Let $W_t$ a standard Brownian motion, with $t\geq0$. Determine:
(a)$P(W_2>0|W_1>0)$;
(b)$P(W_2-W_1>0|W_1=0)$.
For point (b) I try to do this:
$P(W_2-W_1>0|W_1=0)=P(W_2-W_1-W_1>0-0|W_1=0)=$(by indipendent increments)$=P(W_2-2W_1>0)=P(Z>0)=1-P(Z\leq0)$, where $Z=W_2-2W_1$.
W_1 and W_2 are two Normal variable (for Brownian motion) so also Z is a Normal variable with $\mu$(=mean value)=$1*\mu_2+(-2)*\mu_1=1*0-2*0=0$, where $\mu_i$ is mean value of $W_i$, and $\sigma^2$(=variance)=$(1)^2*\sigma^2_2+(-2)^2*\sigma^2_1=1*t+4*t=5t$, where $\sigma^2_i$ is variance of $W_i$.
So, $1-P(Z\leq0)=1-\int_{-\infty}^{0}{\left(\frac{1}{\sqrt{2\pi5t}}\right)e^0dx}=1-\left(\frac{1}{\sqrt{10\pi t}}\right)$
Is it correct?
For point (a) I don't know how do, so I hope that someone can help me. Thank you so much.