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I got a question about the construction of Brownian motions. Just to be all on the same page, for me a Brownian motion on a certain probability space $(\Omega, \mathcal{A}, \mathbb{P})$ is a stochastic process $(B_t)_{t \in [0,\infty)}$, i.e. each $B_t$ is a real valued random variable on $\Omega$ such that

(i) $B_0 = 0$ a.s.

(ii) $B(\omega) : [0,\infty) \to \mathbb{R}$ is continuous for almost all $\omega \in \Omega$

(iii) The increments are independent, i.e. let $0\leq t_1< t_2 <...<t_n$, then $B_{t_n} - B_{t_{n-1}},\dots,B_{t_2}-B_{t_1}$ are independent

(iv) For all $0 \leq s < t$ we have $B_t - B_s \sim \mathcal{N}(0,t-s)$.

Now my question is: If I have a sequence of real valued random variables, say $(A_n)_{n \in \mathbb{N}_0}$, which satisfy (i) as well as (iii) and (iv) for all $t,s \in \mathbb{N}_0$, would it be possible to construct a Brownian motion $B$ such that

$B_n = A_n$ for all $n \in \mathbb{N}_0$.

My idea would be to interpolate linearly between $A_n$ and $A_{n+1}$ to obtain values for $B_t$ with $t \in (n, n+1)$. I.e.

$B_t = (1 - (\lceil t \rceil - t)) B_{\lfloor t \rfloor} + (\lceil t \rceil - t) B_{\lceil t \rceil}$

if we consider $\lceil n \rceil = n = \lfloor n \rfloor$ for integer values. Does that work or am I missing something here?

student7481
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    That doesn't work. Getting a process with continuous paths is much more involved. – Kavi Rama Murthy Apr 03 '22 at 12:28
  • Do you mean, that given such an $A_n$ I won't find such a Brownian motion $B_n$? Or do you mean, that my construction ain't working, but there is another way to get such a process $B_n$? – student7481 Apr 03 '22 at 13:11
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    Your construction will meet the requirements of a BM for $t\in\mathbb N_0$. Now take $s,t\in (n,n+1)$ with $s<t$. Is the increment $B_t-B_s$ independent of $B_s$ ? – Kurt G. Apr 03 '22 at 14:02
  • @KurtG. In that case $B_t - B_s$ and $B_s$ will both depend on $B_n$ and $B_{n+1}$ and thus (iv) won't be satisfied. Okay, so I was indeed missing something. Now I'm still wondering if such a $B_n$ exists. – student7481 Apr 03 '22 at 15:34
  • Such a $B_n$ ( $t=n$ exclamation mark ) *does* exist. Hint : draw i.i.d. standard normals $X_i$ and sum them up. – Kurt G. Apr 03 '22 at 15:37
  • @KurtG. I'm still a little bit confused; I think I have not expressed myself so clearly. My sequence $A_n$ meets the requirements for a Brownian motion for $t \in \mathbb{N}0$. Now I'm wondering if we could find a continuation $(A_t){t \geq 0}$ of the sequence $(A_n){n \in \mathbb{N}}$ such that $(A_t){t \geq 0}$ is a Brownian motion. – student7481 May 01 '22 at 20:06
  • Are you going in circles ? You can find $A_n$ or $B_n$ that satisfy the requirements of a BM at $t\in\mathbb N$ . The linear interpolation cannot be a BM. Discussed amply above. I thought you had agreed on that. Now you are wondering if we could find a continuation of $A_n$ . I don't think so when randomness only occurs at $t\in\mathbb N$. – Kurt G. May 02 '22 at 06:32

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