Canonical Drift Under a Fixed Alternative

Why Fixed-Alternative Drift Matters

At an interim look, a standardized test statistic combines the evidence observed so far into a scale-free quantity. Under the null hypothesis, its expected value is typically zero. Under a fixed alternative, its expected value moves away from zero as information accumulates. The rate of that movement matters in group-sequential design: it describes how the average statistic changes across looks, without saying what any particular trial will observe.

The key relationship is simple but easy to mis-scale. In the canonical parameterization, the expected cumulative numerator is the effect multiplied by information. Standardizing divides that numerator by the square root of information, leaving a mean proportional to the square root of information. The effect and information must refer to the same parameterization; pairing a standardized effect with information defined for an unstandardized parameter gives the wrong drift.

Here, “fixed” means that the same effect \(\theta\) applies throughout the observations accumulated at each look. Information is cumulative and positive. In the normal canonical model, \(U_k\) has a normal distribution with the mean and variance stated above, so \(Z_k\) has variance one and mean \(\theta\sqrt{I_k}\). In other settings this relationship may be an approximation justified by a canonical or large-sample model; its accuracy depends on the model and its assumptions.

The sign of \(\theta\) sets the direction of the drift. If the statistic is oriented so that positive values support an increase, a positive effect gives positive drift and a negative effect gives negative drift. Under \(H_0\), \(\theta=0\), so the drift is zero. A zero mean under the null does not imply that every observed statistic is zero; it describes the center of its sampling distribution.

Deriving the Square-Root Relationship

The derivation follows directly from the definition of \(Z_k\), once the scale of \(U_k\) is specified. For a fixed effect, the expected numerator grows in proportion to information. Dividing by \(\sqrt{I_k}\), rather than by \(I_k\), produces the square-root relationship.

This calculation uses the fact that \(I_k\) is a specified, nonrandom information level for the look. It is not a data-dependent quantity inside the expectation in this model. The result also explains why equal increases in information do not produce equal increases in expected \(Z_k\): the mean follows the square root of cumulative information. If information is multiplied by four, the magnitude of the drift is multiplied by two, provided the effect stays fixed.

The same result can be written using information fractions. If \(I_K\) is the information at a final look and \(t_k=I_k/I_K\), then \(I_k=t_k I_K\), so

Thus the expected statistic at an interim look is the final-look drift multiplied by the square root of that look’s information fraction. This is a statement about means under a fixed effect, not a claim that the observed interim statistic must equal a particular fraction of its eventual value.

A Normal-Mean Derivation

A normal-mean model shows explicitly how the effect and information scales fit together. Suppose \(X_1,\ldots,X_n\) are independent observations with \(X_i\sim N(\mu,\sigma^2)\), where the variance \(\sigma^2\) is known and observations are measured in fixed outcome units. Test \(H_0:\mu=\mu_0\), and consider a fixed alternative \(\mu=\mu_0+\Delta\). The cumulative score evaluated at \(\mu_0\) is

Under the stated alternative, each \(X_i\) has mean \(\mu_0+\Delta\). Independence and the stated variance therefore give

The canonical effect is \(\theta=\Delta\) in this score parameterization. Standardizing the score yields

The units provide a useful check. If the outcome is measured in, for example, millimetres, then \(\Delta\) is in millimetres and \(I_n=n/\sigma^2\) is in inverse squared millimetres. Their product \(\Delta\sqrt{I_n}\) is dimensionless, as a standardized statistic’s mean must be. No special interpretation of the measurement units is needed beyond using the same units consistently.

Equivalent Scales and a Common Source of Error

The same standardized statistic can be expressed using a dimensionless effect, but the information and numerator must be rescaled with it. In the normal-mean example, define \(\eta=\Delta/\sigma\), and use \(V_n=\sum_{i=1}^n(X_i-\mu_0)/\sigma\). Then \(E(V_n)=n\eta\), \(\operatorname{Var}(V_n)=n\), and the corresponding information is \(J_n=n\). Consequently, \(V_n/\sqrt{J_n}\) has mean \(\eta\sqrt{J_n}\). This equals the original expression because

The two parameterizations are equivalent: the score-scale version pairs \(\theta=\Delta\) with \(I_n=n/\sigma^2\), while the dimensionless version pairs \(\eta=\Delta/\sigma\) with \(J_n=n\). In both, the product of the effect and the square root of information is the same. A calculation that combines \(\eta\) with \(I_n\), or \(\theta\) with \(J_n\), generally mixes scales.

Assumptions, Diagnostics, and Limitations

Before applying the formula, check that its premises match the analysis. The information must be cumulative and positive; the numerator must be standardized using that same information; and the effect must be fixed across the observations contributing to the look. The canonical model also requires that the numerator’s expected value be \(\theta I_k\). In the normal-mean derivation, independent observations with known common variance give this relationship exactly. For other models, the canonical relationship may be a working approximation rather than an exact finite-sample identity.

Several checks can reveal a scale or model problem:

• Check the units. The effect times the square root of information must be dimensionless. If it is not, the effect and information may come from incompatible parameterizations.
• Check the standardization. If the numerator has variance \(I_k\), division by \(\sqrt{I_k}\) gives unit variance. Dividing by \(I_k\) instead does not produce the stated standardized statistic.
• Check the direction. A signed effect should produce drift with the same sign, unless the test statistic has been defined in the opposite direction.
• Check the effect across looks. If the effect changes over time, one constant \(\theta\) cannot in general be pulled out of the cumulative expectation. The mean then depends on the effects and information contributions at the separate periods.
• Distinguish mean from observation. A realized \(Z_k\) can fall above or below its expected value. The drift describes the center under a specified alternative, not a guaranteed path or an observed-data diagnostic by itself.

The main alternative to the canonical score scale is often a directly standardized effect scale, as in \(\eta=\Delta/\sigma\) with information \(n\) for the normal-mean example. Neither scale is inherently preferable for calculating drift; consistency is essential. One can also describe progress by sample size in this example because information is proportional to \(n\). In general, however, information—not elapsed calendar time or a raw count alone—is the quantity in the canonical formula.

The formula has limits. It does not determine the probability of crossing a monitoring boundary, the joint distribution of statistics at multiple looks, or the behavior of a procedure under a misspecified model. Those questions require more than the means derived here. Also, if nuisance parameters are estimated, observations are dependent, or the effect varies across participants or time, the information and drift relationship must be justified for that particular model rather than assumed automatically.

Common Mistake: Mixing Effect and Information Scales

A frequent error is to take a dimensionless effect such as \(\Delta/\sigma\) and multiply it by the square root of information \(n/\sigma^2\) from the score parameterization. The resulting number may be arithmetically straightforward, but it does not represent the mean of the specified standardized statistic. The correction is to match each effect with its own numerator and information: use \(\Delta\) with \(n/\sigma^2\), or use \(\Delta/\sigma\) with \(n\). Then verify that both routes give the same drift. Another mistake is to say that drift doubles whenever information doubles; it grows by \(\sqrt{2}\), not by 2, when the effect is fixed.