= Brownian scaling
{c}
{title2=$(B_{ct})_{t\geq0}\overset d=(\sqrt c B_t)_{t\geq0}$}
For every $c>0$, the <stochastic process> $\widetilde B_t=c^{-1/2}B_{ct}$ is standard <Brownian motion>. Its <Brownian increments> remain <independent> and have <variance> $c^{-1}c(t-s)=t-s$; continuity and the zero starting value are preserved. Equivalently,
$$
(B_{ct})_{t\geq0}\overset d=(\sqrt c B_t)_{t\geq0}.
$$
Equality here is of whole-process distributions, including every <finite-dimensional distribution>. It is not an equality of the two expressions along a particular realization. With $c=4$, speeding time up fourfold requires dividing displacement by two to recover the standard law. The figure applies exactly these coordinate changes to four finite samples; the shared colors indicate reused samples, while the dashed lines show pointwise <standard deviations>, not bounds on entire <sample paths>.
\Image[/brownian-paths-scaling.png]
{title=Seeded Gaussian samples before and after Brownian scaling with c equal to four}
The same square-root normalization explains how a <simple random walk> approaches <Brownian motion>. Let $S_k=\sum_{j=1}^k\xi_j$, where the <independent> steps take values $+1$ and $-1$ with equal <probability>. Then $R_n(t)=S_{\lfloor nt\rfloor}/\sqrt n$ has <variance> $\lfloor nt\rfloor/n$. Its <linear interpolation> is
$$
Y_n(t)=\frac{S_k+(nt-k)\xi_{k+1}}{\sqrt n},\qquad k=\lfloor nt\rfloor,
$$
and $\sup_{0\leq t\leq1}|Y_n(t)-R_n(t)|\leq1/\sqrt n$. <Donsker's theorem> says $Y_n\Rightarrow B$ in $C[0,1]$ with the <uniform norm>: this is <convergence in distribution> of random paths. The interpolation bound vanishes, but the theorem does not assert pointwise convergence of an arbitrarily fixed walk to an independently chosen <Brownian motion>. The usefulness is that many-step discrete motion has a universal continuous limit.<Probability: Theory and Examples (fifth edition) by Rick Durrett>[Theorem 8.1.4]{ref}
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