Suppose that we have a Wiener process $x(t)$. Also we know that the random variable $x(h)$ ($h>0$) follows the $\mathcal{N}(0,h)$ normal distribution. Assume that we obervation the stochastic process $x(t)$ at $t=h$, the exactly value is unknown but we know that $|x(h)|<\rho$. I believe that the posterior probability distribution $x(h)$ conditioned on $|x(h)|<\rho$ will not be $\mathcal{N}(0,h)$ anymore. In this case, we need to use the Bayes' rule to calcuate the posterior probability. \begin{equation} f(x(h)||x(h)|<\rho) = \frac{f(|x(h)|<\rho|x(h))f(x(h))}{f(|x(h)|<\rho)} \end{equation} $f(x(h))$ is a normal distribution. $f(|x(h)|<\rho|x(h))$ may be calculated by Kolmogorov backward equation. But I do not know how to specify the boundary conditions. In addition, I do not know how to compute $f(|x(h)|<\rho)$.
Some approximations can be made, such as $x(h)$ conditioned on $|x(h)|<\rho$ follows a uniform distribution between $[-\rho,\rho]$ or trucated normal distribution $[-\rho,\rho]$.
I want to know how to exactly calculate the distribution $f(x(h)||x(h)|<\rho)$ even though it may be computational expensive.
$$\begin{aligned}P(B_h\leq x||B_h|<\rho)&:=\frac{P(B_h\leq x,|B_h|<\rho)}{P(|B_h|<\rho)}\\ &=\frac{\Phi(\max(\min(x,\rho),-\rho)/\sqrt{h})-\Phi(-\rho/\sqrt{h})}{2\Phi(\rho/\sqrt{h})-1} \end{aligned}$$ The derivative wrt $x$ within the $\rho$-radius ball is: $$\frac{d}{dx}P(B_h\leq x||B_h|<\rho)=\frac{\phi(x/\sqrt{h})}{\sqrt{t}(2\Phi(\rho/\sqrt{h})-1)},\,x \in (-\rho,\rho)$$