The Brownian Motion Representation

From Standardized Statistics to a Cumulative Process

At interim looks, the standardized test statistic is convenient for reporting evidence, but it can obscure how evidence accumulates. Its denominator changes with information, so a difference between two standardized statistics is not the score contribution from the data collected between those looks. A cumulative-score scale makes the accumulation explicit and reveals a simple representation in terms of Brownian motion.

Suppose there are \(K\) interim looks, with positive, increasing information levels \(I_1<\cdots<I_K\). Write \(Z_k\) for the information-standardized statistic at look \(k\). Under the canonical joint normal model, assume that under a fixed alternative the mean of \(Z_k\) is \(\theta\sqrt{I_k}\), and for \(i\leq j\), its covariance with \(Z_j\) is \(\sqrt{I_i/I_j}\). Here \(\theta\) is the fixed canonical drift: it is zero under the null and has a fixed value under the alternative.

Definition: The cumulative score at look \(k\), on the information scale, is \(X_k=\sqrt{I_k}Z_k\). The Brownian motion representation of the canonical model is \(X_k=W(I_k)+\theta I_k\), or, equivalently, \(Z_k=\{W(I_k)+\theta I_k\}/\sqrt{I_k}\). Here \(W(t)\) is standard Brownian motion: it starts at zero, has independent increments, and an increment over an interval of length \(h\) has distribution \(N(0,h)\).

The key distinction is between the cumulative process \(X_k\) and the standardized statistic \(Z_k\). The cumulative process has variance proportional to information; standardizing by \(\sqrt{I_k}\) gives each \(Z_k\) variance one. The representation is an equality in joint distribution at the information times, not a claim that every study has a physically observed, continuous Brownian path between its scheduled looks.

Why the Representation Holds

The canonical joint model supplies the means and covariances needed to identify the distribution of the cumulative scores. Multiplying \(Z_k\) by \(\sqrt{I_k}\) gives

E(X_k)=\theta I_k,\qquad \operatorname{Cov}(X_i,X_j) =\sqrt{I_iI_j}\sqrt{\frac{I_i}{I_j}} =I_i=\min(I_i,I_j),\quad i\leq j.

A standard Brownian motion evaluated at the increasing times \(I_1,\ldots,I_K\) has mean zero and covariance \(\operatorname{Cov}\{W(I_i),W(I_j)\}=\min(I_i,I_j)\). Adding the deterministic drift \(\theta I_k\) changes the mean to \(\theta I_k\) without changing the covariance. Since both vectors are jointly normal, matching their means and covariance matrices establishes that \((X_1,\ldots,X_K)\) has the same joint distribution as \((W(I_1)+\theta I_1,\ldots,W(I_K)+\theta I_K)\).

The representation also describes increments. Put \(I_0=0\), \(X_0=0\), and \(\Delta I_k=I_k-I_{k-1}\). Then

X_k-X_{k-1} \sim N(\theta\Delta I_k,\Delta I_k), \qquad k=1,\ldots,K,

and these increments are independent. To see why, the Brownian part of each increment is an increment over a distinct, nonoverlapping information interval. Under the canonical joint normal model, disjoint increments have zero covariance; joint normality then makes them independent. The drift adds the fixed mean \(\theta\Delta I_k\) to each increment, so it does not affect their independence. This is the cumulative-scale form of the independent-increments structure.

Under \(H_0\), \(\theta=0\), so the cumulative score is standard Brownian motion evaluated at the information levels. Under a fixed alternative, the mean of the cumulative process increases linearly as \(\theta I\). Consequently, the mean of the standardized statistic is \(\theta\sqrt{I}\), while its variance remains one.

Worked Example: Converting Standardized Statistics to Cumulative Scores

Consider a hypothetical three-look trial with information levels \(I_1=100\), \(I_2=225\), and \(I_3=400\). Suppose the observed standardized statistics are \(Z_1=1.20\), \(Z_2=1.60\), and \(Z_3=1.90\). Convert them to cumulative scores and describe the increments.

At each look, use \(X_k=\sqrt{I_k}Z_k\):

X_1=\sqrt{100}(1.20)=12.0,\qquad X_2=\sqrt{225}(1.60)=24.0,\qquad X_3=\sqrt{400}(1.90)=38.0.

The cumulative-score increments are \(12.0\), \(24.0-12.0=12.0\), and \(38.0-24.0=14.0\). The corresponding information increments are \(100\), \(125\), and \(175\). Under a fixed alternative with \(\theta=0.05\), the model assigns these increments means \(0.05(100)=5.0\), \(0.05(125)=6.25\), and \(0.05(175)=8.75\), and variances \(100\), \(125\), and \(175\), respectively. Thus the observed increments are realizations, not their expected values. Their standardized-statistic means at the three looks are \(0.05\sqrt{100}=0.50\), \(0.05\sqrt{225}=0.75\), and \(0.05\sqrt{400}=1.00\).

The arithmetic changes scale but not the evidence recorded at a look: for example, \(X_2/\sqrt{I_2}=24.0/15=1.60\), the stated \(Z_2\). The cumulative scale is useful for seeing what was added between looks; the standardized scale is useful for comparing evidence in standard-error units.

Interpreting Drift and Dependence

The fixed drift assumption means the same \(\theta\) applies at every look. It does not mean that the expected increase in the cumulative score is the same at every look: that expected increase is \(\theta\Delta I_k\), which depends on the additional information. Under a positive alternative, the mean cumulative score moves upward with information; under a negative alternative, it moves downward. The Brownian component continues to fluctuate in either case.

Standardization creates dependence across looks because earlier data contribute to later statistics. For \(i\leq j\), the representation gives

\operatorname{Cov}(Z_i,Z_j) =\frac{\min(I_i,I_j)}{\sqrt{I_iI_j}} =\sqrt{\frac{I_i}{I_j}}.

The drift does not change this covariance. Thus the representation provides one way to understand the correlation structure: cumulative scores share the Brownian path up to the earlier information time, while the later cumulative score also includes additional increments. The standardized statistics are not independent observations, even though the new increments on the cumulative scale are independent.

Worked Example: Checking a Two-Look Joint Distribution

Suppose a hypothetical monitoring plan has \(I_1=64\), \(I_2=144\), and fixed alternative drift \(\theta=0.015\). Find the means and covariance of the cumulative scores, and then the means and correlation of the standardized statistics.

The cumulative means are \(\theta I_1=0.015(64)=0.960\) and \(\theta I_2=0.015(144)=2.160\). Their variances are \(64\) and \(144\), and their covariance is the earlier information, \(64\). Equivalently, the first increment has distribution \(N(0.960,64)\); the second has information increase \(144-64=80\), so its distribution is \(N(0.015(80),80)=N(1.20,80)\). The cumulative mean at the second look is the sum \(0.960+1.20=2.160\).

Dividing the cumulative means by the square roots of information gives \(E(Z_1)=0.960/8=0.120\) and \(E(Z_2)=2.160/12=0.180\). Each standardized statistic has variance one. Their covariance, and therefore their correlation, is

\operatorname{Corr}(Z_1,Z_2) =\frac{64}{\sqrt{64(144)}} =\frac{64}{96} =\frac{2}{3}\approx 0.667.

This checks both scales: the cumulative scores have covariance \(64\), while the standardized statistics have correlation \(0.667\). The positive drift changes their means but not their correlation.

Using the Information-Time Scale

The representation is especially helpful when interpreting a rule expressed in terms of standardized statistics. For example, a fixed upper threshold \(Z_k\geq b\) at information \(I_k\) is equivalent at that look to a cumulative-score threshold \(X_k\geq b\sqrt{I_k}\). Across information time, that boundary grows like \(\sqrt{I}\), whereas the alternative mean of \(X(I)\) grows linearly as \(\theta I\). This is a scale conversion, not by itself a method for choosing or calibrating a sequential boundary.

Worked Example: Translating a Standardized Threshold

Suppose a hypothetical design uses an upper standardized-statistic threshold of \(b=2.00\) at information levels \(I=100\) and \(I=400\). What are the equivalent cumulative-score thresholds? Under a fixed alternative with \(\theta=0.03\), compare each threshold with the cumulative-score mean.

The cumulative threshold is \(b\sqrt{I}\). At \(I=100\), it is \(2.00\sqrt{100}=20.0\). The alternative mean is \(\theta I=0.03(100)=3.00\). At \(I=400\), the threshold is \(2.00\sqrt{400}=40.0\), while the alternative mean is \(0.03(400)=12.0\).

These calculations place the threshold and the mean on the same cumulative-score scale; they do not give the probability of crossing the threshold, since that also depends on the process variance and the joint path across looks. Standardizing verifies the conversion: \(20.0/\sqrt{100}=2.00\) and \(40.0/\sqrt{400}=2.00\).

Assumptions, Diagnostics, and Alternatives

The Brownian representation is exact for the stated canonical joint normal model. Its use therefore requires a jointly normal score vector, a common fixed drift under the alternative, increasing information levels, and the canonical covariance structure. The information levels must be positive for standardized statistics to be defined. In applications where the canonical model is an approximation, the representation is approximate as well; its accuracy depends on whether the underlying score statistics are adequately approximated by that model.

A practical check is to compare the planned information levels and model-implied means and covariances with those used to form the interim statistics. In particular, the implied cumulative-score covariance for looks \(i\) and \(j\) must equal the smaller information level. The increment variances must equal the increases in information. A decreasing or repeated information level, an inconsistent covariance pattern, or evidence that the effect changes across looks signals that the simple representation may not apply as written. Random information requires additional care: the formula should not be treated as an unconditional fixed-time representation without assumptions that justify conditioning on or modeling that information.

The main alternatives are to work directly with the vector of standardized statistics or with its independent cumulative increments. The standardized vector is familiar and keeps each statistic on a unit-variance scale, but its dependence is less visually transparent. Increment coordinates make the new contribution at each look explicit, but require tracking the cumulative sum to recover a look-specific statistic. The Brownian representation unifies these views: it expresses cumulative evidence as a process indexed by information, while retaining the same finite-look joint distribution.

A common mistake is to say that the \(Z_k\) statistics themselves are independent because the score increments are independent. That is incorrect: \(Z_k\) includes all information accumulated through look \(k\), so it shares earlier increments with subsequent statistics. Independence applies to the disjoint increments \(X_k-X_{k-1}\), not to the cumulative scores or their standardized versions. Another pitfall is to call \(W(I)+\theta I\) standard Brownian motion. Only \(W\) is standard Brownian motion; adding the drift changes the mean and produces Brownian motion with drift.

Key takeaway: Under the canonical joint normal model, multiplying each standardized statistic by the square root of its information gives a cumulative process with mean \(\theta I\) and Brownian covariance \(\min(I_i,I_j)\). The fixed-alternative representation is \(X(I)=W(I)+\theta I\), interpreted at the planned information times.

Check Your Understanding

Use the representation to answer each question. Solutions follow each prompt.

If \(I=81\) and \(Z=1.50\), what is the cumulative score? Solution: \(X=\sqrt{81}(1.50)=9(1.50)=13.5\).
Under a fixed alternative with \(\theta=0.02\), what are the mean and variance of an increment over an information increase of \(50\)? Solution: Its distribution is \(N(\theta\Delta I,\Delta I)=N(0.02(50),50)=N(1.00,50)\), so the mean is \(1.00\) and the variance is \(50\).
For information levels \(I_1=25\) and \(I_2=100\), what is the correlation of the two standardized statistics? Solution: \(\sqrt{I_1/I_2}=\sqrt{25/100}=0.500\).
Does the fixed drift change the covariance between cumulative scores? Solution: No. The deterministic term \(\theta I\) changes the mean only; the covariance remains \(\min(I_i,I_j)\).
Why does independent-increment structure not make the standardized statistics independent? Solution: Each later cumulative score includes the earlier increments. Therefore the standardized statistics share data-derived components, giving correlation \(\sqrt{I_i/I_j}\) for \(i\leq j\), rather than zero.