Brownian Limits For Integrated Processes

Why Brownian Motion Appears in Integrated Series

A random walk can look erratic from one period to the next, yet its large-sample behavior has a simple description. When the time axis is compressed to the interval from zero to one and the levels are scaled by the square root of the sample size, the whole path is approximated by Brownian motion. This is a functional approximation: it describes the evolving path, not just the distribution of the endpoint.

That path-level result matters because many statistics for integrated series use sums of levels, squared levels, or products of levels and innovations. After the appropriate scaling, such sums become integrals involving Brownian motion. The resulting random variables are generally not normally distributed. The purpose here is to derive this connection and show how it changes the large-sample behavior of an illustrative regression statistic.

Definition: Let $u_t$ be innovations that are independent and identically distributed, with $E[u_t]=0$, $E[u_t^2]=\sigma^2$, and $0<\sigma^2<\infty$. A random walk with initial value $y_0=0$ is $y_t=\sum_{j=1}^{t}u_j$. It is an integrated process because its level accumulates the innovations over time. Brownian motion $W(r)$, $0\leq r\leq1$, is a continuous, mean-zero Gaussian process with $W(0)=0$ and covariance $E[W(r)W(s)]=\min(r,s)$.

The Functional Central Limit Approximation

The ordinary central limit theorem describes the scaled sum at one chosen time. The functional central limit theorem, also called the invariance principle or Donsker’s theorem, describes the entire sequence of partial sums at once. Define the step process $S_n(r)=\sum_{t=1}^{\lfloor nr\rfloor}u_t/(\sigma\sqrt{n})$, for $0\leq r\leq1$. Here $\lfloor nr\rfloor$ is the greatest integer not exceeding $nr$; thus $S_n(r)$ uses the innovations observed by the fraction $r$ of the sample.

Theorem: Under the independent and identically distributed innovation assumptions above, $$ S_n(\cdot)\ \Rightarrow\ W(\cdot),\qquad 0\leq r\leq1. $$ Here $\Rightarrow$ denotes weak convergence of processes: the distribution of the scaled path converges to the distribution of Brownian motion. The step paths may be viewed as elements of $D[0,1]$, the space of right-continuous functions with left limits, with its usual Skorokhod topology. Since the limit $W$ is continuous, the result also supports the usual continuous-path interpretations of the limits below.

The scaling has two parts. The time index $t$ is replaced by the fraction $r=t/n$, so the observed path is compressed to $[0,1]$. The level is divided by $\sigma\sqrt n$, because a sum of $t$ mean-zero innovations has variance $t\sigma^2$, and therefore has a typical magnitude proportional to $\sigma\sqrt t$. The endpoint $S_n(1)$ is the familiar standardized sum; the theorem strengthens that endpoint result to the whole path.

Worked Example: Scaling a Short Random Walk

Suppose a hypothetical four-period random walk starts at $y_0=0$ and has observed innovations $u_1=1,\ u_2=1,\ u_3=-1,\ u_4=2$. For illustration, take the scaling value $\sigma=1$. Find the scaled path at the sample grid points.

The partial sums are $y_0=0,\ y_1=1,\ y_2=2,\ y_3=1,\ y_4=3$. With $n=4$, the level divisor is $\sigma\sqrt n=1\cdot2=2$. Thus, at $r=0,\frac14,\frac12,\frac34,1$, the scaled levels are $$ \frac{1}{2}(y_0,y_1,y_2,y_3,y_4)=(0,\ 0.5,\ 1,\ 0.5,\ 1.5). $$

This finite path is not itself Brownian motion, and four observations do not establish an asymptotic approximation. The theorem says that, under its assumptions, the distribution of the entire scaled path approaches the distribution of $W(r)$ as $n$ grows.

From Paths to Integrals

Once the scaled level path converges to $W$, sums of functions of the levels often become ordinary integrals of functions of $W$. For example, with $y_0=0$, $$ \frac{1}{\sigma^2 n^2}\sum_{t=1}^{n}y_{t-1}^{\,2} = \frac{1}{n}\sum_{t=1}^{n} \left(\frac{y_{t-1}}{\sigma\sqrt n}\right)^2 \ \Rightarrow\ \int_0^1 W(r)^2\,dr. $$

The right-hand side is random: different Brownian paths produce different areas under their squared paths. The integral is finite with probability one. This is an example of a Brownian-motion functional, meaning a random quantity defined from the path of $W$. Path convergence and the continuous mapping theorem justify the limiting integral; the sum on the left is the corresponding discrete Riemann sum.

Worked Example: Approximating an Integral of Squared Levels

Use the hypothetical four-period path from the previous example. Its scaled levels at the left endpoints of the four intervals are $0,\ 0.5,\ 1,\ 0.5$. Calculate the discrete approximation to the Brownian integral of the squared path, and verify the scaling using the unscaled levels.

With interval width $1/4$, the left-endpoint Riemann sum is $$ \frac14\left(0^2+0.5^2+1^2+0.5^2\right) =\frac14(0+0.25+1+0.25)=0.375. $$

The unscaled lagged levels are $y_0,y_1,y_2,y_3=0,1,2,1$, so their squared sum is $0+1+4+1=6$. The scaled statistic is $$ \frac{1}{\sigma^2n^2}\sum_{t=1}^{4}y_{t-1}^{\,2} =\frac{6}{1^2\cdot4^2}=0.375. $$ The values agree exactly because this statistic is the same left-endpoint sum written in scaled units. For a long random walk, its distribution is approximated by that of $\int_0^1W(r)^2\,dr$, not by a constant equal to this short-sample value.

A different kind of sum pairs each lagged level with its next innovation. Such sums are not ordinary Riemann sums, because the innovation is a new random increment. In the limit, the corresponding object is the stochastic integral $\int_0^1W(r)\,dW(r)$, defined using the increments of Brownian motion and the lagged path value. For this particular integral, a simple identity gives a complete justification.

Use the discrete level update.
For the random walk, $y_t=y_{t-1}+u_t$, so $y_t^2-y_{t-1}^2=2y_{t-1}u_t+u_t^2$.

Sum across time.
Summing the identity and rearranging yields $\sum_{t=1}^n y_{t-1}u_t=\frac12(y_n^2-y_0^2-\sum_{t=1}^n u_t^2)$.

Apply the endpoint limit and law of large numbers.
The functional central limit theorem gives $y_n/(\sigma\sqrt n)\Rightarrow W(1)$, while $n^{-1}\sum_{t=1}^n u_t^2$ converges in probability to $\sigma^2$. With $y_0=0$, dividing the identity by $\sigma^2n$ gives the limit $\frac12(W(1)^2-1)$.

Identify the stochastic integral.
For Brownian motion, the Itô identity is $\int_0^1W(r)\,dW(r)=\frac12(W(1)^2-1)$. Hence $\frac{1}{\sigma^2n}\sum_{t=1}^n y_{t-1}u_t\Rightarrow\int_0^1W(r)\,dW(r)$.

A Regression Statistic With a Nonstandard Limit

Consider the no-intercept regression $\Delta y_t=\gamma y_{t-1}+u_t$, where $\Delta y_t=y_t-y_{t-1}$. Under the random-walk model, the population coefficient is $\gamma=0$. The ordinary least-squares slope is $$ \widehat{\gamma} =\frac{\sum_{t=1}^n y_{t-1}\Delta y_t}{\sum_{t=1}^n y_{t-1}^{\,2}} =\frac{\sum_{t=1}^n y_{t-1}u_t}{\sum_{t=1}^n y_{t-1}^{\,2}}. $$

The numerator is of order $n$, while the denominator is of order $n^2$. Combining the two Brownian limits therefore gives $$ n\widehat{\gamma} \ \Rightarrow\ \frac{\int_0^1W(r)\,dW(r)} {\int_0^1W(r)^2\,dr}. $$

The factor $n$ is important: the coefficient estimator contracts at rate $n$, rather than the usual square-root-of-$n$ rate for many estimators based on stationary data. A corresponding illustration is the conventional slope statistic formed by dividing $\widehat{\gamma}$ by its estimated standard error. If the error variance estimate converges to $\sigma^2$, its standard error is asymptotically $\sigma/\sqrt{\sum y_{t-1}^2}$, of order $1/n$. Consequently, the statistic has the limit $$ \frac{\int_0^1W(r)\,dW(r)} {\sqrt{\int_0^1W(r)^2\,dr}}. $$ This expression is a ratio of Brownian functionals. It is not justified to compare it with an ordinary t distribution merely because it is a regression statistic. This derivation explains the source of nonstandard behavior; it does not specify a unit-root testing procedure or its critical values.

Worked Example: Computing the Finite-Sample Slope and Its Scaling

For the hypothetical innovations $1,1,-1,2$, calculate the no-intercept slope in the regression of $\Delta y_t$ on $y_{t-1}$, and compare its scaling to the Brownian-function approximations.

The lagged levels are $0,1,2,1$, and the innovations are $1,1,-1,2$. Thus the numerator is $0(1)+1(1)+2(-1)+1(2)=1$, while the denominator is $0^2+1^2+2^2+1^2=6$. Therefore $$ \widehat{\gamma}=\frac{1}{6}=0.1667,\qquad n\widehat{\gamma}=4\left(\frac16\right)=0.6667. $$

For this path, the discrete approximation to the numerator functional is $\sum w_{t-1}\Delta w_t=0(0.5)+0.5(0.5)+1(-0.5)+0.5(1)=0.25$, where the scaled path values are $w=(0,0.5,1,0.5,1.5)$. The denominator approximation from the previous example is $0.375$. Their ratio is $0.25/0.375=0.6667$, matching $n\widehat{\gamma}$. These short-sample values illustrate the scaling identity; they are not critical values or evidence about a real data set.

Interpretation and a Common Pitfall

The nonstandard limits arise because the regressor is a persistent level built from past innovations. Its size grows with the sample, and its observations are strongly dependent across time. The regression numerator and denominator therefore accumulate along a random path; they do not behave like sums of independent, identically distributed terms with a fixed regressor distribution. Brownian functionals retain this path dependence in the limit.

A common mistake is to reason, “The slope is estimated by least squares, so its t-statistic must be approximately t-distributed under the null.” The correction is to check the regressor’s stochastic behavior and the estimator’s rate before applying conventional reference distributions. For a random-walk level, the scaled slope and its statistic have Brownian-functional limits. Their distributions require unit-root-specific asymptotic analysis rather than the standard regression approximation.

The exact Brownian functional depends on the statistic and on the deterministic specification. Including a constant or trend changes how the path enters the statistic; it does not license using the same limiting distribution without checking. The central lesson is that the functional central limit theorem supplies the path approximation, and the statistic’s algebra determines which Brownian functional—and therefore which limiting distribution—results.

Key takeaway: A random walk scaled by $\sigma\sqrt n$ converges as a process to standard Brownian motion. Sums involving its levels then converge to Brownian integrals or ratios of such integrals, explaining why integrated-series statistics can have nonstandard limits.

Check Your Understanding

Answer each question using the assumptions and scaling developed above.

Why is a random walk’s level divided by $\sigma\sqrt n$ in the functional central limit approximation?
State the limiting process for $S_n(r)$ and explain what is meant by convergence of the whole path.
Why does the scaled sum of squared lagged levels converge to an integral involving $W(r)^2$?
Use the discrete identity for $y_t^2-y_{t-1}^2$ to identify the limit of $(\sigma^2n)^{-1}\sum_{t=1}^n y_{t-1}u_t$.
Why does the no-intercept slope statistic in the random-walk regression not automatically have an ordinary t distribution?