$S_t$ is a geometric Brownian motion, i.e., $dS_t = \mu S_t dt + \sigma S_t dW_t.$
Does the following hold:
$$ \mathbb{E}_t (S_T \mathbb{1}_{\left\{ \tau (\delta) < T \right\}}) = \mathbb{E}_t (S_T) \mathbb{E}_t (\mathbb{1}_{\left\{ \tau (\delta) < T \right\}}) $$ where $\tau$ is a stopping time and defined as $$\tau (\delta) = \inf\left\{0<t<T: S_t < \delta\right\}? $$
I only have some rough intuitions regarding the independence between $S_T$ and $\mathbb{1}_{\tau (\delta) <T }$: $S_T$ is at the end point and its distribution is independent of the path leading to $S_T$, while $\mathbb{1}_{\tau (\delta) <T }$ is pretty much dependent on the path, therefore, I thought the equation holds.