Why Absolute Value Needs a Correction
Itô’s formula applies directly to twice continuously differentiable functions. The absolute-value function \(x\mapsto |x-a|\), however, has a corner at the level \(a\), so its ordinary second derivative does not capture what happens when Brownian motion visits that level. Tanaka’s formula makes the missing term explicit: it is the local time accumulated at \(a\). The formula is not an ordinary chain rule, nor is its correction a Stratonovich correction.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) with respect to the filtration: in particular, \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has law \(N(0,t-s)\) for \(s<t\). This filtration condition is essential for the Itô integral’s martingale and isometry properties. Throughout, the Brownian motion starts at \(B_0=0\). Define \(\operatorname{sgn}(x)=1\) for \(x>0\), \(\operatorname{sgn}(x)=-1\) for \(x<0\), and \(\operatorname{sgn}(0)=0\).
For a progressively measurable process \(H\), the Itô integral \(\int_0^t H_s\,dB_s\) is defined here for \(H\in\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) for each finite \(T\). Its isometry is \(\mathbb{E}[(\int_0^tH_s\,dB_s)^2]=\mathbb{E}\int_0^tH_s^2\,ds\). We will also use that such an integral is a martingale: for \(s\le t\), its increment has conditional expectation zero given \(\mathcal{F}_s\). This follows first for adapted step integrands from independence and zero mean of Brownian increments, and then for \(\mathcal{L}^2\) integrands by approximation and the isometry.
Tanaka’s Formula and Its Proof
Proof. Fix \(a\) and a finite time horizon \(T\). For \(\varepsilon>0\), set \(f_\varepsilon(x)=\sqrt{x^2+\varepsilon^2}\). Its derivatives are $$ f_\varepsilon'(x)=\frac{x}{\sqrt{x^2+\varepsilon^2}},\qquad f_\varepsilon''(x)=\frac{\varepsilon^2}{(x^2+\varepsilon^2)^{3/2}}. $$ The function \(f_\varepsilon\) is twice continuously differentiable, so Itô’s formula, applied to \(B_t-a\), gives $$ f_\varepsilon(B_t-a)=f_\varepsilon(-a)+\int_0^t f_\varepsilon'(B_s-a)\,dB_s +\frac12\int_0^t f_\varepsilon''(B_s-a)\,ds. $$ In particular, the second-derivative correction is explicitly \(\frac12\int_0^t f_\varepsilon''(B_s-a)\,ds\); it does not disappear as \(\varepsilon\) decreases.
First, \(\big|f_\varepsilon(x)-|x|\big|\le\varepsilon\), since \(0\le\sqrt{x^2+\varepsilon^2}-|x|\le\varepsilon\). Thus the left side tends to \(|B_t-a|\), and \(f_\varepsilon(-a)\) tends to \(|a|\). Next, \(f_\varepsilon'(x)\to\operatorname{sgn}(x)\) for \(x\ne0\), and \(|f_\varepsilon'(x)|\le1\). The set of times at which \(B_s=a\) has Lebesgue measure zero almost surely: indeed, for \(s>0\), \(B_s\) has a density, so \(\mathbb{P}(B_s=a)=0\); Tonelli’s theorem gives $$ \mathbb{E}\left[\int_0^T\mathbf{1}_{\{B_s=a\}}\,ds\right] =\int_0^T\mathbb{P}(B_s=a)\,ds=0. $$ The value at \(s=0\) does not affect this time integral. Dominated convergence on \([0,T]\times\Omega\) therefore gives $$ \mathbb{E}\int_0^T\left|f_\varepsilon'(B_s-a)-\operatorname{sgn}(B_s-a)\right|^2\,ds\longrightarrow0. $$ By the Itô isometry, the stochastic integrals converge in \(L^2(\mathbb{P})\). By the stated smoothing construction, the last term converges in probability to \(L_t^a\). Taking limits in the displayed Itô identity yields the claimed equality in probability, hence almost surely for each fixed \(t\). Both sides have continuous versions: the Itô integral has continuous paths, and the constructed local time is continuous. Applying the fixed-time equality on the nonnegative rational times, taking a countable intersection of full-probability events, and using continuity proves the identity simultaneously for all \(t\ge0\).
Finally, \(H_s=\operatorname{sgn}(B_s-a)\) is progressively measurable as a Borel function of the adapted continuous process \(B_s\), and \(|H_s|\le1\). Consequently \(\mathbb{E}\int_0^T H_s^2\,ds\le T\), so the integral is defined in \(\mathcal{L}^2\). It is a true martingale, not merely a local martingale: the square-integrability bound and the Itô integral martingale property above give integrability and \(\mathbb{E}[\int_s^tH_u\,dB_u\mid\mathcal{F}_s]=0\). \(\square\)
Reading the Terms at a Fixed Level
The stochastic-integral term records the signed Brownian increments on either side of the level: its integrand is \(1\) above \(a\), \(-1\) below \(a\), and \(0\) at \(a\). The local-time term records the correction needed at the corner. The choice \(\operatorname{sgn}(0)=0\) is convenient and does not change the integral if another value bounded by one is assigned at zero: the set of times \(B_s=a\) has zero Lebesgue measure almost surely, so the corresponding integrands agree almost everywhere in time and probability.
Worked Example: The Smoothing Kernel and Its Normalization
Check the mass and value at the origin of the correction kernel for \(\varepsilon=1\). From the derivatives in the proof, the kernel is $$ k_\varepsilon(x)=\frac12 f_\varepsilon''(x) =\frac{\varepsilon^2}{2(x^2+\varepsilon^2)^{3/2}}. $$ At \(\varepsilon=1\), this gives \(k_1(x)=1/[2(x^2+1)^{3/2}]\), so \(k_1(0)=1/2\). To compute its total mass, use the antiderivative $$ \int\frac{\varepsilon^2}{2(x^2+\varepsilon^2)^{3/2}}\,dx =\frac{x}{2\sqrt{x^2+\varepsilon^2}}. $$ At the endpoints \(+\infty\) and \(-\infty\), the antiderivative tends respectively to \(1/2\) and \(-1/2\); their difference is \(1\). Thus the kernel has unit mass, not mass \(1/2\) or \(2\), as required by the stated local-time normalization.
Check: \(f_\varepsilon''\) has total mass \(2\), so the factor \(\tfrac12\) in Itô’s formula makes the correction kernel’s mass \(1\). Omitting that factor would give the wrong normalization.
Worked Example: The Sign Integral at Level Two
Take \(a=2\), and write \(M_t=\int_0^t\operatorname{sgn}(B_s-2)\,dB_s\). Its integrand is progressively measurable and bounded by one, so it belongs to \(\mathcal{L}^2\). The Itô integral martingale property gives \(\mathbb{E}[M_t]=0\); the isometry gives $$ \mathbb{E}[M_t^2]=\mathbb{E}\int_0^t\operatorname{sgn}^2(B_s-2)\,ds=t. $$ For the last equality, the square of the sign is \(1\) off the level and \(0\) on it, and the time spent at level \(2\) is zero almost surely by the density-and-Tonelli argument in the proof.
At \(t=4\), these calculations give \(\mathbb{E}[M_4]=0\) and \(\mathbb{E}[M_4^2]=4\), so the variance is \(4-0^2=4\). Tanaka’s formula at this level reads \(|B_4-2|=2+M_4+L_4^2\). The initial constant is \(|B_0-2|=|0-2|=2\).
Check: at \(t=0\), the integral and local time are both zero, and the formula reads \(|0-2|=2\), as required. The variance calculation is consistent with the isometry and uses the actual horizon \(t=4\), not a bound independent of \(t\).
Worked Example: Comparing the Levels Minus One and One
Write Tanaka’s formula once at \(a=-1\) and once at \(a=1\), then subtract the second identity from the first. Since \(|B_0+1|=1\) and \(|B_0-1|=1\), their initial constants cancel, giving $$ |B_t+1|-|B_t-1| =\int_0^t\big(\operatorname{sgn}(B_s+1)-\operatorname{sgn}(B_s-1)\big)\,dB_s +L_t^{-1}-L_t^1. $$ For a real value \(x\), the difference of signs is \(2\) if \(-1<x<1\), \(1\) if \(x=-1\) or \(x=1\), and \(0\) if \(x<-1\) or \(x>1\). For example, at \(x=-1\) it is \(0-(-1)=1\), and at \(x=1\) it is \(1-0=1\). Brownian motion spends zero time at either endpoint, so changing these two boundary values does not change the Itô integral. Therefore the identity can also be written $$ |B_t+1|-|B_t-1|=2\int_0^t\mathbf{1}_{\{-1<B_s<1\}}\,dB_s+L_t^{-1}-L_t^1. $$ The integrand in this last integral is bounded and progressively measurable, hence square-integrable on every finite interval.
Check: at \(t=0\), the left side is \(|1|-|-1|=1-1=0\). On the right, the integral and both local times start at zero, also giving \(0\). The strict interval in the indicator is valid because the boundary times have zero Lebesgue measure almost surely.
Why the Correction Matters
A common mistake is to apply the ordinary chain rule and write \(|B_t|=\int_0^t\operatorname{sgn}(B_s)\,dB_s\), omitting local time. The integral on the right is a true martingale by the integrability argument in the theorem, so it has expectation zero. But \(B_1\) has the standard normal density, and direct integration gives $$ \mathbb{E}|B_1|=\frac{2}{\sqrt{2\pi}}\int_0^\infty x e^{-x^2/2}\,dx =\frac{2}{\sqrt{2\pi}}\cdot1 =\sqrt{\frac{2}{\pi}}>0. $$ The proposed identity would therefore imply \(\mathbb{E}|B_1|=0\), a contradiction. Tanaka’s formula supplies the missing term \(L_t^0\). This correction comes from the singular behavior near the corner of the absolute-value function as its smooth approximations become sharp; it is distinct from the \(\tfrac12 f''\,dt\) correction for a smooth function in Itô’s formula.
The equality is under \(\mathbb{P}\), and the integral is an Itô integral. It does not assert a pathwise ordinary chain rule, nor does it replace the Itô integral by a Stratonovich integral. The level \(a\) can be any fixed real number, including \(a=0\); the initial term is always \(|B_0-a|=|a|\).
Check Your Understanding
For each question, use the local-time normalization and sign convention stated above.
- Write Tanaka’s formula for \(|B_t+3|\), including its initial constant and the level at which local time is evaluated.
- Why does \(\operatorname{sgn}(B_s-a)\) belong to \(\mathcal{L}^2\) on every finite time interval?
- At the fixed level \(a=0\), what does the Itô isometry give for the second moment of \(\int_0^t\operatorname{sgn}(B_s)\,dB_s\), and why?
- For the smoothing \(f_\varepsilon(x)=\sqrt{x^2+\varepsilon^2}\), compute \(f_\varepsilon'(x)\) and \(f_\varepsilon''(x)\), and identify the correction term in Itô’s formula.
- In the comparison of levels \(-1\) and \(1\), why may the sign difference at the two endpoints be replaced by zero inside the stochastic integral?