The Itô–Tanaka Formula for Convex Functions

From Curvature to Local-Time Correction

For a twice continuously differentiable function, Itô’s formula weights the second derivative by one half of the elapsed time along the path. A convex function need not have an ordinary second derivative: its slope can increase by a jump, or increase continuously in a way that is not described by a density. Its second derivative is then naturally a measure. The Itô–Tanaka formula pairs that measure with Brownian local time, giving a correction that includes both concentrated and spread-out curvature.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix a deterministic starting point $x\in\mathbb{R}$ and set $X_t=x+B_t$. Use the symmetric Brownian local time $L_t^a$, normalized by the occupation-density formula $\int_{\mathbb{R}}g(a)L_t^a\,da=\int_0^t g(X_s)\,ds$ for nonnegative Borel $g$. With this normalization, the positive-part Tanaka formula has local-time term $\tfrac12 L_t^a$.

Definition: If $f:\mathbb{R}\to\mathbb{R}$ is finite and convex, its distributional second derivative is the locally finite positive Borel measure $\mu=f''$ characterized by $\int_{\mathbb{R}}\varphi(a)\,\mu(da)=\int_{\mathbb{R}}f(a)\varphi''(a)\,da$ for every smooth compactly supported test function $\varphi$. Write $f'_-(z)$ for the left derivative of $f$ at $z$. Convexity ensures that this derivative exists and is locally bounded.

The measure $\mu$ records all the increases of the slope. For example, if the slope jumps at a point, $\mu$ has an atom there; if the slope changes continuously with density $q$, then $\mu(da)=q(a)\,da$. The local-time integral $\int_{\mathbb{R}}L_t^a\,\mu(da)$ is pathwise finite: for each fixed finite time, local time is continuous in $a$ and vanishes outside the range visited by $X$, while $\mu$ is finite on compact sets.

The Itô–Tanaka Formula

Theorem: For every finite convex $f:\mathbb{R}\to\mathbb{R}$, every deterministic $x$, and every $t\ge0$, $$ f(X_t)=f(x)+\int_0^t f'_-(X_s)\,dB_s+\frac12\int_{\mathbb{R}}L_t^a\,\mu(da) $$ almost surely. The stochastic integral is a continuous local martingale, defined by localization: for each $R>|x|$, its integrand stopped when $X$ exits $(-R,R)$ is progressively measurable and belongs to $\mathcal{L}^2$ on every finite time interval. The local martingale need not be a true martingale without further integrability.

Proof. We first establish the representation needed on a compact interval $[a,b]$. For $z\in[a,b]$, define $$ g(z)=f(a)+f'_-(a)(z-a)+\int_{[a,b)}(z-y)^+\,\mu(dy), $$ where $r^+=\max(r,0)$. The measure in this integral is finite because it is locally finite and $[a,b]$ is compact. The right derivative of the integral at $z$ is $\mu([a,z])$; at $z=a$ this is $\mu(\{a\})$. The convex derivative relation is $f'_+(z)=f'_-(a)+\mu([a,z])$ for $a\le z<b$, where $f'_+$ denotes the right derivative. Thus $g$ and $f$ have the same right derivative on $[a,b)$, and both equal $f(a)$ at $a$. A convex function is absolutely continuous on a compact interval, and its increments are the integrals of its almost-everywhere derivative. Hence $g=f$ on $[a,b]$, including $b$ by continuity. This proves the representation.

Now fix $R>|x|$, put $a=-R-1$, $b=R+1$, and let $\tau_R=\inf\{s\ge0:|X_s|\ge R\}$, with $\inf\varnothing=\infty$. For each $y\in[a,b)$, apply the positive-part Tanaka formula to $X_{t\wedge\tau_R}-y$: $$ (X_{t\wedge\tau_R}-y)^+=(x-y)^++\int_0^{t\wedge\tau_R}\mathbf{1}_{\{X_s>y\}}\,dB_s+\frac12 L_{t\wedge\tau_R}^y. $$ This is the earlier positive-part formula, with the same symmetric local-time normalization. Integrate this identity against $\mu(dy)$ on $[a,b)$, and add the affine term in the compact-interval representation. Stochastic Fubini applies here: $\mu([a,b))<\infty$, the indicators are bounded and progressively measurable, and their squared integrals over $[0,t]\times\Omega\times[a,b)$ are at most $t\mu([a,b))$. Approximation by simple functions and the Itô isometry therefore interchange the measure integral and the stochastic integral in $L^2(\mathbb{P})$.

The representation’s stochastic-integrand term is $f'_-(X_s)$ while $X_s\in[-R,R]$. This follows from $f'_-(z)=f'_-(a)+\mu([a,z))$, which is the left-derivative version of the slope relation; the positive-part integrand uses $\mathbf{1}_{\{z>y\}}$. Consequently the integrated identity is the claimed formula stopped at $\tau_R$. Since $X$ has continuous paths, $\tau_R\to\infty$ almost surely as $R\to\infty$. Every finite time interval is therefore covered by these stopped identities, proving the formula. For the integrability assertion, $f'_-$ is bounded on each compact interval, so the stopped integrand has $\mathbb{E}\int_0^T\mathbf{1}_{\{s\le\tau_R\}}f'_-(X_s)^2\,ds<\infty$. The resulting unstopped integral is a local martingale; the argument establishes only its localized martingale property, not a global integrability bound. $\square$

The factor $\tfrac12$ comes from the positive-part Tanaka formula. It is tied to the stated local-time normalization; changing that normalization changes the coefficient. When $\mu$ has a density $q$, the occupation-density formula gives $\int L_t^a q(a)\,da=\int_0^t q(X_s)\,ds$, recovering the usual Itô correction. When $\mu$ has atoms, local time at those points supplies the corresponding corrections.

Worked Applications

Worked Example: Quadratic Curvature

Take $f(z)=z^2$, so $X_t=x+B_t$. Its derivatives are $f'(z)=2z$ and $f''(z)=2$, hence $\mu(da)=2\,da$. The formula gives $$ X_t^2=x^2+\int_0^t 2X_s\,dB_s+\int_{\mathbb{R}}L_t^a\,da =x^2+\int_0^t 2X_s\,dB_s+t. $$ The last equality follows from the occupation-density formula with $g(a)=1$. The stochastic integrand is in $\mathcal{L}^2$ on $[0,T]$, since $\mathbb{E}\int_0^T4X_s^2\,ds=4x^2T+2T^2<\infty$; its integral is therefore a true martingale, with conditional expectation of each future increment equal to zero.

Check: Itô’s formula has $f'=2z$, $f''=2$, and correction $\tfrac12 f''\,dt=dt$, giving the same identity. Taking expectations gives $\mathbb{E}[X_t^2]=x^2+t$, since the stochastic integral has mean zero. This also agrees with $X_t=x+B_t$ and $\mathbb{E}[B_t^2]=t$.

Worked Example: Exponential Curvature

Take $f(z)=e^z$. Then $f'(z)=e^z$, $f''(z)=e^z$, and $\mu(da)=e^a\,da$. The formula reads $$ e^{X_t}=e^x+\int_0^t e^{X_s}\,dB_s+\frac12\int_{\mathbb{R}}e^aL_t^a\,da =e^x+\int_0^t e^{X_s}\,dB_s+\frac12\int_0^t e^{X_s}\,ds. $$ Here the stochastic integral is in $\mathcal{L}^2$ on $[0,T]$, because $\mathbb{E}\int_0^T e^{2X_s}\,ds=e^{2x}\int_0^T e^{2s}\,ds<\infty$. Thus it is a true martingale.

Check: Using the normal moment formula, $\mathbb{E}[e^{X_t}]=e^{x+t/2}$. Taking expectations on the last display gives $e^x+\tfrac12\int_0^t e^{x+s/2}\,ds=e^x+(e^{x+t/2}-e^x)=e^{x+t/2}$, as required.

Worked Example: A Smooth Convex Function with Nonconstant Curvature

Let $f(z)=\sqrt{1+z^2}$. Direct differentiation gives $$ f'(z)=\frac{z}{\sqrt{1+z^2}},\qquad f''(z)=\frac{1}{(1+z^2)^{3/2}}. $$ The second derivative is positive, so $f$ is convex, and its curvature measure has density $(1+a^2)^{-3/2}$. The formula is $$ \sqrt{1+X_t^2}=\sqrt{1+x^2}+ \int_0^t\frac{X_s}{\sqrt{1+X_s^2}}\,dB_s+ \frac12\int_{\mathbb{R}}\frac{L_t^a}{(1+a^2)^{3/2}}\,da. $$ The stochastic integrand is bounded in absolute value by $1$, so it belongs to $\mathcal{L}^2$ on every finite time interval and its integral is a true martingale.

Check: The occupation-density formula turns the correction into $\tfrac12\int_0^t(1+X_s^2)^{-3/2}\,ds$. In ordinary Itô’s formula the correction is exactly $\tfrac12 f''(X_s)\,ds=\tfrac12(1+X_s^2)^{-3/2}\,ds$. Thus the curvature-measure expression agrees with the classical formula when the curvature has a density.

What the Formula Does—and Does Not—Say

The formula identifies where convexity enters: $f$ itself contributes the left-derivative stochastic integral, and its positive curvature measure contributes a nonnegative local-time term. The curvature is measured in the space variable $a$, not in time. Atoms of $\mu$ weight local time at individual levels; continuous parts of $\mu$ average local time over levels. This is why a measure-valued second derivative is the natural object even when a classical second derivative does not exist.

A common mistake is to replace the curvature measure by an ordinary density without checking that one exists. For instance, for $f(z)=|z|$, the distributional second derivative is $2\delta_0$, not a function that can be integrated against $da$. Treating it as though it were a smooth density and writing $\int L_t^a f''(a)\,da$ misses the correction. The measure form instead gives $\tfrac12\int L_t^a\,2\delta_0(da)=L_t^0$, consistent with the specified local-time normalization.

Another mistake is to call the stochastic integral in the theorem a martingale without checking integrability. Localization guarantees a local martingale, while a true martingale requires additional conditions, such as $\mathbb{E}\int_0^T f'_-(X_s)^2\,ds<\infty$. The worked examples verify that condition individually. For a general convex function, the theorem alone does not provide it.

Check Your Understanding

Use the stated local-time normalization and distinguish the measure-valued correction from the stochastic integral.

For $f(z)=z^4$, find the density of its curvature measure and write the local-time correction in the formula.
Explain why a jump in the slope of a convex function produces an atom in its distributional second derivative.
For $f(z)=e^z$, identify the condition that makes the stochastic integral a true martingale on a fixed finite time interval.
Why does replacing a curvature atom by a density with respect to $da$ give the wrong correction?
State the occupation-density identity that turns a curvature measure with density $q$ into the classical Itô correction.

The Itô–Tanaka Formula for Convex Functions

What You'll Learn