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(I asked this question on MSE but I think it might have more success here)

Good day,

I was going over some exercises and I stumbled upon a question that, for its solution, requires me to find/simplify $$ \tilde{\Bbb{E}}[S_T|\mathcal{F}_t] $$ in terms of $S_t$ where $$ S_t=S_0Y_t+Y_t\int^t_0\frac{a}{Y_s}ds $$ $$ dY_t=rY_tdt+\sigma Y_td\tilde{W}_t$$ $$ \ Y_t=exp \left( \sigma\tilde{W}_t+(r-0.5\sigma^2)t \right) $$ $$ dS_t=rS_tdt+\sigma S_t d\tilde{W}_t +adt$$

$\tilde{\Bbb{P}}$ is the risk neutral measure.

$Y_t$ is a GBM and thus I think the first term is easy to deal with, but the 2nd one with the integral is a bit of a mystery to me. Do I have to take the $Y_T$ inside the integral and play with the exponential form of the GBM? Any help would be appreciated.

In essence, how do I find the following? $$ \tilde{\Bbb{E}}[Y_T\int^T_0\frac{a}{Y_s}ds|\mathcal{F}_t] $$

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Let $$Z_t=Y_t\int_0^t\frac{a}{Y_s}ds$$ Then $Z_0=0$.
We differentiate $Z_t$ and obtain $$dZ_t=\int_0^t\frac{a}{Y_s}dsdY_t+Y_t\frac{a}{Y_t}dt=\int_0^t\frac{a}{Y_s}ds(rY_tdt+\sigma Y_td\tilde{W_t})+adt$$ $$=rY_t\int_0^t\frac{a}{Y_s}dsdt+\sigma Y_t\int_0^t\frac{a}{Y_s}dsd\tilde{W_t}+adt$$ Then $$dZ_t=rZ_tdt+\sigma Z_td\tilde{W_t}+adt$$ We have $$d(e^{-rt}Z_t)=e^{-rt}(\sigma Z_td\tilde{W_t}+adt)$$ Thus, $$e^{-rt}Z_t=\int_0^te^{-rs}\sigma Z_sd\tilde{W_s}+a\int_0^te^{-rs}ds$$ $$=\int_0^te^{-rs}\sigma Z_sd\tilde{W_t}-\frac{a}{r}(e^{-rt}-1)$$ So $$Z_T=e^{rT}\int_0^Te^{-rs}\sigma Z_sd\tilde{W_s}-\frac{a}{r}(1-e^{rT})$$ and we can express the desired expectation with quantities known at time $t$ $$\mathbb{E}_t[Z_T]=e^{rT}\int_0^te^{-rs}\sigma Z_sd\tilde{W_s}-\frac{a}{r}(1-e^{rT})$$

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