Suppose $Z(t)$ is a standard Brownian motion process with $Z(0)=0$, then calculate: $P(Z(3)>Z(2)>0)$
I have the following, but unsure if my rationale is correct:
$Z(3)=X_1+X_2+X_3, Z(2)=X_1+X_2$
Then $P(Z(3)>Z(2)>0)=P(X_1+X_2+X_3>X_1+X_2>0)=P(X_3>0)\cdot P(X_1+X_2>0)$
Finally we have $P(X_3>)\cdot P(Z(2)>0)=\frac{1}{4}$
Is this correct?
As Henrik suggested you have to use independent increments and the fact, that they are normal distributed. We have $\{Z(3)>Z(2)\}=\{Z(3)-Z(2)>0\}$, where $Z(3)-Z(2)\sim \mathcal{N}(0,1)$. Moreover, $\{Z(2)>0\}=\{Z(2)-Z(0)>0\}$, since $Z(0)=0$. Again, $Z(2)-Z(0)\sim\mathcal{N}(0,2)$.
We have $\{Z(3)-Z(2)>0\}\cap\{Z(2)>0\}=\{Z(3)>Z(2)>0\}$. Using independence of the increments of Brownian Motion, we get
$$P(Z(3)>Z(2)>0)= P(\{Z(3)-Z(2)>0\}\cap\{Z(2)>0\})=P(Z(3)-Z(2)>0)P(Z(2)>0)$$
Using the distribution above, I leave it to you to calculate this numerically.