# Tanaka-Meyer formula

I have a simple question about Tanaka-Meyer formula, I am having difficulty applying it. Let $X$ be a continous martingale vanishing at zero. From Tanaka-Meyer formula it holds $$d|X_t| = sgn(X_t)dX_t+d \Lambda^X_t(0)$$ where $\Lambda^X_t(0)$ is the local time accumulated by the process $X$ at the origin.

I am interested in the process $Z = X^2$. From Itô's formula we have $$dZ_t = 2X_t dX_t + d \langle X \rangle_t.$$ Since obviously $Z = |Z|$ applying Tanaka-Meyer should have the same differential, however $$d|Z_t| = 2sgn(Z_t) X_t dX_t + sgn(Z_t)d \langle X \rangle_t + d \Lambda^Z_t(0) ,$$ and $$dZ^+_t = 2*\mathbb{1}_{[Z_t>0]} X_t dX_t + \mathbb{1}_{[Z_t>0]}d \langle X \rangle_t + {1 \over 2} d \Lambda^Z_t(0) .$$

Applying the expected value to processes $|Z|$ and $Z^+$, I have \begin{align} \mathbb{E}[\int_0^T sgn(X^2_t)d\langle X \rangle_t +\Lambda^Z_t(0) ] = \mathbb{E}[\int_0^T \mathbb{1}_{[X^2_t>0]}d\langle X \rangle_t +{1 \over 2}\Lambda^Z_t(0)] \end{align}

which feels like that $\Lambda^Z_t(0) = 0 ~ a.s.$ Where is my reasoning wrong? I have been staring at it longer than I like to admit.

• Why shouldn't $\Lambda^Z_t(0)$ be zero? I think one can verify this directly for $X_t$ a Brownian motion, for instance. – Nate Eldredge Mar 6 '18 at 4:22

It looks to me like nothing's wrong; the local time $\Lambda^Z_t(0)$ of $Z$ at zero is identically zero. This makes a certain amount of intuitive sense, because $Z$ should have "bounded variation at zero".
If we follow the definitions in "The Pedestrian's Guide to Local Time" by Björk, local time $\Lambda^Z_t(x)$ is cadlag in $x$ and satisfies \begin{align*}\int_0^\epsilon \Lambda^Z_t(x)\,dx &= \int_0^t 1_{\{0 \le Z_s \le \epsilon\}}\,d\langle Z \rangle_s \\ &= \int_0^t 1_{\{0 \le Z_s \le \epsilon\}} 4 Z_s d\langle X \rangle_s \\ &\le 4\epsilon \int_0^t 1_{\{-\sqrt{\epsilon} \le X_s \le \sqrt{\epsilon}\}} d\langle X \rangle_s \\ &= 4\epsilon \int_{-\sqrt{\epsilon}}^{\sqrt{\epsilon}} \Lambda_t^X(x)\,dx \end{align*} Hence, we have $\inf_{x \in (0,\epsilon)} \Lambda^Z_t(x) \le 4 \int_{-\sqrt{\epsilon}}^{\sqrt{\epsilon}} \Lambda_t^X(x)\,dx$. As $\epsilon \downarrow 0$ the right side goes to zero (because $\Lambda_t^X$ is cadlag and hence bounded in $x$). So $\liminf_{x \downarrow 0} \Lambda_t^Z(x) \le 0$, which forces $\Lambda_t^Z(0) = 0$ because it is nonnegative and cadlag.
Alternatively, use Corollary 1.9 in Revuz & Yor and $d\langle Z \rangle_t = 4 X_t^2 d \langle X \rangle_t$ (as Eldredge did) to write \begin{align*} \Lambda_t^Z(0) &= \lim_{\epsilon \downarrow 0} \frac{1}{\epsilon} \int_0^t 1_{\{ 0 \le Z_s < \epsilon\} } d \langle Z \rangle_s \\ &= \lim_{\epsilon \downarrow 0} \frac{1}{\epsilon} \int_0^t 1_{\{ -\sqrt{\epsilon} < X_s < \sqrt{\epsilon} \}} 4 X_s^2 d \langle X \rangle_s \\ &\le 4 \lim_{\epsilon \downarrow 0} \int_0^t 1_{ \{-\sqrt{\epsilon} < X_s < \sqrt{\epsilon}\} } d \langle X \rangle_s \\ &\le 4 ( \Lambda_t^X(0) + \Lambda_t^{-X}(0)) \lim_{\epsilon \downarrow 0} \sqrt{\epsilon} = 0 \quad \text{a.s.} \end{align*} which implies $\Lambda_t^Z(0) = 0$ a.s.