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Brownian Motion: Definition and Construction · Tutorial 814 of 994

Fourth-Moment Bounds for Random-Walk Tightness

Learn how finite fourth moments yield uniform increment bounds for interpolated random walks and how those bounds establish tightness by controlling moduli of continuity.

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What You'll Learn

  • Express an interpolated path increment as a weighted sum of independent walk increments.
  • Derive the fourth-moment formula for a sum of independent centered increments.
  • Prove a uniform fourth-moment bound proportional to the square of the time increment.
  • Use dyadic grids and Markov’s inequality to control moduli of continuity.
  • Identify why finite-dimensional convergence alone does not establish tightness.
  • Avoid omitting cross terms when calculating fourth moments.

From Increment Moments to Path Control

To compare random walks with continuous paths, it is not enough to understand their values at a fixed finite set of times. A path could make large excursions between those times. Tightness in the space of continuous functions rules out that possibility by controlling how rapidly paths can vary on short time intervals. For polygonally interpolated, diffusively rescaled walks, a fourth-moment estimate gives exactly the increment control needed for this purpose.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((\xi_k)_{k\ge1}\) be independent, identically distributed real random variables with \(\mathbb{E}\xi_1=0\), \(\mathbb{E}\xi_1^2=\sigma^2<\infty\), and \(\mathbb{E}\xi_1^4=\beta<\infty\). Write \(S_0=0\) and \(S_k=\sum_{r=1}^k\xi_r\) for \(k\ge1\). For each positive integer \(n\), define a continuous, linearly interpolated path on \([0,\infty)\) by \(\widetilde S_n(t)=S_k+(nt-k)\xi_{k+1}\) when \(k=\lfloor nt\rfloor\), and set \(W_n(t)=n^{-1/2}\widetilde S_n(t)\). Thus \(W_n\) has state space \(\mathbb{R}\), continuous time parameter \(t\ge0\), and \(W_n(0)=0\). The goal is a bound uniform in \(n\) and in the time points.

Definition: For \(T>0\), let \(C[0,T]\) be the space of real-valued continuous functions on \([0,T]\), with distance \(\|f-g\|_\infty=\sup_{0\le t\le T}|f(t)-g(t)|\). A family of probability laws on \(C[0,T]\) is tight if for each \(\varepsilon>0\) there is a compact set \(K\subset C[0,T]\) such that every law in the family assigns probability at least \(1-\varepsilon\) to \(K\). The modulus of continuity of \(f\in C[0,T]\) is \(\omega_f(\delta)=\sup\{|f(t)-f(s)|:0\le s,t\le T,\ |t-s|\le\delta\}\).

The Fourth-Moment Bound

The key calculation is the fourth moment of a weighted sum. The weights arise because an interval can begin or end partway through an interpolation segment. For independent centered variables, terms involving an unpaired first power vanish, but terms involving two squared increments do not vanish; those cross terms are essential.

Proposition: Under the assumptions above, for every positive integer \(n\) and all \(s,t\ge0\), \(\mathbb{E}|W_n(t)-W_n(s)|^4\le C_4|t-s|^2\), where \(C_4=\max\{\beta,3\sigma^4\}\). In particular, the constant does not depend on \(n,s,\) or \(t\).

Proof. It is enough to consider \(s\le t\), since the absolute increment is unchanged when the endpoints are exchanged. Put \(h=t-s\). The unscaled interpolated increment \(\widetilde S_n(t)-\widetilde S_n(s)\) can be written as \(\sum_r w_r\xi_r\), where \(w_r\) is the length of the intersection of \([ns,nt]\) with \([r-1,r]\). Only finitely many weights are nonzero, each lies in \([0,1]\), and their sum is the length of \([ns,nt]\): \[ \sum_r w_r=nh. \] For any finite set of deterministic weights, independence and centering give \[ \mathbb{E}\left(\sum_r w_r\xi_r\right)^4 =\beta\sum_r w_r^4+6\sigma^4\sum_{r<q}w_r^2w_q^2. \] Indeed, in the expansion of the fourth power, a term with an index appearing just once has expectation zero by independence and \(\mathbb{E}\xi_r=0\). The remaining terms either use one index four times, contributing \(\beta w_r^4\), or two indices twice each; the latter have six orderings and contribute \(6\sigma^4w_r^2w_q^2\). Now \[ \beta\sum_r w_r^4+6\sigma^4\sum_{r<q}w_r^2w_q^2 \le C_4\left(\sum_r w_r^2\right)^2, \] because the square on the right expands as \(\sum_r w_r^4+2\sum_{r<q}w_r^2w_q^2\), and \(C_4\ge\beta\), \(2C_4\ge6\sigma^4\). Since \(0\le w_r\le1\), \(w_r^2\le w_r\), so \(\sum_r w_r^2\le nh\). Dividing by \(n^2\) for the rescaling \(W_n=n^{-1/2}\widetilde S_n\) gives \[ \mathbb{E}|W_n(t)-W_n(s)|^4 \le n^{-2}C_4(nh)^2=C_4h^2. \] If \(s=t\), both sides are zero, so the bound holds in that case too. \(\square\)

The proof covers intervals shorter than a mesh interval as well as intervals spanning many mesh points. In particular, it does not assume that either endpoint lies on the walk’s time grid. The exponent \(2\) on the right is what allows a dyadic argument to control all small increments at once.

Worked Example: An Increment Crossing Two Partial Segments

Take independent Rademacher increments, meaning \(\mathbb{P}(\xi_r=1)=\mathbb{P}(\xi_r=-1)=\tfrac12\) for each \(r\). Then \(\mathbb{E}\xi_r=0\), \(\sigma^2=1\), and \(\beta=1\), so \(C_4=\max\{1,3\}=3\). For \(n=4\), calculate the fourth moment of \(W_4(\tfrac78)-W_4(\tfrac18)\). The interval in the unscaled time coordinate is \([4/8,28/8]=[1/2,7/2]\), so its weights on \(\xi_1,\xi_2,\xi_3,\xi_4\) are respectively \(\tfrac12,1,1,\tfrac12\). The rescaled increment is therefore \(\tfrac12(\tfrac12\xi_1+\xi_2+\xi_3+\tfrac12\xi_4)\).

For these weights, the sum of fourth powers is \((\tfrac12)^4+1^4+1^4+(\tfrac12)^4=\tfrac1{16}+1+1+\tfrac1{16}=\tfrac{17}{8}\). The sum of pairwise products of squared weights is \(\tfrac14\cdot1+\tfrac14\cdot1+\tfrac14\cdot\tfrac14+1\cdot1+1\cdot\tfrac14+1\cdot\tfrac14 =\tfrac12+\tfrac1{16}+1+\tfrac12=\tfrac{33}{16}\). Thus the fourth moment before the outside factor \(\tfrac12\) is \(\tfrac{17}{8}+6\cdot\tfrac{33}{16}=\tfrac{17}{8}+\tfrac{99}{8}=\tfrac{29}{2}\). The outside factor contributes \((\tfrac12)^4=\tfrac1{16}\), giving \[ \mathbb{E}|W_4(\tfrac78)-W_4(\tfrac18)|^4=\tfrac{29}{32}. \] Check: The time difference is \(\tfrac78-\tfrac18=\tfrac34\), so the proposition bounds this moment by \(3(\tfrac34)^2=\tfrac{27}{16}\). Since \(\tfrac{29}{32}\le\tfrac{54}{32}=\tfrac{27}{16}\), the direct calculation satisfies the bound.

Worked Example: A Non-Rademacher Walk at Mesh Times

Let each increment equal \(\sqrt{2}\) with probability \(\tfrac14\), equal \(-\sqrt{2}\) with probability \(\tfrac14\), and equal \(0\) with probability \(\tfrac12\), independently across steps. Its mean is \(\tfrac14\sqrt{2}+\tfrac14(-\sqrt{2})=0\); its variance is \(\tfrac14(2)+\tfrac14(2)=1\); and its fourth moment is \(\tfrac14(4)+\tfrac14(4)=2\). Take \(n=5\) and compare \(W_5(3/5)\) with \(W_5(0)\). This interval contains three full increments, so its fourth moment is the fourth moment of \((\xi_1+\xi_2+\xi_3)/\sqrt{5}\).

Using the proved expansion, the unscaled sum has fourth moment \(3\beta+6\binom{3}{2}\sigma^4=3(2)+6(3)(1)=6+18=24\). After rescaling, the answer is \(24/5^2=\tfrac{24}{25}\). The proposition gives \(C_4=\max\{2,3\}=3\), and its bound at time difference \(\tfrac35\) is \(3(\tfrac35)^2=\tfrac{27}{25}\). Since \(\tfrac{24}{25}\le\tfrac{27}{25}\), the computed moment passes the bound. This example also shows that the constant accounts for the variance cross terms even when the increment distribution is not Rademacher.

A Dyadic Tightness Criterion

Here is the path-space result that turns an increment estimate into tightness. We give the dyadic argument because its summation shows precisely why the moment exponent matters. The result is stated for fourth moments, which is the case needed for the interpolated walks.

Theorem: Let \(T>0\), and let \((Z_a)_{a\in I}\) be a family of random elements of \(C[0,T]\), indexed by any set \(I\). Suppose \(Z_a(0)=0\) almost surely and that, for some finite constant \(C\) independent of \(a,s,t\), \(\mathbb{E}|Z_a(t)-Z_a(s)|^4\le C|t-s|^2\) for all \(a\in I\) and \(s,t\in[0,T]\). Then the laws of \((Z_a)_{a\in I}\) are tight in \(C[0,T]\).

Proof. Fix \(0<\gamma<\tfrac14\), a number \(b>0\), and a nonnegative integer \(m_0\). At level \(m\), use the dyadic grid points \(t_{m,k}=kT2^{-m}\), \(k=0,\ldots,2^m\). Call a level-\(m\) adjacent increment bad if \[ |Z_a(t_{m,k+1})-Z_a(t_{m,k})|>b2^{-\gamma m} \] for at least one \(k\). Markov’s inequality gives, for each adjacent pair, \[ \mathbb{P}\bigl(|Z_a(t_{m,k+1})-Z_a(t_{m,k})|>b2^{-\gamma m}\bigr) \le \frac{C(T2^{-m})^2}{b^4 2^{-4\gamma m}} =\frac{CT^2}{b^4}2^{-(2-4\gamma)m}. \] There are \(2^m\) adjacent pairs at level \(m\), so the union bound gives probability at most \(\frac{CT^2}{b^4}2^{-(1-4\gamma)m}\) that level \(m\) has a bad increment. Summing over \(m\ge m_0\), and using \(1-4\gamma>0\), shows that the probability of any bad increment at any such level is at most \[ \frac{CT^2}{b^4}\, \frac{2^{-(1-4\gamma)m_0}}{1-2^{-(1-4\gamma)}}. \] This bound is uniform over \(a\).

On the event that no level \(m\ge m_0\) has a bad increment, continuity controls increments between grid points too. For \(x\in[0,T]\), let \(x_m\) be the largest level-\(m\) grid point not exceeding \(x\). For \(x<T\), the points \(x_l\) and \(x_{l+1}\) are either equal or adjacent at level \(l+1\), and \(x_l\) tends to \(x\). Continuity and the bounds on adjacent increments therefore imply \[ |Z_a(x)-Z_a(x_m)| \le\sum_{l=m+1}^{\infty}b2^{-\gamma l} =\frac{b2^{-\gamma(m+1)}}{1-2^{-\gamma}}. \] For \(x=T\), the approximation is exact. If \(|x-y|\le T2^{-m}\), the level-\(m\) grid points \(x_m,y_m\) are equal or adjacent, so their values differ by at most \(b2^{-\gamma m}\). Applying the preceding approximation to both endpoints yields \[ \omega_{Z_a}(T2^{-m}) \le b2^{-\gamma m} +\frac{2b2^{-\gamma(m+1)}}{1-2^{-\gamma}} =D_\gamma b2^{-\gamma m}, \qquad D_\gamma=1+\frac{2\cdot2^{-\gamma}}{1-2^{-\gamma}}. \] Consequently, for every \(m\ge m_0\), the probability that this modulus bound fails is at most the displayed uniform probability of a bad dyadic increment.

To conclude tightness, fix \(\varepsilon>0\). For each integer \(r\ge1\), choose \(m_r\) sufficiently large that \(D_\gamma 2^{-\gamma m_r}\le1/r\) and the probability bound above, with \(b=1\) and \(m_0=m_r\), is less than \(\varepsilon 2^{-r}\). Define \[ K=\{f\in C[0,T]: f(0)=0,\ \omega_f(T2^{-m_r})\le1/r\text{ for every }r\ge1\}. \] Each condition defining \(K\) is closed in the uniform distance: changing a path by at most \(\eta\) changes its modulus at any fixed scale by at most \(2\eta\). The conditions also make \(K\) equicontinuous, since \(1/r\) tends to zero. The paths are uniformly bounded as well: using the \(r=1\) condition and \(f(0)=0\), divide \([0,T]\) into finitely many intervals of length at most \(T2^{-m_1}\); each step changes a path value by at most \(1\). Thus the Arzelà–Ascoli compactness theorem applies, and \(K\) is compact. By the union bound, each \(Z_a\) lies in \(K\) with probability at least \(1-\sum_{r\ge1}\varepsilon2^{-r}=1-\varepsilon\). This is tightness. \(\square\)

The only compactness theorem used here is Arzelà–Ascoli, the standard characterization that a closed, uniformly bounded, equicontinuous subset of \(C[0,T]\) is compact. The probability estimate preceding it provides the required uniform equicontinuity; it does not assume that the paths are differentiable or that a bound on a finite grid alone controls their behavior between grid points.

Worked Example: A Uniform Modulus Bound for the Rademacher Walks

For the Rademacher walks, the fourth-moment proposition gives \(\mathbb{E}|W_n(t)-W_n(s)|^4\le3|t-s|^2\). Apply the dyadic estimate on \([0,1]\) with \(\gamma=\tfrac18\), \(b=4\), and \(m_0=4\). At level \(m\), the probability of any adjacent increment exceeding \(4\cdot2^{-m/8}\) is at most \[ 2^m\frac{3(2^{-m})^2}{(4\cdot2^{-m/8})^4} =\frac{3}{256}2^{-m/2}. \] The probability that at least one such increment is too large at some level \(m\ge4\) is therefore at most \[ \frac{3}{256}\sum_{m=4}^{\infty}2^{-m/2} =\frac{3}{1024(1-2^{-1/2})} <\frac18. \] For the strict inequality, \(2^{-1/2}<\tfrac34\), since squaring the positive sides gives \(\tfrac12<\tfrac9{16}\). Hence \(1-2^{-1/2}>\tfrac14\), so the displayed fraction is less than \(\tfrac3{256}<\tfrac18\).

On the event that all these adjacent increments satisfy their thresholds, the chaining estimate from the proof gives \(\omega_{W_n}(2^{-4})\le D_{1/8}\,4\,2^{-4/8}\), where \(D_{1/8}=1+2\cdot2^{-1/8}/(1-2^{-1/8})\). Thus, uniformly in \(n\), the probability that the modulus at scale \(2^{-4}\) exceeds this explicit bound is less than \(\tfrac18\). Repeating the argument with finer starting levels makes the probability of a larger modulus arbitrarily small. This is path control, not merely control at a chosen finite collection of times.

Why the Cross Terms Matter

A common mistake is to compute the fourth moment of a sum of independent centered increments by adding only the individual fourth moments. For two independent Rademacher increments, that incorrect calculation gives \(\mathbb{E}(\xi_1+\xi_2)^4=1+1=2\). The correction follows directly from the expansion: \[ \mathbb{E}(\xi_1+\xi_2)^4 =\mathbb{E}\xi_1^4+\mathbb{E}\xi_2^4 +6\mathbb{E}(\xi_1^2\xi_2^2) =1+1+6(1)(1)=8. \] The same formula also explains why one must identify the increment distribution before quoting a numerical fourth moment. For independent, identically distributed centered increments with variance one and fourth moment \(\beta\), the answer for two increments is \(2\beta+6\), not always \(8\). For example, standard normal increments have \(\beta=3\), giving \(2(3)+6=12\).

There is a second important distinction: a uniform fourth-moment bound controls increments of the full interpolated paths, including times between mesh points. A calculation only at mesh times would not by itself establish the needed modulus control for continuous paths. The weights in the proposition handle partial segments, and the dyadic argument extends grid estimates to every time. These tools provide tightness of the rescaled paths on each finite interval; by themselves, they do not identify a limiting path law or prove convergence to Brownian motion.

Check Your Understanding

Use the increment expansion and the dyadic argument to answer the questions.

  1. For independent centered, identically distributed increments with variance \(\sigma^2\) and fourth moment \(\beta\), compute the fourth moment of \(\xi_1+\xi_2\).
  2. Why does the weighted-sum proof use \(\sum_r w_r^2\le\sum_r w_r\)? Which assumption on the interpolation weights ensures this inequality?
  3. For the distribution taking values \(\sqrt2,-\sqrt2,0\) with probabilities \(\tfrac14,\tfrac14,\tfrac12\), respectively, find the fourth moment of the sum of four independent increments.
  4. In the dyadic tightness proof, what restriction on \(\gamma\) makes the sum of bad-increment probabilities over levels finite?
  5. Explain why controlling adjacent increments on dyadic grids at all sufficiently fine levels, together with continuity, controls the modulus between arbitrary time points.