Why a Kink Produces Local Time
For a twice continuously differentiable function, Itô’s formula accounts for curvature through a second-derivative term. A piecewise-linear function has no ordinary curvature away from its corners, but its slope changes abruptly at each kink. The Itô–Tanaka formula records that concentrated change in slope through local time. The coefficient is half the slope jump, with the sign of the jump retained even when the function is not convex.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). On a fixed interval \([0,T]\), consider a continuous process \(X_t=X_0+\int_0^t b_s\,ds+\int_0^t \sigma_s\,dB_s\), where \(X_0\) is \(\mathcal{F}_0\)-measurable, \(b\) and \(\sigma\) are progressively measurable, \(\int_0^T|b_s|\,ds<\infty\) almost surely, and \(\mathbb{E}\int_0^T\sigma_s^2\,ds<\infty\). Each stochastic integral below is an Itô integral of a progressively measurable integrand in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\).
Use the symmetric semimartingale local-time normalization: at a level \(a\), \(L_t^a(X)\) is the local time appearing with coefficient \(\tfrac12\) in the positive-part Tanaka formula. In particular, the established Tanaka formulas are \((X_t-a)^+=(X_0-a)^++\int_0^t\mathbf{1}_{\{X_s>a\}}\,dX_s+\tfrac12L_t^a(X)\) and \(|X_t-a|=|X_0-a|+\int_0^t\big(\mathbf{1}_{\{X_s>a\}}-\mathbf{1}_{\{X_s\le a\}}\big)\,dX_s+L_t^a(X)\). These are standard results from the earlier material on Tanaka’s formula. Here \(dX_s\) in those formulas means \(b_s\,ds+\sigma_s\,dB_s\); when split into two integrals, the stochastic integrand must satisfy the stated \(\mathcal{L}^2\) condition.
The Formula for Several Kinks
Let \(a_1<\cdots<a_n\) be the kink locations of a continuous piecewise-linear function \(f\), and let its successive slopes be \(m_0,m_1,\ldots,m_n\). Write \(\Delta m_j=m_j-m_{j-1}\). Denote the left derivative by \(f'_-(x)\): at a kink, this is the slope on the interval immediately to its left. The choice at a kink matters when specifying the integrand pointwise, so the formula below uses this convention throughout.
Proof. Put \(g(x)=f(X_0)+m_0(x-X_0)+\sum_{j=1}^n\Delta m_j\big((x-a_j)^+-(X_0-a_j)^+\big)\). This function equals \(f\) at \(X_0\). On the interval to the left of \(a_1\), its slope is \(m_0\). Crossing \(a_j\) adds \(\Delta m_j\) to its slope, so its slope on each successive interval is \(m_0+\sum_{k=1}^j\Delta m_k=m_j\). Thus \(g\) and \(f\) have the same value at \(X_0\) and the same slope on every interval between kinks; continuity then gives \(g(x)=f(x)\) for every \(x\). Apply the positive-part Tanaka formula to each hinge \((X_t-a_j)^+\), and use the identity \(X_t-X_0=\int_0^t b_s\,ds+\int_0^t\sigma_s\,dB_s\). The resulting local-time term for the \(j\)-th hinge is \(\tfrac12\Delta m_jL_t^{a_j}(X)\). The coefficient of \(dX_s\) is \(m_0+\sum_j\Delta m_j\mathbf{1}_{\{X_s>a_j\}}\), which equals \(f'_-(X_s)\), including at a kink because the indicator is strict. Splitting \(dX_s\) into its drift and Brownian parts gives the asserted formula. \(\square\)
The representation also explains why the formula applies to nonconvex functions. A convex kink has a nonnegative slope jump and therefore a nonnegative local-time coefficient. A concave kink has a negative slope jump and a negative coefficient. The local-time process itself remains nonnegative; it is the coefficient that carries the sign.
Worked Example: Absolute Value at a Level
Take \(f(x)=|x-a|\). Its slope is \(-1\) to the left of \(a\) and \(1\) to the right, so the jump is \(1-(-1)=2\), and the local-time coefficient is \(\tfrac12(2)=1\). The left derivative is \(-1\) for \(x\le a\) and \(1\) for \(x>a\). Hence $$ |X_t-a|=|X_0-a|+\int_0^t\big(\mathbf{1}_{\{X_s>a\}}-\mathbf{1}_{\{X_s\le a\}}\big)b_s\,ds +\int_0^t\big(\mathbf{1}_{\{X_s>a\}}-\mathbf{1}_{\{X_s\le a\}}\big)\sigma_s\,dB_s+L_t^a(X). $$
For \(X=B\) and \(a=0\), this becomes \(|B_t|=\int_0^t(\mathbf{1}_{\{B_s>0\}}-\mathbf{1}_{\{B_s\le0\}})\,dB_s+L_t^0(B)\). The stochastic integrand is bounded, so its square has expectation integral at most \(T\); the Itô integral is therefore a true martingale, with conditional expectation of each future increment equal to zero. To check the coefficient, the kink has slope change \(2\), giving exactly \(1\) times local time. A claimed coefficient \(2\) would count the full jump rather than half the jump.
Worked Example: Positive Part at a Shifted Kink
Let \(f(x)=(x-1)^+\) and \(X=B\). The slopes are \(0\) for \(x\le1\) and \(1\) for \(x>1\), so the jump is \(1-0=1\). Since \(B_0=0\), \(f(B_0)=0\), and the formula is $$ (B_t-1)^+=\int_0^t\mathbf{1}_{\{B_s>1\}}\,dB_s+\frac12L_t^1(B). $$
The integrand is bounded and progressively measurable, with \(\mathbb{E}\int_0^T\mathbf{1}_{\{B_s>1\}}^2\,ds\le T\), so the integral is a true martingale. The identity also checks at \(t=0\): both sides are zero because \(B_0=0\), the integral over an interval of length zero is zero, and \(L_0^1(B)=0\).
At \(x=1\), the left derivative is \(0\), whereas \(\operatorname{sgn}(x-1)\), with \(\operatorname{sgn}(0)=0\), also happens to be \(0\) for this example. That agreement is specific to this convention and function. In the absolute-value example the left derivative at the kink is \(-1\), not \(0\); assigning a convenient value at the kink should never replace checking the chosen derivative convention.
Worked Example: Two Kinks with Different Slope Jumps
Define \(f(x)=-2x+3(x+1)^++3(x-2)^+\). Its kinks are \(-1\) and \(2\). To the left of \(-1\), both hinge terms vanish and the slope is \(-2\). Between \(-1\) and \(2\), only the first hinge has slope \(1\), so the total slope is \(-2+3=1\). To the right of \(2\), both hinge terms have slope \(1\), so the total slope is \(-2+3+3=4\). Thus the slope jumps are \(1-(-2)=3\) at \(-1\) and \(4-1=3\) at \(2\), giving coefficient \(\tfrac32\) at each kink.
For \(X=B\), \(f(B_0)=f(0)=3\). The left derivative is \(-2\) for \(x\le-1\), \(1\) for \(-1<x\le2\), and \(4\) for \(x>2\). Therefore $$ f(B_t)=3+\int_0^t\big(-2\mathbf{1}_{\{B_s\le-1\}}+\mathbf{1}_{\{-1<B_s\le2\}}+4\mathbf{1}_{\{B_s>2\}}\big)\,dB_s+\frac32L_t^{-1}(B)+\frac32L_t^2(B). $$
The displayed integrand is bounded in absolute value by \(4\), so its squared time integral has expectation at most \(16T\); its Itô integral is a true martingale. As a direct check of the piecewise-linear calculation, \(f(-2)=4\), since \(-2(-2)=4\) and both hinges vanish; \(f(1)=-2+3(2)+0=4\); and \(f(3)=-6+3(4)+3(1)=9\). These values match the slopes: starting from \(f(0)=3\), slope \(1\) gives \(f(1)=4\), and the slope \(4\) from \(2\) to \(3\) gives \(f(3)=f(2)+4\), where \(f(2)=-4+9=5\), hence \(f(3)=9\).
Derivative Conventions and a Common Error
At a kink, different pointwise choices for a derivative can sometimes produce the same Brownian stochastic integral, but that does not make the choices identical. For example, for standard Brownian motion and a fixed level \(a\), \(\mathbb{P}(B_s=a)=0\) for every \(s>0\), since \(B_s\) has a normal density. Consequently, by nonnegativity and integration, \(\mathbb{E}\int_0^T\mathbf{1}_{\{B_s=a\}}\,ds=\int_0^T\mathbb{P}(B_s=a)\,ds=0\). Thus the time spent at \(a\) is zero almost surely. If two bounded Brownian integrands differ only when \(B_s=a\), their difference has expected squared time integral zero, and the Itô isometry shows their stochastic integrals agree. This permits replacing the left-derivative integrand by a sign convention in certain Brownian formulas; it does not justify a false pointwise identity or an unexamined replacement in a drift integral for a general process.
Common mistake: For \(f(x)=|x-a|\), writing the local-time term as \(2L_t^a(X)\) uses the slope jump itself as the coefficient. The correct jump is \(1-(-1)=2\), and the formula uses half of it, so the term is \(L_t^a(X)\). A second possible error is to identify the left derivative of \(|x-1|\) with \(\operatorname{sgn}(x-1)\) at \(x=1\). The left derivative there is \(-1\), while \(\operatorname{sgn}(0)=0\). The correct pointwise integrand is \(-\mathbf{1}_{\{x\le1\}}+\mathbf{1}_{\{x>1\}}\). For standard Brownian motion, replacing it by the sign function inside the Itô integral is valid because the process spends zero Lebesgue time at \(1\); the pointwise identification remains false.
This method is useful whenever a payoff, constraint, or transformation is linear between finitely many thresholds. Find the slopes on the intervals, take each right slope minus its left slope, and multiply by one half to obtain the local-time coefficient. Keep the left derivative in the integral, and use local time at the corresponding level. This procedure separates the ordinary drift and stochastic-integral contributions from the correction concentrated at the kinks.
Check Your Understanding
For each question, retain the left-derivative convention at kink points and use the same local-time normalization as in the formulas above.
- For \(f(x)=|x-3|\), find the slope jump and the coefficient of \(L_t^3(X)\).
- For \(f(x)=2(x+1)^+-5(x-2)^+\), find the slopes on the three intervals and the local-time coefficient at each kink.
- Write the left derivative of \((x-a)^+\), specifying its value at \(x=a\).
- Why does a Brownian stochastic integral not change if its bounded integrand is altered only at the times \(s\) for which \(B_s=a\)? State the integrability fact that justifies the conclusion.
- For a kink whose left slope is \(4\) and right slope is \(-2\), determine the signed local-time coefficient and explain its sign.