Can you construct a baby brownian motion from a brownian motion?

Question

By a baby brownian motion, I mean a brownian motion $X_t$, $t \in [0,1]$. We can construct a brownian motion on $[0,\infty)$ as \begin{align} Y_t = (1+t)X_\frac{t}{1+t} - t X_1, t \in [0, \infty) \end{align} But can we also construct a baby brownian motion from a brownian motion?

Answer 1

I think that the question is interesting and not just a matter of restricting the given BM $Y$ to $[0,1]\,.$ There is rather a relation to the concept of Brownian bridge.

1. Before dealing with this I'd like to verify first that given the "baby" BM $X$ on $[0,1]$ the process $Y_t$ defined by

$$\tag{1} Y_t=(1+t)X_{\textstyle \frac{t}{1+t}}-tX_1 $$ is a BM on $[0,\infty)$ because this was not so obvious to me first. (Clearly, $X$ is the restriction to $[0,1]$ of some BM on $[0,\infty)$ but that is not the point at all.)

From the known property $$ \mathbb E[X_tX_s]=\min(t,s) $$ of Brownian motions it follows easily that $$ \mathbb E[Y_tY_s]=\min(t,s) $$ holds. It is then relatively straightforward to see that $Y$ is a Gaussian process with Gaussian and independent increments that have variance $\mathbb E[(Y_t-Y_s)^2]=t-s\,.$ Therefore $Y$ is a BM.

2. Relation with Brownian bridge. Using the variable transformation $$ s=\frac{t}{1+t}\,,\quad t=\frac{s}{1-s} $$ one can write (1) as \begin{align} Y_{\textstyle \frac{s}{1-s}}=\frac{X_s-sX_1}{1-s} \end{align} so that $$ B_s=(1-s)Y_{\textstyle \frac{s}{1-s}}=X_s-sX_1 $$ is a Brownian bridge on [0,1]. That is, a Gaussian process with expectation zero, $B_0=B_1=0$ almost surely, and $$ E[B_tB_s]=\min(t,s)-ts\,. $$

3. Construction of $X$ from $Y$. Let's now forget how the BM $Y_t$ was constructed from $X\,.$ The question is: given a BM $Y_t$ on $[0,\infty)$ does there exist a BM $X_t$ on $[0,1]$ such that (1) holds for all $t\ge 0\,$? The answer is yes: in the link it is mentioned that $$ B_s=(1-s)Y_{\textstyle\frac{s}{1-s}}\,,\quad s\in[0,1]\,. $$ is a Brownian bridge. Next, with a Gaussian r.v. $Z$ that is independent of $B$ we obtain a BM $X$ on $[0,1]$ by $$ X_s=B_s+sZ $$ which will satisfy $$ B_s=X_s-sX_1\,. $$ Thus $$ (1-s)Y_{\textstyle\frac{s}{1-s}}=X_s-sX_1\,. $$ Transforming $s$ back to $t$ yields (1).