I am wondering if I can show that
For given $x,y\in \mathbb{R}$ there are two stochastic processes $S_t$ and $B_t$ such that $S_t$ and $B_t$ are two one dimensional Brownian motions starting at $x$ and $y$ respectively and $P\{S_t = B_t\}=1$ for $t>>0$.
I thank in advance for any help!
Thanks to @RaphaelB4, I could figure out the proof even for general $d-$dimensional case. Here is my proof.
Let $B_t^x$ and $B_t^y$ be two independent $1-$dimensional Brownian motions with initial points $x$ and $y$ in $\mathbb{R}$ respectively. Now, let \begin{align*} \tau=\inf\{t\geq 0 : B_t^x=B_t^y\} \end{align*}
then note that $\tau<\infty$ a.s. because $1D$ Brownian motion is recurrent. Now, consider $W_t^y=B_t^y\chi_{[0,\tau ]}+B_t^x\chi_{(\tau,\infty)}$ then $W_t^y$ is a $1-$dimensional Brownian motion with initial point $y$ and with $P\{B_t^x=W_t^y \text{ for all $t>\tau$} \}=1.$
Now, we want to construct $d-$dimensional Brownian motion for arbitrary $d\in \mathbb{Z}^+$. Let $x,y\in \mathbb{R}^d$ be given. Then, by the result above, for each $i\in \{1,\dots, d\}$, if $x_i$ and $y_i$ are $i$th entries of vectors $x$ and $y$, there exist two $1-$dimensional Brownian motions $B_t^{x_i}$ and $B_t^{y_i}$ with initial points $x_i$ and $y_i$ respectively such that $P\{B_t^{x_i}=B_t^{y_i} \text{ for all }t\geq\tau_i \}=1$ where $\tau_i=\inf\{t\geq 0 :B_t^{x_i}=B_t^{y_i}\}$. Now, let $B_t^x$ and $B_t^y$ be $\mathbb{R}^d$ valued function such that \begin{align*} B_t^x=\begin{bmatrix} B_t^{x_1}\\ \vdots\\ B_t^{x_d} \end{bmatrix}\hspace{4cm} B_t^y=\begin{bmatrix} B_t^{y_1}\\ \vdots\\ B_t^{y_d} \end{bmatrix} \end{align*}
Now, let $T=\max_{ i} \tau_i$. Noting that $P\{\tau_i<\infty\}=1$ we know that \begin{align*} P\{T<\infty \}=P\left( \bigcap_i \{\tau_i<\infty\} \right)=1. \end{align*} (https://stats.stackexchange.com/a/100576)
Then $B_t^x$ and $B_t^y$ are two $d-$dimensional Brownian motions such that $P\{B_t^x=B_t^y \text{ for all }t>T\}=1$.
