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For every $c>0$, the stochastic process ${\tilde{B}}_t=c^{-1/2}{B}_{ct}$ 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,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.

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 isand ${\sup }_{0\le t\le 1}|Y_n(t)-R_n(t)|\le 1/\sqrt{n}$. 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.^[ref]