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To shorten the notation, let's write $T_t = T(D_t,y_t)$ and $\delta_t = \delta(D_t,y_t)$. There are two ways to show that, in fact, the dynamics of $$ \xi_t = \xi(D_t, y_t,t) = e^{-\int_0^t \delta_s ds}\, T_t $$ is given by $$ \frac{d\xi_t}{\xi_t} = \left( -\delta_t + \frac{\mathscr{L} T_t}{T_t} \right)dt \quad+\quad \text{diffusion terms}. $$ First way ...


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I think you are having trouble differentiating the integral of $\delta$. You should remember the differential notation is just notation for an integral: $A_td B_t = A'_t dB'_t$ just means $\int_0^T A_td B_t = \int_0^T A'_t dB'_t$. In particular, $d\int_0^t A'_s dB'_s = A'_t dB'_t$ is a tautology. So $$ d ( e^{\int_0^t \delta(s,X_s) ds} )= e^{\int_0^t ...



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