# Conditional expectation of an integrated GBM

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? $$\tilde{\Bbb{E}}[Y_T\int^T_0\frac{a}{Y_s}ds|\mathcal{F}_t]$$

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 $$\int_0^t e^{-\sigma W_s}\,ds=\frac{2}{\sigma^2}e^{-\sigma W_t}+\frac{2}{\sigma}\int_0^te^{-\sigma W_s}\,dW_s.)$$