Universality for Random Walks with Finite-Variance Increments

From a Particular Walk to a General Limit

The simple symmetric random walk is one route from discrete motion to Brownian motion, but its increments are not essential. The central ingredients are independent, identically distributed increments, zero mean, and finite positive variance. The variance determines the scale of the Brownian motion that appears in the limit: variance-one increments give standard Brownian motion, while increments with variance \(\sigma^2\) give \(\sigma B\).

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(X_1,X_2,\ldots\) be independent and identically distributed real-valued random variables with \(\mathbb{E}X_1=0\) and \(0<\operatorname{Var}(X_1)=\sigma^2<\infty\). Write \(S_0=0\) and \(S_k=\sum_{r=1}^k X_r\) for \(k\ge1\). For \(t\ge0\), define the polygonally interpolated, diffusively scaled process by \[ W_n(t)=\frac{1}{\sqrt n}\left(S_{\lfloor nt\rfloor}+(nt-\lfloor nt\rfloor)X_{\lfloor nt\rfloor+1}\right). \] Thus \(W_n(k/n)=S_k/\sqrt n\) at the grid points, and between them the path is linear. For each finite \(T\), \(W_n\) is a random element of the space \(C[0,T]\) of continuous real-valued functions, equipped with the uniform norm.

The theorem requires identical distribution as well as independence in this form. The mean-zero condition removes linear drift, and finite positive variance supplies the nonzero, finite diffusive scale. Polygonal interpolation specifies convergence of continuous paths, rather than only convergence of values at a finite set of times.

Why the Brownian Scale Is \(\sigma\)

The normalization \(1/\sqrt n\) is fixed by diffusive scaling. It does not, by itself, make the limiting variance equal to one. To obtain standard Brownian motion, divide the increments by their standard deviation \(\sigma\) as well. The following calculation checks the scale directly at the level of covariances.

Proof. It is enough to take \(s\le t\), since covariance is symmetric. Write \(ns=m+a\), where \(m=\lfloor ns\rfloor\) and \(0\le a<1\). In \(W_n(s)\), the coefficient of each of \(X_1,\ldots,X_m\) is \(1/\sqrt n\), the coefficient of \(X_{m+1}\) is \(a/\sqrt n\), and all later coefficients are zero. In \(W_n(t)\), the first \(m\) increments have coefficient \(1/\sqrt n\). The coefficient of \(X_{m+1}\) is \(b/\sqrt n\) for some \(b\in[a,1]\): it is \(1/\sqrt n\) if \(t\ge(m+1)/n\), and otherwise \(b=(nt-m)\). Since distinct increments are independent and centered, their cross-covariances are zero. Therefore \[ \operatorname{Cov}(W_n(s),W_n(t))=\frac{\sigma^2}{n}(m+ab). \] But \(s=(m+a)/n\), so the difference between this covariance and \(\sigma^2s\) is \(\frac{\sigma^2}{n}a(b-1)\), whose absolute value is at most \(\sigma^2/n\). It tends to zero. Since \(s=\min(s,t)\), the claimed limit follows. \(\square\)

This calculation explains the scale but does not prove convergence in distribution: matching covariances alone is not enough to establish a process limit. Donsker’s theorem supplies that stronger conclusion under its hypotheses. For any fixed \(c>0\), the limiting covariance of \(cB\) is \(c^2\min(s,t)\); hence the covariance calculation agrees with the theorem’s limit \(cB\) when \(c=\sigma\).

Worked Applications

Standardization and a Common Error

For general increments, define \(U_k=X_k/\sigma\). Then the \(U_k\) are still iid, \(\mathbb{E}U_k=0\), and \(\operatorname{Var}(U_k)=1\). Their interpolated walk is exactly \(W_n/\sigma\), so applying the variance-one version of Donsker’s theorem to \(U_k\) yields standard Brownian motion. Conversely, multiplying that standardized walk by \(\sigma\) returns the unstandardized walk and the limit \(\sigma B\). This is why it is important to distinguish the two normalizations.

Common mistake. In the Bernoulli example, one might incorrectly use \(\mathbb{E}Z_k^2=1/3\) as the variance of the centered increment. That calculation ignores the centering. The correct computation is \[ \operatorname{Var}(Z_k-\tfrac13)=\mathbb{E}(Z_k-\tfrac13)^2 =\tfrac13(\tfrac23)^2+\tfrac23(-\tfrac13)^2 =\tfrac4{27}+\tfrac2{27}=\tfrac29. \] Using \(1/3\) would incorrectly predict Brownian scale \(\sqrt{1/3}\) instead of \(\sqrt{2}/3\). The correction is to calculate the variance of the actual centered increments, not the uncentered second moment of the original Bernoulli variables.

Why the Hypotheses Matter

The positive-variance condition rules out a degenerate limit. If the increments are iid, centered, and have variance zero, then \(\mathbb{E}X_1^2=0\), which implies \(X_1=0\) almost surely. Every partial sum is then zero almost surely, so the scaled paths converge to the zero function, not to Brownian motion.

Finite variance is also essential for this Brownian scaling conclusion. For example, let the increments be iid symmetric stable random variables with characteristic function \(\mathbb{E}e^{iuX_k}=e^{-|u|^{3/2}}\). This standard distribution has mean zero and infinite variance. At time \(1\), the characteristic function of \(S_n/\sqrt n\) is \[ \mathbb{E}e^{iuS_n/\sqrt n} =\left(e^{-|u/\sqrt n|^{3/2}}\right)^n =e^{-n^{1/4}|u|^{3/2}}. \] For every \(u\ne0\), this tends to zero, while at \(u=0\) it equals one. The pointwise limit is discontinuous at zero, so it cannot be the characteristic function of a probability distribution; the sequence cannot converge in distribution to a Brownian time-one value. Thus a centered independent-increment model alone does not guarantee the Brownian limit under \(1/\sqrt n\) scaling.

The practical sequence is: center the increments, verify that they are iid, compute their variance, and then choose the scale. If that variance is finite and positive, dividing the partial sums by \(\sigma\sqrt n\) gives standard Brownian motion in the limit; dividing only by \(\sqrt n\) retains the factor \(\sigma\). A limiting distribution is not a stationary distribution or a statement about the exact finite-\(n\) law: it describes convergence in distribution of the rescaled paths.