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.

  • $\begingroup$ Why shouldn't $\Lambda^Z_t(0)$ be zero? I think one can verify this directly for $X_t$ a Brownian motion, for instance. $\endgroup$ Mar 6, 2018 at 4:22

2 Answers 2


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.

  • 1
    $\begingroup$ Thank you very much for your answer. To clarify my confusion, there's a formula which states the relationship between the local time of a martingale and its transformation. Then, due to this formula, I thought this would imply that since the local time of Z is identically zero, so should the local time of X, but I made there very stupid mistake - x at zero is zero, not one. Thank you again. $\endgroup$
    – Johny
    Mar 6, 2018 at 16:04
  • $\begingroup$ The post is closed but I just want to add a precision in the Nate Eldredge's post : The explanation of last equality can be found in Theorem 4.1.6. in "The Pedestrian's Guide to Local Time" by Björk, for those who are interested. (or Th. 3.1.4. for Brownian motion) $\endgroup$
    – Al Bundy
    May 4, 2020 at 15:34

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.


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