Let $X_t$ a brownian motion with $t\geq0$, calculate $P(X_2>0|X_1>0)$.
I did that:
$P(X_2>0|X_1>0)=P((X_2-X_1)+X_1>0|X_1>0)=P(X_2-X_1>0|X_1>0)$ because we know $X_1>0$ for condtionament.
Now $P(X_2-X_1>0|X_1>0)=\frac{P((X_2-X_1>0)\cap(X_1>0))}{P(X_1>0)}$
$X_t$ is a brownian motion, so its increment are independent. For this reason we can write:
$\frac{P((X_2-X_1>0)\cap(X_1>0))}{P(X_1>0)}=\frac{P(X_2-X_1>0)P(X_1>0)}{P(X_1>0)}=P(X_2-X_1>0)=1-P(Z\leq0)$, with $Z\sim N(0,2t)$.
But now I discovered that the passage $P((X_2-X_1)+X_1>0|X_1>0)=P(X_2-X_1>0|X_1>0)$ is totaly wrong.
I don't find any other way to resolve it. So I hope that someone can help me to resolve it.
Thank you.
Note that $Z := X_2 - X_1 \sim \mathcal N (0,1)$ and independent to $Y := X_1 \sim \mathcal N (0,1)$. So just treat them separately as gaussian random variables to calculate $$\Bbb P ( (X_2 - X_1 ) + X_1 > 0 , X_1 >0) = \Bbb P (Z + Y > 0 , Y >0)$$ Furthermore its obvoius how to calculate $\Bbb P (X_1 > 0)$.