Calculate this probability: $$ P(\max_{0\leq t \leq T}B(t)>x | \min_{0\leq t \leq T}B(t)>0) = ? $$
My attempt of solution:
$$ P(\max_{0\leq t \leq T}B(t)>x | \min_{0\leq t \leq T}B(t)>0) = \frac{P(\max_{0\leq t \leq T}B(t)>x \wedge \min_{0\leq t \leq T}B(t)>0)}{ P(\min_{0\leq t \leq T}B(t)>0)} $$
Now I use the fact: $$ \min_{0\leq t \leq T}B(t) \stackrel{\text{d}}{=} -\max_{0\leq t \leq T}(-B(t)) $$ and I have: $$ P(\max_{0\leq t \leq T}B(t)>x | \min_{0\leq t \leq T}B(t)>0) = \frac{P(\max_{0\leq t \leq T}B(t)>x \wedge -\max_{0\leq t \leq T}(-B(t))>0)}{ P(\min_{0\leq t \leq T}B(t)>0)} = \frac{P(\max_{0\leq t \leq T}B(t)>x \wedge \max_{0\leq t \leq T}(-B(t))<0)}{ P(\min_{0\leq t \leq T}B(t)>0)} $$
I don't know what I can do next?
I am not sure if my attempt is correct, I am learning this as well.
Assume we get $\min B(t)$ at point $a$, get $\max B(t)$ at point $b$. $$\min B(t)=B(a), \max B(t)=B(b)$$
Rewrite your problem: $$\frac{P[B(b)>x \cap B(a)>0 ]}{P[B(a)>0]}$$ Consider inequality here: If we have $B(b)-B(a)>x$ and $B(a)>0$, we can imply $B(b)>x$, so I am trying to substitute $B(b)>x$ with $B(b)-B(a)>x$
Hence we have
$$\frac{P[(B(b)-B(a))>x \cap B(a)>0 ]}{P[B(a)>0]}$$
Since $B(b)-B(a)=B(|b-a|)$ is independent of $B(a)$, we just need to calculate $$P[(B(|b-a|))>x]$$ That is, $$P[(N(0,|b-a|)>x]$$