The Lévy–Ciesielski construction of a Brownian motion for $t\in \mathbb R$ (and not only $t\in[0,1]$)

The Lévy–Ciesielski construction of a Brownian motion is based on $$W(t):=\sum_{k=0}^{\infty} A_k s_k(t) \quad \quad \quad\quad \quad \quad\quad \quad \quad (1)$$ for times $$0 \leq t \leq 1$$, where the coefficients $$\left\{A_k\right\}_{k=0}^{\infty}$$ are independent, $$N(0,1)$$ random variables defined on some probability space. (see "Evans, Lawrence C. An introduction to stochastic differential equations. Vol. 82. American Mathematical Soc., 2012.")

The function $$s_k$$ is the $$k$$-th Schauder function: $$s_k(t):=\int_0^t h_k(s) d s \quad(0 \leq t \leq 1)$$ where $$k=0,1,2, \ldots$$ and the family $$\left\{h_k(\cdot)\right\}_{k=0}^{\infty}$$ of Haar functions are defined for $$0 \leq t \leq 1$$ as follows: $$\begin{gathered} h_0(t):=1 \quad \text { for } 0 \leq t \leq 1 \\ h_1(t):=\left\{\begin{array}{lr} 1 & \text { for } 0 \leq t \leq \frac{1}{2} \\ -1 & \text { for } \frac{1}{2} If $$2^n \leq k<2^{n+1}, n=1,2, \ldots$$, we set $$h_k(t):=\left\{\begin{array}{l} 2^{n / 2} \text { for } \frac{k-2^n}{2^n} \leq t \leq \frac{k-2^n+1 / 2}{2^n} \\ -2^{n / 2} \text { for } \frac{k-2^n+1 / 2}{2^n} In Evan's book, I read that: $$W(t, \omega):=\sum_{k=0}^{\infty} A_k(\omega) s_k(t) \quad(0 \leq t \leq 1) \quad \quad \quad \quad \quad (2)$$ defines a Brownian motion for $$0 \leq t \leq 1$$.

Probably my question is very naive, but I would like to know if there exists for $$W(\cdot)$$ and $$t\in\mathbb R$$ an expression in infinite sum using the Haar system like the (1) and (2). For example, if we let the indices $$k$$ and $$n$$ vary over the whole $$\mathbb Z$$, can we get an expression of the Brownian motion over $$\mathbb R$$? (For the latter, I think not because otherwise, I would have found it written in some book...)

• Shilling proposes "glueing" independent chunks of Brownian motions on $[0,1]$ to obtain Brownian motion on $[0, \infty)$, which he does not write down explicitly iirc. This seems to be some technical step that is not particularly enlightening, perhaps that's why it is skipped. I don't think that just letting $k$ and $n$ be any integers would work just out of the box, but I don't remember the details of construction rn to check it out Apr 10, 2023 at 22:27