Say we have a one-dimensionless Ito process, $Z$, but our $f$ isn't smooth, and are trying got figure out the processes of $f(Z)$. I just want to focus on the special case where $Z$ is a $(\mu,\sigma)$ Brownian motion on $(\Omega, \mathcal{F}, P)$. If $f(x)=|x|$, then Tanaka's formula tells us $|Z_t|=\int_{0}^{t} sgn(Z)dZ+\sigma^2l(t,0)$, $t\geq 0$, where $\frac{1}{2\epsilon}\int_0^t 1_{\{|Z(s)|\leq\epsilon\}}ds\rightarrow l(t,0)$, as $t\downarrow 0$.
Is there a similar formula for a $f(x)=max(0,x)=(x)^{+}$?
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2 Answers
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The censoring function is the same as the absolute value plus the identity.
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$\begingroup$ Do you mean use $max(0,x)=\frac{x+|x|}{2}$? $\endgroup$ Jul 12, 2019 at 18:13
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$\begingroup$ @DrunkDeriving yes, let me know if you can't get the answer from there $\endgroup$– BananachJul 12, 2019 at 18:58
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I found it in a few papers: $$(Z_T)^+=(Z_0)^++\int_0^T 1_{\{Z_s>0\}} dZ_s+\frac{1}{2}l(T,0)$$