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Let $\tau_s = \inf\{ t\,:\,\int_0^t X^2_u\,du > s\}$, and let $x_s = \int_0^s X_u\,dw$ ($X_t$ is some stochastic process and $w$ is a Standard Brownian Motion. I won't define $X_t$ for now since I don't think the specific form is important). The book I'm looking at uses a "change of variables" to conclude $$ \int_{\tau_s}^{\tau_t} e^{i\lambda x_u} X_u^2\,du = \int_s^t e^{i\lambda x_{\tau_u}}\,du. $$

Now I'm probably just missing some basic insight into stopping times (I never really investigated them too much, I'm kind of just picking up stochastic stuff where I need it), but I'm trying to see why this change of variables is true. I've been mostly pondering, so no need to give me a full rigorous proof or anything, but some hints/insights or possible a source might be helpful.

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    $\begingroup$ If $\beta, g : [0,\infty) \to [0,\infty)$ are Borel measurable such that $\beta$ and $\beta g$ are both locally $L^1$ and $\tau_t = \inf\{u : \int_0^u \beta(r) \,d r > t\}$, then $$ \int_0^{\tau_t} \beta(u)g(u) \,d u = \int_0^t g(\tau_u)\, d u.$$ This is given as problem 6.12 in Ethier, Kurtz Markov Processes: Characterization and Convergence. Now define $\beta (u) = X_u^2$ and $g(u) = e^{i\lambda x_u}$. $\endgroup$ – Sayantan Apr 4 at 6:51

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