous, nonnegative, and increasing in $t$, as is immediate from its integral definition. The identity on the right will establish convergence; it is not an assumption about a pre-existing local time.

Theorem: For each fixed $a\in\mathbb{R}$ and $T<\infty$, $A^{a,\varepsilon}$ converges as $\varepsilon\downarrow0$ in $L^2$ for the uniform norm on $[0,T]$. Its limit has a continuous, nonnegative, increasing version $L_t^a$, and $$ |B_t-a|=|B_0-a|+\int_0^t \operatorname{sgn}(B_s-a)\,dB_s+L_t^a,\qquad 0\le t\le T, $$ where $\operatorname{sgn}(x)=1$ for $x>0$, $-1$ for $x<0$, and $0$ for $x=0$. The integral is an Itô integral of a progressively measurable $\mathcal{L}^2$ integrand.

Proof. Write $M_t^\varepsilon=\int_0^t f_\varepsilon'(B_s-a)\,dB_s$. Since the integrand is progressively measurable and bounded by $1$, its expected squared integral on $[0,T]$ is at most $T$. The Itô integral is therefore a square-integrable martingale: its conditional expectation satisfies $\mathbb{E}[M_t^\varepsilon\mid\mathcal{F}_s]=M_s^\varepsilon$ for $s\le t$, and the Itô isometry gives $\mathbb{E}[(M_t^\varepsilon)^2]\le T$.

For each $x\ne0$, $f_\varepsilon'(x)\to\operatorname{sgn}(x)$. For $s>0$, $B_s$ has a normal density, so $\mathbb{P}(B_s=a)=0$. Consequently $\mathbb{E}\int_0^T\mathbf{1}_{\{B_s=a\}}\,ds=\int_0^T\mathbb{P}(B_s=a)\,ds=0$. Thus $f_\varepsilon'(B_s-a)\to\operatorname{sgn}(B_s-a)$ for almost every $(s,\omega)$ under product measure $ds\,d\mathbb{P}$. The squared difference is bounded by $4$. Dominated convergence gives $$ \mathbb{E}\int_0^T \left|f_\varepsilon'(B_s-a)-\operatorname{sgn}(B_s-a)\right|^2ds\longrightarrow0. $$ The limiting sign integrand is progressively measurable and bounded, hence belongs to $\mathcal{L}^2$. Define $M_t^a=\int_0^t\operatorname{sgn}(B_s-a)\,dB_s$. The Itô isometry proves $M_T^\varepsilon\to M_T^a$ in $L^2$.

For convergence uniformly over time, use the standard Doob $L^2$ maximal inequality, whose proof is not developed here: for a square-integrable martingale $N$ starting at zero, $\mathbb{E}\sup_{t\le T}|N_t|^2\le4\mathbb{E}|N_T|^2$. Applied to the difference of the two Itô integrals, this inequality and the Itô isometry show that $\mathbb{E}\sup_{t\le T}|M_t^\varepsilon-M_t^a|^2\to0$.

Now use the smoothed identity. Since $|f_\varepsilon(x)-|x||\le\varepsilon$, the difference between the two endpoint terms $f_\varepsilon(B_t-a)-f_\varepsilon(B_0-a)$ and $|B_t-a|-|B_0-a|$ is bounded uniformly over $t\le T$ by $2\varepsilon$. The martingale terms converge in $L^2$ for the uniform norm, as just shown. Hence $A^{a,\varepsilon}$ converges in that norm to $$ L_t^a=|B_t-a|-|B_0-a|-M_t^a. $$ The limiting process is continuous because Brownian paths and the Itô integral have continuous versions. To verify monotonicity and nonnegativity, take a sequence $\varepsilon_n\downarrow0$ along which the uniform convergence holds almost surely; such a sequence exists by the $L^2$ convergence and Markov’s inequality. Each $A^{a,\varepsilon_n}$ is nonnegative and increasing, so its uniform limit has both properties. Rearranging the displayed definition of $L_t^a$ gives the asserted identity. $\square$

Examples and Checks on the Limit

Worked Example: The Approximation at the Zero Level

Take $a=0$ and $B_0=0$. Write the smoothed process explicitly and identify its limit from the theorem.

The kernel formula gives $$ A_t^{0,\varepsilon} =\int_0^t\frac{\varepsilon^2}{2(B_s^2+\varepsilon^2)^{3/2}}\,ds. $$ In the smoothed Itô identity, $f_\varepsilon(B_0)=f_\varepsilon(0)=\varepsilon$, so $$ A_t^{0,\varepsilon} =\sqrt{B_t^2+\varepsilon^2}-\varepsilon -\int_0^t\frac{B_s}{\sqrt{B_s^2+\varepsilon^2}}\,dB_s. $$ As $\varepsilon\downarrow0$, the first two terms converge uniformly in $t$ to $|B_t|$, and the integral converges in the uniform-in-time $L^2$ sense to $\int_0^t\operatorname{sgn}(B_s)\,dB_s$. Therefore the limit is $L_t^0=|B_t|-\int_0^t\operatorname{sgn}(B_s)\,dB_s$. Check: at $t=0$, both $A_0^{0,\varepsilon}$ and $L_0^0$ are zero, as required by their definitions.

Worked Example: A Nonzero Level

Let $a=2$, still with $B_0=0$. Compute the initial smoothed endpoint and write the limiting local-time identity.

The initial endpoint is $f_\varepsilon(B_0-a)=f_\varepsilon(-2)=\sqrt{4+\varepsilon^2}$. The correction term is $$ A_t^{2,\varepsilon} =\int_0^t\frac{\varepsilon^2}{2((B_s-2)^2+\varepsilon^2)^{3/2}}\,ds. $$ The smoothed identity is $$ A_t^{2,\varepsilon} =\sqrt{(B_t-2)^2+\varepsilon^2}-\sqrt{4+\varepsilon^2} -\int_0^t\frac{B_s-2}{\sqrt{(B_s-2)^2+\varepsilon^2}}\,dB_s. $$ The theorem gives $L_t^2=|B_t-2|-2-\int_0^t\operatorname{sgn}(B_s-2)\,dB_s$. Check: substituting $t=0$ into the limit gives $|0-2|-2-0=0$. The initial endpoint is therefore necessary; omitting it would give the wrong starting value.

Worked Example: The Expected Limit at Zero

For $a=0$ and $B_0=0$, compute the expectation of the approximants and the expectation of their limit at a fixed time $T>0$.

The stochastic integral in $A_T^{0,\varepsilon}=\sqrt{B_T^2+\varepsilon^2}-\varepsilon-M_T^\varepsilon$ is square-integrable and has mean zero. Indeed, it is a square-integrable Itô-integral martingale starting at zero, so $\mathbb{E}M_T^\varepsilon=0$. Thus $$ \mathbb{E}A_T^{0,\varepsilon} =\mathbb{E}\sqrt{B_T^2+\varepsilon^2}-\varepsilon. $$ The bound $0\le\sqrt{x^2+\varepsilon^2}-|x|\le\varepsilon$ shows that this expectation converges to $\mathbb{E}|B_T|$. Since $B_T$ has density $(2\pi T)^{-1/2}e^{-x^2/(2T)}$, $$ \mathbb{E}|B_T| =\frac{2}{\sqrt{2\pi T}}\int_0^\infty x e^{-x^2/(2T)}\,dx =\frac{2T}{\sqrt{2\pi T}} =\sqrt{\frac{2T}{\pi}}. $$ For $T=1$, the limit is exactly $\sqrt{2/\pi}$. Check: the same expectation follows from the $L^2$ convergence, which implies convergence of expectations by Cauchy–Schwarz. The direct normal-density calculation confirms the value independently.

What the Construction Establishes

The result constructs local time at each fixed level as the limit of the smoothed occupation terms $\int_0^t\rho_\varepsilon(B_s-a)\,ds$. Its normalization is determined by the unit mass of $\rho_\varepsilon$, and the proof shows convergence uniformly in time in $L^2$, not merely convergence at a single time. The limit is continuous, nonnegative, and increasing in time.

The limiting identity is Tanaka’s identity for the absolute value. Its correction term arises because the second derivatives of the smooth approximations concentrate at the corner. This construction is for a fixed level $a$; it does not by itself prove joint continuity in $t$ and $a$, or the occupation density formula. Those require additional arguments.

Common mistake: it is not valid to discard the Itô correction on the grounds that $f_\varepsilon$ approaches $|x|$. The second derivative does not converge to zero: $\rho_\varepsilon$ becomes taller and narrower while keeping total mass one. Nor is convergence of the endpoint functions alone enough. The stochastic integrals must also converge, which is why the proof checks the squared integrands and applies the Itô isometry and Doob’s inequality.

Check Your Understanding

Use the kernel normalization and the convergence proof to check each answer.

Differentiate $f_\varepsilon(x)=\sqrt{x^2+\varepsilon^2}$ and identify its Itô correction.
Why does $\mathbb{P}(B_s=a)=0$ for each $s>0$ help prove convergence of the stochastic integrands?
Which estimate controls the difference of the stochastic integrals uniformly over $0\le t\le T$?
For a general level $a$ with $B_0=0$, what is the initial endpoint in the smoothed Itô identity?
Why does the unit mass of $\rho_\varepsilon$ matter for the local-time normalization?