# How to show Tanaka's formula is the Doob-Meyer decomposition?

In Wikipidia it's said that the Tanaka's formula is the explicit Doob–Meyer decomposition of the submartingale $$\vert B(t)\vert$$.

In general we have that Tanaka's formula could be written as:

$$\vert B(t)-a\vert =\vert a\vert+ \int_0^t {\text{sgn}}(B(s)-a)dB(s)+\mathbb P\lim_{\epsilon\to0}\frac{1}{2\epsilon}\int_0^t1_{(a-\epsilon,a+\epsilon)}(B(s))ds$$

I can see that since $$\vert{\text{sgn}}(B(s)-a)\vert \leq 1$$ we have that $$\mathbb E\int_0^t \vert{\text{sgn}}(B(s)-a)\vert^2ds<\infty,$$

and hence $$M_t=\int_0^t {\text{sgn}}(B(s)-a)dB(s)$$ is a continuous martingale.

Then I should be able to show that $$\mathbb P\lim_{\epsilon\to0}\frac{1}{2\epsilon}\int_0^t1_{(a-\epsilon,a+\epsilon)}(B(s))ds$$ (the local time) is a predictable, rigt continuous increasing processes.

This may be straightforward but I am not able to see the latter. Could you please give me some hint? Thanks in advance.

• It's not too difficult to show that the limit, call it $L(t)$, exists in the $L^2$ sense, and clearly $L(s)\le L(t)$ a.s. if $s<t$. Also, $L(t)$ is adapted to the Brownian filtration. And $t\mapsto L(t)$ has a continuous version because the stochastic integral has one. Commented Feb 11, 2020 at 22:55
• @JohnDawkins I've followed you until the continuity part, to which stochastic integral do you refer? The integral in the local time is a Riemann integral Commented Feb 12, 2020 at 10:11
• @RScrlli You need to read it the other way round: $$\mathbb{P}\lim_{\epsilon} \frac{1}{2\epsilon} \int_0^t 1_{(a-\epsilon,a+\epsilon)}(B(s)) \, ds = |B_t-a| - |a| - \int_0^t \text{sgn}(B_s-a) \, dB_s.$$ JohnDawkins was saying that the right-hand side has a continuous version and, hence, the left-hand side as well.
– saz
Commented Feb 12, 2020 at 15:17
• I get it! Thanks both for the help Commented Feb 12, 2020 at 17:35