Consider Ito's lemma in the following standard version $$h(W_t) = h(W_0) + \int_0^t \nabla h(W_s) dW_s + \frac{1}{2} \int_0^t \Delta h(W_s) ds.$$

I am asking myself under which conditions, the deterministic time $t$ can be replaced by $t \wedge \tau$, where $\tau$ is a stopping time. Does anybody have an idea?


2 Answers 2


If (as is customary) the stochastic integral is understood to be continuous in $t$ (a.s.) then the equality holds for all $t$ simultaneously, with probability 1. As such, $t$ can be replaced throughout by any non-negative random variable and the a.s. equality will persist.

  • $\begingroup$ Yes, this is much easier. $\endgroup$
    – zhoraster
    Jan 9, 2020 at 21:18

Under the same conditions under which the Itô formula is valid. Indeed, the process $X_s = W_{\tau \wedge s}$ is an Itô process with stochastic differential $dX_s = \mathbf{1}_{[0,\tau]}(s) dW_s$ (see e.g. our book with Yuliya Mishura, Theorem 8.4). Then, using the Itô formula and this "locality" property once more, $$ h(X_t) = h(X_0) + \int_0^t \nabla h(X_s)\mathbf{1}_{[0,\tau]}(s) dW_s + \frac{1}{2} \int_0^t \Delta h(X_s) \mathbf{1}_{[0,\tau]}(s)^2 ds \\ = h(W_0) + \int_0^{t \wedge \tau} \nabla h(W_s) dW_s + \frac{1}{2} \int_0^{t \wedge \tau} \Delta h(W_s) ds. $$


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