# Stretching the Brownian Motion in $[0,1]$ to get another Brownian Motion in $[0,t]$

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?

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 $$W_s$$.
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. $$W_t = \sqrt{T} B_{t/T}, \qquad t \in [0,T].$$