Diffusive Scaling of Symmetric Random Walks

Why Scale Time and Space Together?

A simple symmetric random walk takes steps of size $1$ in either direction. After $m$ steps its variance is $m$, so its typical displacement has size comparable to $\sqrt{m}$. To observe the walk over a time interval of order $1$, speed up its clock by a factor of $n$; to keep its displacement on an order-one scale, shrink each spatial step by a factor of $\sqrt{n}$. This is called diffusive scaling.

The scaling makes the walk's mean and covariance comparable to those of standard Brownian motion. It does not, by itself, say that the walk is Gaussian or establish convergence of its distributions or paths. The calculations here are exact moment calculations for each fixed scaling parameter $n$, with attention to the integer step counts that underlie the rescaled process.

Definition and Moment Formulas

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(\varepsilon_k)_{k\ge1}$ be independent random variables with $\mathbb{P}(\varepsilon_k=1)=\mathbb{P}(\varepsilon_k=-1)=\tfrac12$. Define $S_0=0$ and $S_m=\sum_{k=1}^m\varepsilon_k$ for integers $m\ge1$. For each positive integer $n$, use the filtration $\mathcal{F}_m=\sigma(\varepsilon_1,\ldots,\varepsilon_m)$, with $\mathcal{F}_0=\{\varnothing,\Omega\}$, and define the rescaled process by

W^{(n)}(t)=\frac{1}{\sqrt{n}}S_{\lfloor nt\rfloor},\qquad t\ge0.

Here $\lfloor x\rfloor$ is the greatest integer no larger than $x$. For fixed $n$, time is continuous, the sample paths are constant between successive times $k/n$, and the state space is the countable set $S_n=n^{-1/2}\mathbb{Z}$. Its time-$t$ information is $\mathcal{F}^{(n)}_t=\mathcal{F}_{\lfloor nt\rfloor}$. The factor $1/\sqrt n$ is applied to the sum of the steps, not separately to its variance.

Proposition: For every positive integer $n$ and all $s,t\ge0$, $$ \mathbb{E}[W^{(n)}(t)]=0,\qquad \operatorname{Var}(W^{(n)}(t))=\frac{\lfloor nt\rfloor}{n},\qquad \operatorname{Cov}(W^{(n)}(s),W^{(n)}(t))=\frac{\lfloor n\min(s,t)\rfloor}{n}. $$

Proof. For every $k$, symmetry gives $\mathbb{E}[\varepsilon_k]=0$, and $\varepsilon_k^2=1$ gives $\operatorname{Var}(\varepsilon_k)=1$. If $k\ne \ell$, independence gives $\mathbb{E}[\varepsilon_k\varepsilon_\ell]=\mathbb{E}[\varepsilon_k]\mathbb{E}[\varepsilon_\ell]=0$, so $\operatorname{Cov}(\varepsilon_k,\varepsilon_\ell)=0$. Set $a=\lfloor ns\rfloor$ and $b=\lfloor nt\rfloor$. Then $S_a$ and $S_b$ are sums of $a$ and $b$ centered steps, respectively. Their expectations are zero. Their variances are $a$ and $b$, because the individual variances add and all distinct-step covariances vanish. Their covariance is the sum of the covariances between steps in the two sums: a step contributes $1$ if it occurs in both sums and $0$ otherwise. Thus $\operatorname{Cov}(S_a,S_b)=\min(a,b)$. Since $x\mapsto\lfloor x\rfloor$ is nondecreasing, $\min(a,b)=\lfloor n\min(s,t)\rfloor$. Multiplying each sum by $1/\sqrt n$ multiplies its variance and covariance by $1/n$, proving all three formulas. $\square$

The proposition gives a useful way to read a covariance matrix at fixed times: its $(i,j)$ entry counts how many random-walk steps have contributed to both observations, divided by $n$. In particular, the variance at time $t$ is the number of steps taken by that time divided by $n$.

Consequence: For all $s,t\ge0$, $$ 0\le \min(s,t)-\operatorname{Cov}(W^{(n)}(s),W^{(n)}(t))<\frac{1}{n}. $$

Proof. Put $x=n\min(s,t)$. The defining property of the floor gives $0\le x-\lfloor x\rfloor<1$. Dividing by $n>0$ gives the stated bound after using the covariance formula. $\square$

Thus the covariance is within $1/n$ of $\min(s,t)$ at every pair of fixed times. This is an exact numerical bound on covariance, not a claim that the process is Brownian motion. A covariance calculation alone does not establish that a process has Gaussian finite-dimensional distributions or that its paths converge.

Worked Covariance Calculations

Worked Example: Three Observation Times with n Equal to 12

Take $n=12$ and observe the walk at $t_1=\tfrac14$, $t_2=\tfrac23$, and $t_3=1$. First calculate the numbers of steps: $\lfloor12\cdot\tfrac14\rfloor=3$, $\lfloor12\cdot\tfrac23\rfloor=8$, and $\lfloor12\cdot1\rfloor=12$. Therefore all three means are zero, and the covariance between two observations is the smaller of their step counts divided by $12$. The full covariance matrix is

\begin{aligned} \operatorname{Cov}\bigl((W^{(12)}(t_1),W^{(12)}(t_2),W^{(12)}(t_3))\bigr) &= \begin{bmatrix} 3/12&3/12&3/12\\ 3/12&8/12&8/12\\ 3/12&8/12&12/12 \end{bmatrix}\\ &= \begin{bmatrix} \tfrac14&\tfrac14&\tfrac14\\ \tfrac14&\tfrac23&\tfrac23\\ \tfrac14&\tfrac23&1 \end{bmatrix}. \end{aligned}

For example, the covariance of the first and third observations is $3/12=\tfrac14$, because the first three steps occur in both sums. The diagonal entries are the variances: $3/12=\tfrac14$, $8/12=\tfrac23$, and $12/12=1$. These agree with the formula $\operatorname{Var}(W^{(12)}(t_i))=\lfloor12t_i\rfloor/12$, providing a check on the matrix.

Worked Example: Covariance of Two Overlapping Increments

Take $n=5$, and compare the increments from time $0$ to $\tfrac12$ and from time $\tfrac15$ to $\tfrac45$. Since $\lfloor5\cdot\tfrac12\rfloor=2$, $\lfloor5\cdot\tfrac15\rfloor=1$, and $\lfloor5\cdot\tfrac45\rfloor=4$, the increments are

W^{(5)}(\tfrac12)-W^{(5)}(0)=\frac{\varepsilon_1+\varepsilon_2}{\sqrt5}, \qquad W^{(5)}(\tfrac45)-W^{(5)}(\tfrac15)=\frac{\varepsilon_2+\varepsilon_3+\varepsilon_4}{\sqrt5}.

Only $\varepsilon_2$ appears in both increments. Each increment has mean zero. In the variance of the first, the two step variances sum to $1+1=2$, and division by $5$ gives $2/5$. In the variance of the second, the three step variances sum to $1+1+1=3$, and division by $5$ gives $3/5$. In their covariance, every product of distinct steps has expectation zero, while $\mathbb{E}[\varepsilon_2^2]=1$, so the covariance is $1/5$.

Check: The covariance formula for the increments gives the same answer: their shared step-count interval is from step $2$ through step $2$, which has one step, so the covariance is $1/5$. The positive covariance is due to that shared step; the increments are not independent.

Worked Example: Adjacent Increments with n Equal to 10

Take $n=10$ and the three times $t_1=\tfrac3{10}$, $t_2=\tfrac7{10}$, and $t_3=\tfrac9{10}$. The step counts are $\lfloor10t_1\rfloor=3$, $\lfloor10t_2\rfloor=7$, and $\lfloor10t_3\rfloor=9$. The increments over the first and second time intervals are therefore

W^{(10)}(\tfrac7{10})-W^{(10)}(\tfrac3{10}) =\frac{\varepsilon_4+\varepsilon_5+\varepsilon_6+\varepsilon_7}{\sqrt{10}}, \qquad W^{(10)}(\tfrac9{10})-W^{(10)}(\tfrac7{10}) =\frac{\varepsilon_8+\varepsilon_9}{\sqrt{10}}.

The first increment has four terms, so its mean is $0$ and its variance is $4/10=2/5$. The second has two terms, so its mean is $0$ and its variance is $2/10=1/5$. No step occurs in both sums. Every cross-product in their covariance therefore involves distinct independent centered variables and has expectation zero, giving covariance $0$.

Check: The first increment uses steps $4,5,6,7$, while the second uses steps $8,9$; the sets have empty intersection. Their covariance is consequently $0$, as computed directly. The endpoint count also checks: the two increments together contain six steps, and $9-3=6$.

Why the Floor Matters

The rescaled process is a step function, not a process that takes a new random-walk step at every real-valued time. In general, its variance at time $t$ is $\lfloor nt\rfloor/n$, not exactly $t$. If $nt$ is an integer, these values agree; otherwise the floor must be retained. The difference is always nonnegative and strictly smaller than $1/n$.

For increments, the same counting principle is useful. If $0\le s\le t$, then

W^{(n)}(t)-W^{(n)}(s) =\frac{1}{\sqrt n}\sum_{k=\lfloor ns\rfloor+1}^{\lfloor nt\rfloor}\varepsilon_k, \qquad \operatorname{Var}(W^{(n)}(t)-W^{(n)}(s)) =\frac{\lfloor nt\rfloor-\lfloor ns\rfloor}{n}.

If the upper step count equals the lower step count, the sum has no terms and the increment is zero. Otherwise it contains exactly $\lfloor nt\rfloor-\lfloor ns\rfloor$ independent centered steps, which proves the variance formula by adding their variances. Two increments have covariance equal to the number of step indices in both sums divided by $n$. Disjoint sets of step indices give covariance zero; overlapping sets give positive covariance.

Common mistake. At $n=5$ and $t=\tfrac12$, one might write $\operatorname{Var}(W^{(5)}(\tfrac12))=\tfrac12$ by replacing $\lfloor nt\rfloor/n$ with $t$. The actual step count is $\lfloor5\cdot\tfrac12\rfloor=\lfloor\tfrac52\rfloor=2$, so the variance is $2/5$. Substitution into the defining formula confirms it: $W^{(5)}(\tfrac12)=(\varepsilon_1+\varepsilon_2)/\sqrt5$, whose variance is $(1+1)/5=2/5$. The unrounded expression gives the Brownian covariance value at that time, not the exact variance of this finite-$n$ walk.

Diffusive scaling is useful because it puts the walk's time and displacement on scales where its moments resemble those of Brownian motion. The calculations also show precisely what is being compared: an observation at time $t$ uses $\lfloor nt\rfloor$ steps, and its covariance with another observation is determined by their shared steps. Establishing more than these moment identities requires additional arguments; the identities themselves do not say that the rescaled walk has a Gaussian law.

Check Your Understanding

For each question, identify the relevant integer step counts before calculating.

For $n=8$, find the mean and variance of $W^{(8)}(\tfrac38)$. Verify the variance by writing the random-walk sum explicitly.
For $n=6$, calculate $\operatorname{Cov}(W^{(6)}(\tfrac13),W^{(6)}(\tfrac56))$. Which steps contribute to both observations?
For $n=5$, write $W^{(5)}(\tfrac45)-W^{(5)}(\tfrac15)$ as a sum of scaled steps and find its variance.
For $n=10$, compare the variance at $t=\tfrac14$ with $t$. What is the difference, and does it satisfy the strict upper bound $1/n$?
Explain why two increments with disjoint step-index sets have covariance zero, and why overlapping step-index sets need not have covariance zero.

Diffusive Scaling of Symmetric Random Walks

What You'll Learn