The Central Limit Theorem at Fixed Brownian Times

From a Random Walk to One Brownian Marginal

A symmetric random walk takes steps of size \(1\) or \(-1\) with equal probability. After \(m\) steps, its typical displacement is of order \(\sqrt m\), not \(m\). Diffusive scaling uses this square-root growth for space and scales the number of steps in proportion to elapsed time. At one fixed time, the central limit theorem then identifies the limiting distribution of the displacement. This is a statement about a single marginal distribution; it does not assert convergence of an entire path.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((\xi_k)_{k\ge1}\) be independent random variables with \(\mathbb{P}(\xi_k=1)=\mathbb{P}(\xi_k=-1)=\tfrac12\). Define \(S_0=0\) and \(S_m=\sum_{k=1}^m\xi_k\) for integers \(m\ge1\). For each positive integer \(n\), define the rescaled walk \((X_n(t))_{t\ge0}\), with state space \(\mathbb{R}\) and continuous time index \(t\ge0\), by

Here \(\lfloor nt\rfloor\) is the greatest integer not exceeding \(nt\). Thus at time \(t\), the walk has taken \(m_n=\lfloor nt\rfloor\) steps. For \(t>0\), \(m_n\) grows without bound and \(m_n/n\) approaches \(t\). At \(t=0\), the walk has taken no steps and \(X_n(0)=0\).

The standard iid central limit theorem used here says that if the summands are independent and identically distributed with mean \(0\), variance \(1\), and finite variance, then their sum divided by the square root of the number of summands converges in distribution to a standard normal random variable. This is the classical central limit theorem; its general proof is a standard result of probability theory. The work here is to apply it with the correct number of steps and spatial scaling. Convergence in distribution means convergence of distribution functions at every continuity point of the limiting distribution function.

Proof and Scaling Details

Proof. Fix \(t>0\), and write \(m_n=\lfloor nt\rfloor\). Since \(t>0\), \(m_n\to\infty\), and for all sufficiently large \(n\), \(m_n\ge1\). The increments \(\xi_k\) have mean \(0\), variance \(1\), and are iid. The iid central limit theorem therefore gives \(S_{m_n}/\sqrt{m_n}\) converging in distribution to a standard normal variable. For these \(n\), set \(c_n=\sqrt{m_n/n}\). Then

For completeness, the deterministic scale change preserves the stated limit as follows. Let \(F_n\) be the distribution function of \(S_{m_n}/\sqrt{m_n}\), and let \(\Phi\) be the standard normal distribution function. For any real \(x\), \(c_n>0\) and

Because \(c_n\to\sqrt t>0\), \(x/c_n\to x/\sqrt t\). Given \(\varepsilon>0\), for all sufficiently large \(n\), \(x/\sqrt t-\varepsilon < x/c_n < x/\sqrt t+\varepsilon\). Monotonicity of distribution functions bounds \(F_n(x/c_n)\) between \(F_n(x/\sqrt t-\varepsilon)\) and \(F_n(x/\sqrt t+\varepsilon)\). The central limit theorem gives limits \(\Phi(x/\sqrt t-\varepsilon)\) and \(\Phi(x/\sqrt t+\varepsilon)\) for those bounds. Letting \(\varepsilon\) decrease to zero and using continuity of \(\Phi\), we obtain \(F_n(x/c_n)\to\Phi(x/\sqrt t)\). This is the distribution function of a centered normal variable with variance \(t\). At \(t=0\), \(X_n(0)=S_0/\sqrt n=0\) for every \(n\), so the limit is the constant zero. By the defining marginal law of standard Brownian motion, \(B_t\) is centered normal with variance \(t\) for \(t>0\), and \(B_0=0\). Thus the limiting law agrees with that of \(B_t\). \(\square\)

The distinction between \(m_n\) and \(nt\) is important: the walk takes an integer number of steps, while \(nt\) need not be an integer. The floor causes no change in the limiting variance because \(0\le nt-\lfloor nt\rfloor<1\), and hence \(0\le t-m_n/n<1/n\). The proof does not require the floor to vanish; it requires only that \(m_n/n\to t\) and, for positive \(t\), that \(m_n\to\infty\).

Proof. Put \(m=\lfloor nt\rfloor\). Each increment has \(\mathbb{E}[\xi_k]=\tfrac12(1)+\tfrac12(-1)=0\) and \(\operatorname{Var}(\xi_k)=\mathbb{E}[\xi_k^2]-(\mathbb{E}[\xi_k])^2=1-0=1\). Linearity of expectation gives \(\mathbb{E}[S_m]=0\). Independence makes the covariances between distinct increments zero, so \(\operatorname{Var}(S_m)=\sum_{k=1}^m\operatorname{Var}(\xi_k)=m\), including \(m=0\), when the empty sum is zero. Scaling by \(1/\sqrt n\) gives \(\mathbb{E}[X_n(t)]=0\) and \(\operatorname{Var}(X_n(t))=m/n=\lfloor nt\rfloor/n\). The bound on the floor gives \(\lfloor nt\rfloor/n\to t\). \(\square\)

Worked Fixed-Time Calculations

Why the Hypotheses Matter

The iid assumptions are substantive. For example, let \(\xi_k=Z\) for every \(k\), where \(Z\) is a single random variable with \(\mathbb{P}(Z=1)=\mathbb{P}(Z=-1)=\tfrac12\). Each increment separately is centered with variance \(1\), but the increments are dependent. At \(t=1\), the rescaled sum is \(X_n(1)=\sqrt n\,Z\). For every fixed \(K>0\), \(\mathbb{P}(|X_n(1)|>K)=1\) whenever \(n>K^2\). The distributions therefore do not converge to a finite normal law. Having the right one-step mean and variance does not replace independence.

Centering matters as well. If instead the iid steps have mean \(a\ne0\) and finite variance, then the mean of their uncentered rescaled sum at fixed \(t>0\) is \(a\lfloor nt\rfloor/\sqrt n\), which grows in magnitude like \(|a|t\sqrt n\). This is not the centered Brownian scaling in the theorem. A different centering and, depending on the question, a different scaling would be needed.

Common mistake. A frequent error is to replace \(\sqrt n\) by \(\sqrt{\lfloor nt\rfloor}\) and then report a standard normal limit without accounting for the change of scale. That normalization gives a standard normal limit for \(t>0\), but it is not the process \(X_n(t)=S_{\lfloor nt\rfloor}/\sqrt n\). For the latter, the multiplier is \(\sqrt{\lfloor nt\rfloor/n}\), which tends to \(\sqrt t\); omitting it incorrectly predicts variance \(1\) at every positive time. For example, at \(t=\tfrac12\), the exact variance is \(\lfloor n/2\rfloor/n\to1/2\), not \(1\).

The result is deliberately limited to one fixed time at a time. It identifies the limiting law of \(X_n(t)\) with the Brownian marginal \(B_t\); it does not establish the joint law at multiple times or control fluctuations of the paths between times. Those require additional arguments beyond the ordinary one-dimensional central limit theorem.