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Brownian Motion: Definition and Construction · Tutorial 812 of 994

Joint Convergence of Scaled Walks on Finite Time Grids

Learn how disjoint walk-increment blocks yield joint Gaussian limits on finite time grids, and how their cumulative sums produce the Brownian covariance matrix.

Advanced 10 min read

What You'll Learn

  • Define a diffusively scaled simple symmetric walk at arbitrary real times using integer-part rounding.
  • Partition its increments into disjoint blocks determined by an ordered time grid.
  • Verify the multivariate central limit theorem conditions for the block vector.
  • Convert independent limiting increments into cumulative Brownian values.
  • Compute and check finite-grid covariance matrices, including when times do not align with the step grid.
  • Recognize why convergence on finite grids does not by itself establish convergence of entire paths.

From Disjoint Blocks to Joint Limits

A central limit theorem at one time describes one scaled walk value. A finite time grid asks for more: the values at several times are dependent, because each later walk value includes the increments that formed the earlier ones. The useful reorganization is to divide the walk into disjoint blocks between successive grid times. The block sums have a joint Gaussian limit with independent coordinates; adding those coordinates back together gives the joint limit of the walk values.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space carrying independent random variables \((\xi_k)_{k\ge1}\) with \(\mathbb{P}(\xi_k=1)=\mathbb{P}(\xi_k=-1)=\tfrac12\). Set \(S_0=0\) and \(S_r=\sum_{k=1}^r\xi_k\) for integers \(r\ge1\). For each positive integer \(n\), define the scaled walk at real time \(t\ge0\) by \(W^{(n)}(t)=n^{-1/2}S_{\lfloor nt\rfloor}\). The integer part matters: at time \(t\), the walk has made exactly \(\lfloor nt\rfloor\) steps.

Theorem (Joint convergence on a finite time grid): Fix \(m\ge1\) and \(0<t_1<\cdots<t_m<\infty\). As \(n\to\infty\), the vector \((W^{(n)}(t_1),\ldots,W^{(n)}(t_m))\) converges in distribution to a mean-zero Gaussian vector with covariance matrix \(\Sigma\) whose entries are \(\Sigma_{ab}=\min(t_a,t_b)\). Equivalently, its successive increments are independent centered normal variables with variances \(t_1,t_2-t_1,\ldots,t_m-t_{m-1}\).

We use the multivariate Lindeberg–Feller central limit theorem as a standard result; its proof is outside this tutorial. In the form needed here, if a row consists of independent, mean-zero random vectors, the sum of their covariance matrices converges to a matrix \(\Gamma\), and the Lindeberg condition holds in every fixed direction, then the row sums converge in distribution to a centered Gaussian vector with covariance \(\Gamma\). The Lindeberg condition requires that, for every \(\varepsilon>0\) and fixed vector \(a\), the sum of the expectations of \((a^\mathsf{T}Y)^2\mathbf{1}_{\{|a^\mathsf{T}Y|>\varepsilon\}}\) over the row tends to zero. We verify these conditions for the disjoint blocks rather than treating cumulative walk values as independent.

Proof. Put \(r_{n,0}=0\), \(r_{n,\ell}=\lfloor nt_\ell\rfloor\) for \(1\le\ell\le m\), and \(b_{n,\ell}=r_{n,\ell}-r_{n,\ell-1}\). For each fixed \(\ell\), \(b_{n,\ell}/n\to\Delta_\ell\), where \(\Delta_1=t_1\) and \(\Delta_\ell=t_\ell-t_{\ell-1}\) for \(\ell\ge2\). Every \(\Delta_\ell\) is positive, so for all sufficiently large \(n\), every block has positive length.

For \(1\le k\le r_{n,m}\), let \(Y_{n,k}\) be the \(m\)-dimensional vector with coordinate \(\xi_k/\sqrt n\) in the unique block \(\ell\) satisfying \(r_{n,\ell-1}<k\le r_{n,\ell}\), and zero in all other coordinates. For fixed \(n\), these vectors are independent and mean zero. Their row sum \(Z_n=\sum_{k=1}^{r_{n,m}}Y_{n,k}\) has coordinate \(\ell\) equal to the scaled sum of the increments in block \(\ell\). Because the blocks use disjoint indices, the total covariance matrix is diagonal: its \(\ell\)th diagonal entry is \(b_{n,\ell}/n\), and every off-diagonal entry is zero. Thus this matrix converges to \(\operatorname{diag}(\Delta_1,\ldots,\Delta_m)\).

To check Lindeberg, fix \(a=(a_1,\ldots,a_m)^\mathsf{T}\) and \(\varepsilon>0\). Each \(Y_{n,k}\) belongs to one block, so \(|a^\mathsf{T}Y_{n,k}|\le \max_\ell|a_\ell|/\sqrt n\), since \(|\xi_k|=1\). For all sufficiently large \(n\), this bound is at most \(\varepsilon\). Every indicator in the Lindeberg sum is then zero, so the condition holds. The multivariate central limit theorem gives \(Z_n\Rightarrow Z\), where \(Z\) is centered Gaussian with independent coordinates and variances \(\Delta_1,\ldots,\Delta_m\).

Finally, \(W^{(n)}(t_j)=\sum_{\ell=1}^j Z_{n,\ell}\) exactly: the first \(j\) blocks contain precisely the increments numbered \(1\) through \(r_{n,j}\). Applying the continuous mapping theorem to this fixed linear summation map gives convergence of the cumulative vector. Its \(a\)th and \(b\)th coordinates, for \(a\le b\), share the first \(a\) limiting increments. Their covariance is therefore \(\sum_{\ell=1}^a\Delta_\ell=t_a=\min(t_a,t_b)\). Symmetry gives the same formula when \(b<a\). This proves the claimed joint limit and its covariance. \(\square\)

Reading the Covariance from the Blocks

The limiting block vector has independent coordinates. The limiting walk values do not: the value at a later time contains all the earlier blocks. More generally, if \(G_\ell\) denotes the limiting increment over block \(\ell\), then the value at grid time \(t_j\) is \(G_1+\cdots+G_j\). Two such sums share exactly the blocks up to the earlier time. This is why the covariance is the earlier time, rather than zero.

The theorem identifies the joint law at any fixed finite grid with the corresponding Brownian motion values. This is a statement about vectors of values, not a statement that entire random paths converge. To obtain path convergence one must also control behavior between grid points; finite-grid convergence alone does not provide that control.

Worked Example: Two Times at One-Quarter and Three-Quarters

Take \(t_1=\tfrac14\) and \(t_2=\tfrac34\). For \(n\) divisible by \(4\), the first block has \(n/4\) steps and the second has \(3n/4-n/4=n/2\) steps. The scaled block variances are \((n/4)/n=\tfrac14\) and \((n/2)/n=\tfrac12\). The block limit is therefore a pair of independent centered normals with these variances. Summing the blocks gives the limiting grid values, whose covariance matrix is

$$ \begin{pmatrix} \frac14 & \frac14\\[2pt] \frac14 & \frac14+\frac12 \end{pmatrix} = \begin{pmatrix} \frac14 & \frac14\\[2pt] \frac14 & \frac34 \end{pmatrix}. $$

Check: the first value has variance \(\tfrac14\); the second has variance \(\tfrac14+\tfrac12=\tfrac34\); and their covariance is the variance of their shared first block, \(\tfrac14\). These entries agree with \(\min(t_a,t_b)\) for each pair of grid times.

Worked Example: Three Equally Spaced Times

Take \(t_1=\tfrac12\), \(t_2=1\), and \(t_3=\tfrac32\). For even \(n\), the three block lengths are \(n/2\), \(n-n/2=n/2\), and \(3n/2-n=n/2\). Dividing each length by \(n\), each limiting block variance is \(\tfrac12\). The limiting values are \(G_1\), \(G_1+G_2\), and \(G_1+G_2+G_3\), with independent blocks of variance \(\tfrac12\). Their covariance matrix is

$$ \begin{pmatrix} \frac12 & \frac12 & \frac12\\ \frac12 & \frac12+\frac12 & \frac12+\frac12\\ \frac12 & \frac12+\frac12 & \frac12+\frac12+\frac12 \end{pmatrix} = \begin{pmatrix} \frac12 & \frac12 & \frac12\\ \frac12 & 1 & 1\\ \frac12 & 1 & \frac32 \end{pmatrix}. $$

Check: each diagonal entry is the sum of the variances of the blocks in that value: \(\tfrac12\), \(1\), and \(\tfrac32\). For the second and third values, both include the first two blocks, so their covariance is \(\tfrac12+\tfrac12=1\). Every matrix entry is \(\min(t_a,t_b)\).

Worked Example: A Grid That Does Not Align with the Step Count

Consider \(t_1=\tfrac13\), \(t_2=\tfrac23\), and \(t_3=1\), and take \(n=10\) to see the integer-part effect. The step counts are \(\lfloor10/3\rfloor=3\), \(\lfloor20/3\rfloor=6\), and \(\lfloor10\rfloor=10\). Thus the three disjoint block lengths are \(3-0=3\), \(6-3=3\), and \(10-6=4\). The scaled block variances are \(\tfrac3{10},\tfrac3{10},\tfrac4{10}=\tfrac25\). At this finite \(n\), the covariance matrix of the scaled walk values is

$$ \begin{pmatrix} \frac3{10} & \frac3{10} & \frac3{10}\\ \frac3{10} & \frac6{10} & \frac6{10}\\ \frac3{10} & \frac6{10} & \frac{10}{10} \end{pmatrix} = \begin{pmatrix} \frac3{10} & \frac3{10} & \frac3{10}\\ \frac3{10} & \frac35 & \frac35\\ \frac3{10} & \frac35 & 1 \end{pmatrix}. $$

As \(n\to\infty\), the block lengths divided by \(n\) tend to \(\tfrac13,\tfrac13,\tfrac13\). The limiting covariance matrix is consequently

$$ \begin{pmatrix} \frac13 & \frac13 & \frac13\\ \frac13 & \frac23 & \frac23\\ \frac13 & \frac23 & 1 \end{pmatrix}. $$

Check: at \(n=10\), the second diagonal entry is \((3+3)/10=\tfrac35\), while the third is \((3+3+4)/10=1\). In the limit, those same cumulative sums use block proportions \(\tfrac13\) each, giving \(\tfrac23\) and \(1\). The finite-\(n\) values need not equal their limits; the floor errors divided by \(n\) vanish.

Common Pitfall: Treating Cumulative Values as Independent

A tempting but incorrect calculation is to treat the values at two different times as independent because the walk's increments are independent. For the grid \(t_1=\tfrac14,t_2=\tfrac34\), this would assign covariance zero. But the first value uses the first block, and the second value uses that same block plus the next block. Their covariance is the variance of the shared block. At the limiting scale, that variance is \(\tfrac14\), not zero.

The same error can be seen before taking a limit. When \(n\) is divisible by \(4\), the first walk value is \(n^{-1/2}\sum_{k=1}^{n/4}\xi_k\) and the second is \(n^{-1/2}\sum_{k=1}^{3n/4}\xi_k\). In the covariance of these sums, terms from distinct increments have covariance zero, while each of the \(n/4\) shared increments has variance \(1\). Therefore the covariance is \((n/4)/n=\tfrac14\). Independence applies to disjoint increment blocks, not to cumulative walk values.

For any finite-grid calculation, first compute the block lengths \(b_{n,\ell}=\lfloor nt_\ell\rfloor-\lfloor nt_{\ell-1}\rfloor\). Their scaled limits determine the independent Gaussian increments. Then form cumulative sums to obtain the values and their covariances. This order of operations preserves the dependence structure and makes the role of integer-part rounding explicit.

Check Your Understanding

Use the disjoint-block representation to answer each question. When computing a covariance, identify which blocks the two values share.

  1. For the grid \(t_1=\tfrac13,t_2=1\), what are the two limiting block variances, and what is the limiting covariance matrix of the values?
  2. For \(t_1=\tfrac12,t_2=\tfrac32\), explain why the covariance of the two limiting values is \(\tfrac12\), even though the second value contains an additional increment.
  3. At \(n=12\) and times \(t_1=\tfrac14,t_2=\tfrac34\), compute the two block lengths and the covariance matrix of the scaled walk values.
  4. Which multivariate central limit theorem condition is checked by bounding \(|a^\mathsf{T}Y_{n,k}|\) by \(\max_\ell|a_\ell|/\sqrt n\)?
  5. Why does convergence on every fixed finite grid not, by itself, prove convergence of whole paths?