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I'm running a simulation of a Standard Brownian Motion by limit of a Symmetric Standard Random Walk $\{S_n ,n\geq 1\}$ and $S_n=\sum_{k=1}^n X_k$, where $$P(X_k =-1)=P(X_k =1)=\frac{1}{2},$$ and interpolate linearly between integer points in this way: $$S(t)=S_{[t]}+(t-[t])(S_{[t]+1}-S_{[t]})$$ using the Donsker's Invariance Principle: $$Z_n(t)=\frac{S(nt)}{\sqrt{n}},$$ then $\{ Z_n : n\geq1\}$ converges in distribution to a standard brownian motion $\{B(t):t\in[0,1]\}$.

This is working fine for $t\in[0,1]$ but, what if I want Brownian Motion in $[0,10]$, for example $\{W_t :t\in[0,10]\}$?

I'm thinking about generating a vector $[B_0,B_{1/n}, B_{2/n},\dots ,B_{n-1/n},B_{1}]$ that contains the position of BM in some points of [0,1] and just assignate $W_0=B_0,W_{10/n}=B_{1/n},\dots,W_{10}=B_1$. My question: Is this correct? Theroretically, this is also a BM? It's some kind of stretched BM in $[0,1]$ to get a BM in $[0,10]$. If is not, then do you have any ideas of how could I get a SBM in $[0,10]$ using the same Random Walk?

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If you take a brownian motion $(B_t)_{t\geq 0}$ and $\lambda > 0$, then the process $(W_t)_{t\geq 0}$ defined by $W_t := \sqrt{\lambda} B_{\frac 1 {\lambda} t} $ is again a Brownian Motion.

So, lets suppose you want to sample $W_s$ where $s\in [0,10]$. What can you do? You said you can sample from $(B_t)_{t\in [0,1]}$. Therefore you could sample $B_{\frac s {10}}$ and scale it by ${\sqrt{10}}$. Thus you have a sample of $W_s$.

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Yes, it is possible to strech a Brownian motion. There is the following result (which is not difficult to prove)

Proposition Let $(B_t)_{t \geq 0}$ be a one-dimensional Brownian motion. Then for any $a>0$ the process defined by $$W_t := \frac{1}{\sqrt{a}} B_{at}, \qquad t \geq 0, $$ is a Brownian motion.

If you have a Brownian motion $(B_t)_{t \in [0,1]}$ then you can use this result to strech the Brownian motion to a larger time interval $[0,T]$ by choosing small $a:=1/T$, i.e. $$W_t = \sqrt{T} B_{t/T}, \qquad t \in [0,T].$$

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