Let $B(t)$ be standard Brownian motion. Suppose, I want to calculate $$P(B(1)>4|B(2)=3)$$ How should I calculate this?
If the problem look like $$P(B(2)>4|B(1)=3)$$ then I know how to do that using independent increment argument, but not the above problem.
From https://www.probabilitycourse.com/chapter5/5_3_2_bivariate_normal_dist.php we have the theorem that if $X$ and $Y$ are jointly normal random variables with parameters $\mu_X$, $\sigma^2_X$, $\mu_Y$, $\sigma^2_Y$, and $\rho$, then \begin{align} \mathbb E[Y\mid X=x] &= \mu_Y + \rho\sigma_Y\frac{x-\mu_X}{\sigma_X}\\ \mathrm{Var}(Y\mid X=x) &= (1-\rho^2)\sigma^2_Y. \end{align} Here $X=B(1)$ and $Y=B(2)$, so $\mu_X=\mu_Y=0$, $\sigma^2_X = 1$, $\sigma^2_Y=2$, and $\rho = 1/\sqrt 2$. Hence \begin{align} \mathbb E[Y\mid X=3] &= (1/\sqrt 2)\sqrt 2\frac{3-0}{1} = 3\\ \mathrm{Var}(Y\mid X=3) &= (1-1/2)\cdot 2 = 1. \end{align} Conditioned on $\{B(2)=3\}$ we have that $B(1)$ is normally distributed with the above parameters. So we compute \begin{align} \mathbb P(B(1)>4\mid B(2)=3) &= \int_4^\infty\frac1{\sqrt{2\pi}} e^{-\frac12\left(x-3\right)^2}\ \mathsf dx\\ &= \frac{1}{2} \text{erfc}\left(\frac{1}{\sqrt{2}}\right). \end{align}