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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: $$ \begin{gathered} h_0(t):=1 \quad \text { for } 0 \leq t \leq 1 \\ h_1(t):=\left\{\begin{array}{lr} 1 & \text { for } 0 \leq t \leq \frac{1}{2} \\ -1 & \text { for } \frac{1}{2}<t \leq 1 \end{array}\right. \end{gathered} $$ If $2^n \leq k<2^{n+1}, n=1,2, \ldots$, we set $$ h_k(t):=\left\{\begin{array}{l} 2^{n / 2} \text { for } \frac{k-2^n}{2^n} \leq t \leq \frac{k-2^n+1 / 2}{2^n} \\ -2^{n / 2} \text { for } \frac{k-2^n+1 / 2}{2^n}<t \leq \frac{k-2^n+1}{2^n} \\ 0 \text { otherwise. } \end{array}\right. $$ 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...)

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    $\begingroup$ 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 $\endgroup$
    – Esgeriath
    Apr 10, 2023 at 22:27

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