I've tried calculating it as $Pr[W(3)-W(1)<-1] + 1 - Pr[W(3)-W(1)<1] =0.23975 + 1- 0.76025 =47.95\%$
But, this doesn't seem to be correct. Please help
2026-04-05 02:39:44.1775356784
Suppose $W(t)$ is a standard Brownian motion. Calculate the probability of $1<W(1)<W(3)−1$
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This is about the independence of $W(1)$ and $W(3)-W(1)$. $$ \begin{align} P(W(3)-W(1)>1 \ \mathrm{and} \ W(1)>1)&=P(W(3)-W(1)>1) P(W(1)>1)\\ &=P(W(2)>1)P(W(1)>1)\\ &=P(\sqrt 2 W(1)>1) P(W(1)>1)\\ &=\left(1-\Phi(\frac1{\sqrt 2}) \right) (1-\Phi(1)) \end{align} $$ where $$ \Phi(x) = \frac 1{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{t^2}2} dt. $$