Math-1692 · mathematics · Probability

Math-1692

Concatenated bounded Brownian pieces

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Statement

Definitions and mathematical statement

For each $k\in\mathbb Z$, let $B^k$ be a real standard Brownian motion started at zero and let $T_k$ be a nonnegative finite stopping time for its filtration. Suppose the stopped pairs $(T_k,(B^k_t)_{0\le t\le T_k})$ are independent across $k$, and that almost surely both $\sum_{k\ge1}T_k$ and $\sum_{k\le-1}T_k$ diverge.

Define seam times by $S_0=0$ and $S_{k+1}-S_k=T_k$. Concatenate the stopped pieces to obtain the continuous process $X$ on $\mathbb R$ determined by $X_0=0$ and \[ X_{S_k+t}-X_{S_k}=B^k_t\qquad(0\le t<T_k). \] The divergence assumptions make the concatenation cover both time directions. Zero-length pieces may be ignored.

Assume there is a deterministic constant $c<\infty$ such that $T_k<c$ almost surely for every $k$. Must $X$ be two-sided Brownian motion in the following precise sense used by the source: does there exist a possibly random time $S\in\mathbb R$ such that the two processes \[ (X_{S+t}-X_S)_{t\ge0},\qquad(X_{S-t}-X_S)_{t\ge0} \] are independent standard Brownian motions?

The anchor $S$ is not required to be the distinguished seam at zero or a stopping time. The stopped pieces need not have the same distribution. The hypothesis is an almost-sure common bound on their durations, rather than a bound on moments or merely finiteness of each duration.

Short English statement

Does concatenating independent Brownian pieces of uniformly bounded stopping-time lengths produce two-sided Brownian motion after a suitable random time shift?

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