Probability that a Wiener process is negative at 2 given that it was positive at 1 Let $W_t$ be a standard Wiener process, i.e., with $W_0=0$.  If $W_1>0$, what is the probability that $W_2<0$?
This is my attempt: we want to determine the conditional probability
$$\mathbb P(W_2 | W_1>0) = \frac{\mathbb P(W_2<0 \cap W_1>0)}{\mathbb P(W_1>0)}.$$
The denominator is easily computed equal to $1/2$. It remains to find $\mathbb P(W_2<0 \cap W_1>0)$.
Now, it is a well known fact that $(W_1, W_2)$ is jointly normally distributed, so we can write 
$$ \mathbb P(W_2<0 \cap W_1>0) = \mathbb P((W_1, W_2)\in ((0, \infty) \times (-\infty, 0)).$$
Also, it is not difficult to determine the covariance matrix of the random vector $(W_1, W_2)$, so the joint normal density can be completely determined. However, from this point on I don't know how to proceed. It seems an integration on the IVth quadrant of this density is necessary, but I was wondering if there is a less pedantic, more intelligent method.
 A: I think I finally found a very simple and nice solution, which uses almost no calculation. Here it is. 
First, we can write $W_2 = (W_2-W_1) + W_1$, and if we write $X=W_2-W_1$, $Y=W_1$, from the data of the problem we see that $X$ and $Y$ are both standard normal random variables, and independent with resepct to each other. Therefore, the problem asks for the computation of the conditional probability
$$\mathbb P(X+Y<0 | Y>0) = \frac{\mathbb P(X+Y<0\cap Y>0)}{\mathbb P(Y>0)}.$$
Now, ${\mathbb P(Y>0)}=1/2$ is immediate since $Y$ is standard normal. To compute the probability $\mathbb P(X+Y<0\cap Y>0)$, let us argue as follows: consider the joint density of $(X, Y)$ on a cartesian diagram labeled $(X,Y)$. Therefore, we are asking for the measure of the set obtained as the intersection of the sets $(X+Y<0\cap Y>0)$. This set corresponds to the lower half of the IInd quadrant on the Cartesian plane, and its measure is thus $1/8$. It follows that the measure of $\mathbb P(X+Y<0 | Y>0)$ is $\frac{1/8}{1/2}$, which is thus equal to $1/4$.
