Consider the event: $A_n:= \{S(t_n) >z_{n-1}S(t_{n-1})\}$ with $z_{n-1}=e^{\sqrt{t_n-t_{n-1}}}, t_k \in [0,T)$.
$S(t_n) $ is an geometric Brownian motion, meaning: $S(t_n) = \exp\left( \left(\mu - \frac{\sigma^2}{2} \right)t_n + \sigma W(t_n)\right)$
How can I argue that the events $A_k$ are independent?
The random variables $X_n$ defined as $$X_n:=\ln\frac{S(t_n)}{S(t_{n-1})}=(\mu-0.5\sigma^2)(t_{n}-t_{n-1})+\sigma(W_{t_n}-W_{t_{n-1}})$$ depend only on the Brownian increments $\Delta W_{t_n}:=W_{t_n}-W_{t_{n-1}}$, which are independent between each other by definition. The following equivalence holds: $$\{S(t_n)>S(t_{n-1})z_{n-1}\}=\{X_n>\sqrt{t_n-t_{n-1}}\}$$ Since $X_n$ and $X_k$ are independent $\forall n \neq k$, we have that $$P(X_n>\sqrt{t_n-t_{n-1}},X_k>\sqrt{t_k-t_{k-1}})=P(X_n>\sqrt{t_n-t_{n-1}})P(X_k>\sqrt{t_k-t_{k-1}})$$