Given $f\in C^2$ with compact support and $\tau$ be a stopping time with $E^x[\tau]<\infty$. Suppose $$Y_t=x+\int_0^t u(s,\omega)ds+\int_0^t v(s,\omega)dB_s(\omega), $$ with bounded $u,v$. Then by Ito formula with fixed $t$, I have
$$ f(Y_t)=f(Y_0)+\int_0^t (u\frac{\partial f}{\partial x}+\frac{1}{2}v^2\frac{\partial^2 f }{\partial^2 x})\,ds +\int_0^t v\frac{\partial f}{\partial x}\, dB_s, $$ My question is, how could I derive the following from the above identity that $$ E^x[f(Y_\tau)]=f(x)+E^x\bigg[\int_0^\tau (u\frac{\partial f}{\partial x}+\frac{1}{2}v^2\frac{\partial^2 f }{\partial^2 x})\,ds\bigg] +E^x\bigg[\int_0^\tau v\frac{\partial f}{\partial x}\, dB_s\bigg]. $$ In other words, how to change the deterministic $t$ into an integrable stopping time?
I think I got it. Please let me know if it has some problems.
For fixed integer $k$, we consider $$Z_t=x+\int_0^t 1_{\{s<\tau\}}u(s,\omega)ds+\int_0^t 1_{\{s<\tau\}}v(s,\omega)dB_s(\omega),t\in [0,k],$$ then Ito formula to $f(Z_t)$ implies $$ f(Z_k)=f(Y_{k\wedge \tau})=f(Y_0)+\int_0^{k\wedge \tau}(u\frac{\partial f}{\partial x}+\frac{1}{2}v^2\frac{\partial^2 f }{\partial^2 x})\,ds +\int_0^{k\wedge \tau} v\frac{\partial f}{\partial x}\, dB_s, $$ taking expectation and pass to limit (due to the dominant convergence, $\tau<\infty$ a.s., and the boundedness of the integrand), we get the desired result.