I have a question on the first hitting processes of Brownian motion.
A Brownian particle starting at $x_{t=0}=0$ and subjected to a Gaussian white noise$\xi(t)$, the motion of the particle will be a normal diffusion process. If we set a wall at $x=L>0$, the particle will hit the wall for the first time at some moment $\tau$. The PDF of the first hitting time $\tau$ also could be calculated.
My question is: among all the possible processes, if we only concern the processes that assured the particle hit the wall at time $\tau$, the noise $\tilde{\xi}(t)$ acting on the particle during these asssured-hitting processes is clearly not Gaussian white noise anymore. Since $\frac{1}{\gamma}\int_{0}^{\tau}\tilde{\xi}(t)dt=L$ for these processes. So, how could we describe this noise $\tilde{\xi}(t)$, such as the average $<\tilde{\xi}(t)>$ and the self-correlation $<\tilde{\xi}(t)\tilde{\xi}(t')>$, or other properties? Is there any books or articles on this?
(Not a complete answer.)
Usual notations are that one considers a Brownian motion $(B_t)$, hence $B_t$ is the position of the particle at time $t$, and, for every $L>0$, $\tau_L=\inf\{t\mid B_t=L\}$. One can then indeed condition, for every positive $t$, on the event $\{\tau_L=t\}$, although the conditioning is not trivial since $P(\tau_L=t)=0$.
More precisely, for every positive $L$ and $t$, consider the PDF $q_L$ of $\tau_L$, and the PDF $p_t$ of $B_t$. Then, for every $s$ in $(0,t)$, using the reflection principle and the Markov property, one sees that the PDF $p_s^{(t,L)}$ of $B_s$ conditioned on $\tau_L=t$ is $$p_s^{(t,L)}(x)=(p_s(x)-p_s(2L-x))\frac{q_{L-x}(t-s)}{q_0(t)}\mathbf 1_{x<L}$$ Likewise, for every $u<s$ in $(0,t)$, the PDF $p_{u,s}^{(t,L)}$ of $(B_u,B_s)$ conditioned on $\tau_L=t$ is $$p_{u,s}^{(t,L)}(x,y)=(p_u(x)-p_u(2L-x))(p_{s-u}(y-x)-p_{s-u}(2L-y+x))\frac{q_{L-y}(t-s)}{q_0(t)}\mathbf 1_{x<L,y<L}$$ Thus, for every $x<L$, the conditional PDF $p_{s\mid u}^{(t,L)}$ of $B_s$ conditioned on $B_u=x$ and $\tau_L=t$ is $$p_{s\mid u}^{(t,L)}(y\mid x)=(p_{s-u}(y-x)-p_{s-u}(2L-y+x))\frac{q_{L-y}(t-s)}{q_{L-x}(t-u)}\mathbf 1_{y<L}$$ One can deduce from the analogue of this formula for $n$ successive times in $(0,t)$ that the process $(B_s)_{0<s<t}$ conditioned on $\tau_L=t$ is indeed a Markov process, but an inhomogeneous one, with transition probabilities $(p_{s\mid u}^{(t,L)})_{0<u<s<t}$.
Unfortunately, deducing even only $E_0(B_s\mid\tau_L=t)$ from these is not direct, it seems, although $$E_0(B_s\mid\tau_L=t)=\int_{-\infty}^Lxp_s^{(t,L)}(x)dx=\frac1{q_0(t)}\int_{-\infty}^Lx(p_s(x)-p_s(2L-x))q_{L-x}(t-s)dx$$ or, equivalently, $$E_0(B_s\mid\tau_L=t)=\frac1{q_0(t)}\int_0^\infty(L-x)(p_s(L-x)-p_s(L+x))q_x(t-s)dx$$