I am solving one problem related to the symmetry property of a Brownian motion.
If we have a Brownian motion, $X_t$, $X_0=0$, $X_1>0$, what is the probability of $X_2 < 0$?
My attempt:
$P(X_2< 0 | X_1 = x) = P(X_2-X_1< -x, X_1 \in x)/P(X_1\in x) = P(X' < -x) = N(-x)$ where I have used the properties of the stationary and increment independence.
Then, the answer is $$ P(X_2<0, X_1 \in x) = \int_0^\infty P(X_2<0|X_1 = x) f_{X_1}(x) dx = \int_0^\infty N(-x)\frac{1}{\sqrt{2\pi}}e^{-0.5x^2}dx $$
But the answer is 1/4. Which step is wrong in the above approach?
You are calculating $P([X_2<0]\cap[X_1>0])$; what you want is $P(X_2<0|X_1>0) = P([X_2<0]\cap[X_1>0])/P(X_1>0)$. This is twice what you have calculated.
If you write $X_1=U$ and $X_2=X_1+V$ (so $U$ is the increment from $X_0$ to $X_1$ and $V$ is the increment from $X_1$ to $X_2$) then $U$ and $V$ are i.i.d. $N(0,1)$ and can be written as $U=R \cos\Theta$ and $V=R\sin\Theta$, with $\Theta$ uniform and independent of $R$. Then your desired conditional probability works out to $P(\Theta<-\pi/4|\Theta\in[-\pi/2,\pi/2]) = (\pi/4)/(\pi)$, using the fact that $\Theta$ is uniformly distributed on $(-\pi,\pi)$.