Hi thanks in advance for any time spent looking at this question,
I am currently studying brownian motion and am stuck on the following question:
$$P(|B_t| \in C | B_s = x) = P(B_t - B_s\in C-x) + P (B_t-B_s \in C+x)$$
where $B_t$ is standard brownian motion and $ C \subseteq [0,\infty) $.
If anyone can help me through this I would appreciate it.
Thanks again for any help.
$$P(|B_t|\in C|B_s=x) = P(B_t \in C|B_s=x) + P(-B_t \in C|B_s=x)$$ This follows because the event on the left is the union of the events on the right, which are mutually exclusive.
Given that $B_s=x$, then the event that $B_t \in C$ is equivalent to the event that $x + (B_t - B_s) \in C$. If this isn't obvious, I can show it step by step.
I guess that $x + (B_t - B_s) \in C$ can be rephrased $B_t - B_s \in C-x$, although I've never seen that notation before.
So, we have $$P(B_t \in C|B_s=x) = P(B_t - B_s \in C-x|B_s=x)$$ Why can we drop the $B_s=x$ behind the conditional bar? I guess it must be the Markov property, though I can't state clearly why.