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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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To compute $\mathbb{E}[Y_T\int_0^Ta Y_s^{-1}\,ds|\mathcal{F}_t]$ you only need to know how to compute $$ \mathbb{E}\left[\int_0^Te^{\sigma(W_T-W_s)}\,ds\,\middle|\,\mathcal{F}_t\right]=\int_0^t\mathbb{E}[e^{\sigma(W_T-W_s)}\,ds\,|\,\mathcal{F}_t]\,ds+\int_t^T\mathbb{E}[e^{\sigma(W_T-W_s)}\,ds\,|\,\mathcal{F}_t]\,ds. $$ We know that $\exp(\sigma W_t-\frac{\sigma^2t^2}{2})$ is a martingale. Therefore $$ \int_0^t\mathbb{E}[e^{\sigma(W_T-W_s)}\,ds\,|\,\mathcal{F}_t]\,ds=e^{\sigma W_t+\frac{\sigma^2(T-t)^2}{2}}\int_0^t e^{-\sigma W_s}\,ds $$ and $$ \int_t^T\mathbb{E}[e^{\sigma(W_T-W_s)}\,ds\,|\,\mathcal{F}_t]\,ds=\int_t^T\mathbb{E}[e^{\sigma(W_T-W_s)}]\,ds. $$ Now $$ \mathbb{E}[e^{\sigma(W_T-W_s)}]=e^{\frac{\sigma^2(T-s)}{2}} $$ and so $$ \int_t^T\mathbb{E}[e^{\sigma(W_T-W_s)}]\,ds=\frac{2}{\sigma^{2}}(e^{\sigma^{2}(T-t)}-1). $$

(Finally, by Ito if you wish $$ \int_0^t e^{-\sigma W_s}\,ds=\frac{2}{\sigma^2}e^{-\sigma W_t}+\frac{2}{\sigma}\int_0^te^{-\sigma W_s}\,dW_s.) $$

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