$P(|B(2)|>|B(1)|)$ for Brownian motion

Question

Let $B(t), t\ge0,$ be a standard Brownian motion with $B(0)=0$. What is $P(|B(2)|>|B(1)|)$? One straightforward method is to consider four cases: both are positive, both are negative, etc. I found it a bit tedious. Is there a simpler method?

Answer 1

Write $X = B(1)$ and $Y = B(2) - B(1)$, and note that $X$ and $Y$ are i.i.d. standard normal variables. Then

\begin{align*} |B(2)| \geq |B(1)| &\quad\iff\quad |X+Y| \geq |X| \\[0.25em] &\quad\iff\quad \begin{cases} \text{$X \geq 0$ and $|X + Y| \geq X$, or} \\ \text{$X \leq 0$ and $|X + Y| \geq -X$.} \end{cases} \\[0.25em] &\quad\iff\quad \begin{cases} \text{$X \geq 0$ and ($X+Y \geq X$ or $-X \geq X+Y$), or} \\ \text{$X \leq 0$ and ($X+Y \geq -X$ or $X \geq X+Y$).} \end{cases} \\[0.25em] &\quad\iff\quad \begin{cases} \text{$X \geq 0$ and ($Y \geq 0$ or $-2X \geq Y$), or} \\ \text{$X \leq 0$ and ($Y \geq -2X$ or $0 \geq Y$).} \end{cases} \end{align*}

The region specified by these conditions corresponds to the union of two infinite-sectors, one given by $0 \leq \Theta \leq \pi-\arctan 2$ and the other by $-\pi \leq \Theta \leq -\arctan 2$. Here, $(R, \Theta)$ refers to the polar coordinates of $(X, Y)$, where $\Theta$ is chosen to lie between $-\pi$ and $\pi$.

Since $\Theta$ is uniformly distributed over the interval between $-\pi$ and $\pi$, it therefore follows that

$$ \mathbf{P}(|B(2)| \geq |B(1)|) = \frac{2(\pi - \arctan 2)}{2\pi} = 1 - \frac{\arctan 2}{\pi}. $$