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Consider a continuous martingale $(X_t)_{t≥0}$ starting at 0 and define the associated Local Time at $a$:

$$L^a_t=\lim_{ϵ→0} {1 \over 2ϵ}∫_0^t 1_{[a−ϵ,a+ϵ]}(X_s)ds.$$

Is it possible for the local time $L^a_t$ to have a derivative in the spatial variable $a$?

Under what conditions is the function $g(a) = E[L^a_t] $ differentiable?

Johny
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    This is not an answer, but you may try the following, when $X$ is the standard Brownian motion. Define $A(y):=|y|$, let ${\psi_{\epsilon}}$ be a smooth approximation to identity and $A_{\epsilon} : = A*\psi_{\epsilon}$. Then one can write $$L^a_t = L^k(\Omega)-\lim_{\epsilon \to 0} \int_0^t A_{\epsilon}'' (X_s-a),d s.$$ Apply Ito's formula to get $$A_{\epsilon}(X_t - a) = A_{\epsilon}(-a) + \int_0^t A_{\epsilon}'(X_s-a) ,d X_s + \frac{1}{2}\int_0^t A_{\epsilon}''(X_s-a) ,d s , a.s.$$ – Sayantan Apr 06 '18 at 03:03
  • Now it may be possible to differentiate this expression wrt $a$. Then one has to check what limit, if any, the differentiated equation has as $\epsilon \to 0$. Compare this with Tanaka's formula https://en.wikipedia.org/wiki/Tanaka's_formula – Sayantan Apr 06 '18 at 03:04

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