Let $X_{t}$ denote the solution to the SDE: $dX_{t}=(\mu -r)X_t dt+\sigma X_t d W_{t}, \ X_{0}=1$
i.e. $X_{t}$ is the process: $X_{t}:=\exp\left\{\left(\mu-r-\frac{\sigma^{2}}{2}\right)t+\sigma W_{t}\right\}$
(here we assume $\mu
Thank you!
Let $X_{t}$ denote the solution to the SDE: $dX_{t}=(\mu -r)X_t dt+\sigma X_t d W_{t}, \ X_{0}=1$
i.e. $X_{t}$ is the process: $X_{t}:=\exp\left\{\left(\mu-r-\frac{\sigma^{2}}{2}\right)t+\sigma W_{t}\right\}$
(here we assume $\mu
Thank you!
Since the stochastic integral process $Y(t)=\int_{0}^{t}X_{s}\,\text{d}W_{s}\ , \quad t \geq 0$ is a martingale and $\tau_{b}$ is an (almost-surely) bounded stopping time (adapted to the same filtration as process $Y$) then by the Optional Stopping Theorem the expected value of the random variable $Y(\tau_{b})$ is equal to that of the random variable $Y(0)$, which is zero.