Whether there exists a measurable function $f(t,x):[0,2]\times \mathbb R \to \mathbb R$ such that $f(t,B_t)=B_1$ for all $t∈(1,2]$?

Question

Whether there exists a measurable function $f(t, x): [0,2] \times \mathbb R \to \mathbb R$ such that $f(t, B_t)=B_1$ for all $t\in (1,2]$? Here $(B_t)_{t\geqslant 0}$ is the standard Brownian motion with natural filtration $(\mathcal F_t)_{t\geqslant 0}$. I tried by using conditional expectation w.r.t. $\mathcal F_1$ but I cannot show the deterministic function $f(t,x)$. I think that it doesnt exists but I dont know how to prove.

Answer 1

There is no such continuous $f$. This follows from a result of F.B. Knight https://cms.math.ca/10.4153/CMB-1997-008-x , which asserts that if the total variation of $[0,2]\ni t\mapsto f(t,B_t)$ is integrable then $f(t,x)=f(t,0)$ for all $(t,x)$.

A more pedestrian approach dispenses with the continuity. I suppose that if $(B_s)_{s\ge 0}$ is any standard Brownian motion and $1<t\le 2$ is fixed, then $f(t,B_t)=B_1$ a.s. Construct a second Brownian motion $\hat B$ from $B$ as follows: $$ \hat B_s:=\cases{B_t-B_{t-s},&$0\le s\le t$;\cr B_s,&$t\le s\le 2$.\cr} $$ Crucially, $\hat B_t=B_t$. Therefore, almost surely, $$ B_t-B_{t-1}=\hat B_1=f(t,\hat B_t) =f(t,B_t)=B_1, $$ which in turn implies that $$ B_t=B_1+B_{t-1}. $$ But the variance of $B_t$ is $t$ while that of $B_1+B_{t-1}$ is $3t-2$. These can be equal only if $t=1$, in violation of our hypothesis. So there isn't even a measurable such $f$.