Suppose $B$, $B_n$ are Brownian motions, and write $B^s$ for $B$ stopped at the first time equals $k$, say. (Similarly $B^s_n$).
I know how to prove the following: if $B_n \to B$ uniformly on compacts in probability then $B^s_n \to B^s$ uniformly on compacts in probability.
The proof is elementary but fiddly. Can anyone provide me with a reference I can quote?
Hi, I am not sure that you need a reference for this as it can easily be proven: $\tau_n \to \tau$ (where $\tau_n$ and $\tau$ are the relevant stopping times) in probability (since for all $\epsilon>0$, on the event $\tau \le T$, there exists $t \in [\tau,\tau+\epsilon]$ such that $B(t)>K$). Now, (always on the event $\tau \le T$) $$ P(\sup_{s \le T} |B^\tau(s)- B_n^\tau(s)|>\epsilon) \le P(\sup_{s \le T} |B(s)- B_n(s)|>\epsilon \cup |\tau - \tau_n|>\epsilon^4 \cup \sup_{s\in [\tau - \epsilon^4, \tau+\epsilon^4]} |B(s)-K|>\epsilon ) $$ which clearly goes to 0 as the probability of each event goes to 0.
Edit : I forgot a fourth event which looks like the third (with $B_n$ instead of $B$), but that still works.
