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Randomization-Based Inference · Tutorial 472 of 501

Studentization With Prognostic Covariates

Compare unadjusted and covariate-adjusted studentized contrasts, including how to estimate a conservative design-based variance without changing treatment assignment.

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What You'll Learn

  • Define unadjusted and baseline-covariate-adjusted Horvitz–Thompson contrasts under sequential assignment.
  • Derive the conditional variance of a centered contribution and a conservative variance bound.
  • Explain why prognostic covariates can improve precision without guaranteeing a gain.
  • Identify conditions needed for studentized, asymptotically valid weak-null inference.
  • Distinguish this approach from sharp-null randomization tests and ordinary fixed-allocation methods.
  • Diagnose unstable probabilities, poor prediction, and invalid covariate use.

Why Studentize a Covariate-Adjusted Contrast?

In a response-adaptive trial, the chance of receiving treatment can depend on the observed history and a participant's baseline covariates. A raw treatment contrast can then be difficult to interpret or calibrate using methods that assume fixed group sizes. Studentization gives the contrast a scale that reflects the actual assignment probabilities. Centering outcomes with a prognostic baseline prediction can further reduce the variability of the contrast when that prediction captures outcome variation unrelated to treatment.

The adjustment described here changes the statistic, not the randomization rule. It does not alter which arm a participant is assigned to, or the probability used by the design. Its purpose is to improve estimation and inference for the average treatment effect, particularly under a weak null that constrains only that average rather than every participant's effect.

Definition: For participant \(i\), let \(Z_i\in\{0,1\}\) indicate treatment assignment, \(Y_i(1)\) and \(Y_i(0)\) be the potential outcomes, and \(\tau_i=Y_i(1)-Y_i(0)\). The finite-population average treatment effect is \(\tau=n^{-1}\sum_{i=1}^n\tau_i\). Let \(H_{i-1}\) denote the observed assignment and outcome history before participant \(i\), and let \(X_i\) be baseline covariates. The known assignment probability is \(\pi_i=P(Z_i=1\mid H_{i-1},X_i)\). A prognostic prediction \(g_i\) is a baseline-outcome prediction specified using information available before \(Z_i\) is assigned.

Assume consistency, no interference between participants, and a correctly specified assignment probability with \(0<\pi_i<1\). For the usual asymptotic justification, assignment probabilities should not approach zero or one too quickly, and the outcomes or residuals should satisfy suitable moment and no-dominating-observation conditions. The predictor \(g_i\) must be available before the current assignment; it may be prespecified or updated using information available at that point, but it must not use the participant's post-assignment outcome.

The Contrast and Its Studentization

The observed outcome is \(Y_i=Z_iY_i(1)+(1-Z_i)Y_i(0)\). Given a predictable baseline prediction \(g_i\), define the inverse-probability-weighted contribution

$$ D_i(g)=\frac{Z_i(Y_i-g_i)}{\pi_i} -\frac{(1-Z_i)(Y_i-g_i)}{1-\pi_i}, \qquad \widehat{\tau}_g=\frac{1}{n}\sum_{i=1}^nD_i(g). $$

The unadjusted contrast is the special case \(g_i=0\) for every participant. Centering is not a subtraction that changes the target: under the assignment mechanism,

$$ E\{D_i(g)\mid H_{i-1},X_i,Y_i(1),Y_i(0)\} =Y_i(1)-Y_i(0)=\tau_i. $$

Thus, conditional on the fixed potential outcomes and the information determining the current assignment, \(\widehat{\tau}_g\) has expectation \(\tau\). Its value for a particular realized assignment may differ substantially from the unadjusted value, even though both target the same average effect.

For weak-null inference, the null is \(H_0:\tau=0\), which allows individual effects \(\tau_i\) to vary. Studentization divides the observed contrast, or its difference from a null value \(\tau_0\), by a design-based standard error. A variance estimate for the contrast is

$$ \widehat V_g= \frac{1}{n^2}\sum_{i=1}^n \left\{ \frac{Z_i(Y_i-g_i)^2}{\pi_i^2} + \frac{(1-Z_i)(Y_i-g_i)^2}{(1-\pi_i)^2} \right\}, \qquad T_g(\tau_0)= \frac{\widehat{\tau}_g-\tau_0}{\sqrt{\widehat V_g}}. $$

The unadjusted statistic \(T_0(\tau_0)\) uses \(g_i=0\) in both the contrast and variance estimate. These formulas use the actual sequential probabilities, not a probability inferred from the final treatment counts.

Why the Variance Estimate Is Conservative

Write \(a_i=Y_i(1)-g_i\) and \(b_i=Y_i(0)-g_i\). Conditional on the information before assignment, \(D_i(g)\) equals \(a_i/\pi_i\) with probability \(\pi_i\), and equals \(-b_i/(1-\pi_i)\) otherwise. Its conditional mean is \(a_i-b_i=\tau_i\). Directly computing the second moment and subtracting the squared mean gives

$$ \begin{aligned} \operatorname{Var}\{D_i(g)\mid H_{i-1},X_i,Y_i(1),Y_i(0)\} &=\frac{a_i^2}{\pi_i}+\frac{b_i^2}{1-\pi_i}-(a_i-b_i)^2. \end{aligned} $$

The final term is nonnegative, so \(a_i^2/\pi_i+b_i^2/(1-\pi_i)\) is an upper bound on this conditional variance. The observed summand in \(\widehat V_g\) has conditional expectation \(a_i^2/\pi_i+b_i^2/(1-\pi_i)\): for example, the treatment part has expectation \(\pi_i a_i^2/\pi_i^2=a_i^2/\pi_i\). Thus \(\widehat V_g\) estimates a conservative bound, not the exact variance. For sequential assignment, the centered contributions are martingale differences; their cross-products have expectation zero under the sequential conditioning, so the corresponding variances add.

A standard large-sample justification requires more than the algebra above. In particular, the estimated bound must be stable relative to its scale, no single contribution may dominate, and a martingale central limit condition must hold. Under such conditions, if the actual variance divided by the estimated-bound scale converges to a value no greater than one, the studentized statistic is asymptotically normal with variance at most one. Using standard normal critical values is then conservative in the limit for two-sided weak-null tests. These are asymptotic claims, not an exact finite-sample guarantee.

Worked Example: Comparing the Two Contrasts

Consider four hypothetical participants assigned with \(\pi_i=0.5\). Participants 1 and 3 receive treatment and have observed outcomes 12 and 15; participants 2 and 4 receive control and have outcomes 8 and 11. Suppose a prespecified baseline prediction gives \(g_i=10,9,12,10\), respectively. These predictions use baseline information only.

With \(g_i=0\), the Horvitz–Thompson contrast is

$$ \widehat{\tau}_0 =\frac{2(12)-2(8)+2(15)-2(11)}{4} =\frac{24-16+30-22}{4} =\frac{16}{4}=4. $$

For the adjusted contrast, the observed residuals are 2, \(-1\), 3, and 1. Therefore,

$$ \widehat{\tau}_g =\frac{2(2)-2(-1)+2(3)-2(1)}{4} =\frac{4+2+6-2}{4} =\frac{10}{4}=2.5. $$

Now use the stated conservative variance estimator. Since \(1/\pi_i^2=1/(1-\pi_i)^2=4\), the unadjusted estimate is

$$ \widehat V_0 =\frac{4(12^2+8^2+15^2+11^2)}{4^2} =\frac{4(554)}{16}=138.5, \qquad \sqrt{\widehat V_0}\approx 11.77. $$

The adjusted estimate is

$$ \widehat V_g =\frac{4(2^2+(-1)^2+3^2+1^2)}{16} =\frac{4(15)}{16}=3.75, \qquad \sqrt{\widehat V_g}\approx 1.936. $$

For testing \(H_0:\tau=0\), the corresponding observed statistics are \(T_0=4/\sqrt{138.5}\approx0.3400\) and \(T_g=2.5/\sqrt{3.75}\approx1.291\). These four participants are too few to justify a normal approximation; the calculations illustrate the statistics, not a calibrated p-value. In particular, the change in the contrast is not an arithmetic error: adjustment can change the realized estimate while preserving its design expectation.

When Prognostic Adjustment Helps

A baseline variable is prognostic when it predicts outcomes, regardless of assigned treatment. If \(g_i\) captures predictable outcome level, the residuals \(Y_i-g_i\) may be much smaller than the outcomes themselves. The variance bound depends on their squared magnitudes, weighted by the assignment probabilities. This is why a useful prognostic prediction can reduce the scale of the studentized statistic's standard error.

A simple hypothetical illustration makes the mechanism clear. Suppose four participants have baseline predictions \(g_i=(10,12,14,16)\), potential control outcomes \((9,11,13,15)\), and potential treatment outcomes \((11,13,15,17)\). Every effect is 2. With \(g_i\) equal to the displayed predictions, the residuals under treatment are all 1 and under control are all \(-1\). At \(\pi_i=0.5\), each participant's variance-bound contribution is \(1^2/0.5+(-1)^2/0.5=4\); for the average, the bound is \(16/4^2=1\). With \(g_i=0\), the bound is \(2\sum_i\{Y_i(1)^2+Y_i(0)^2\}/16=175\). This comparison shows how centering can remove large baseline outcome levels from the bound. Here the actual randomization variance is zero because the effects are constant and the prediction is exactly halfway between the two potential outcomes; the conservative bound remains positive.

The example is deliberately favorable. Prognostic adjustment does not guarantee smaller variance: a poor or poorly scaled prediction can leave large residuals or increase their squared weighted sum. A prediction can be useful even if it is not a treatment-effect model, but it must be defined without using the current participant's unobserved outcome. Covariate selection and prediction should be prespecified where possible; if predictions are updated during the trial, the information used at each step must be available before that participant's assignment.

Worked Example: Using Unequal Sequential Probabilities

In a separate hypothetical trial with three participants, suppose the known probabilities are \(\pi_1=0.25\), \(\pi_2=0.50\), and \(\pi_3=0.75\). The observed assignments and baseline predictions are: participant 1 is treated, with \(Y_1=14\) and \(g_1=10\); participant 2 is control, with \(Y_2=9\) and \(g_2=10\); participant 3 is treated, with \(Y_3=16\) and \(g_3=15\). Assume the listed predictions were available before each assignment.

The three adjusted contributions are \(4/0.25=16\), \(-(-1)/0.50=2\), and \(1/0.75=1.333\). Hence

$$ \widehat{\tau}_g=\frac{16+2+1.333}{3}\approx6.444. $$

The variance estimate uses squared probability denominators. Its three numerator terms are \(4^2/0.25^2=256\), \((-1)^2/0.50^2=4\), and \(1^2/0.75^2\approx1.778\). Thus

$$ \widehat V_g=\frac{256+4+1.778}{3^2}\approx29.09, \qquad \sqrt{\widehat V_g}\approx5.393. $$

The large first term reflects the low treatment probability for participant 1. Replacing the actual probabilities with a common one, or using only the final allocation ratio, would not give the stated design-based calculation.

Diagnostics, Alternatives, and Limits

Before relying on a studentized contrast, check the ingredients rather than only the final statistic. Confirm that each \(\pi_i\) is the probability actually used by the design given the available history, and that it is recorded accurately. Inspect the range of the probabilities and the weighted residual terms: a probability near zero or one can make one participant dominate the estimate or variance. Check that \(g_i\) is baseline-based and temporally available, and compare adjusted and unadjusted residual scales. A large change in the answer or a very large variance contribution from one participant is a warning that approximation and precision may be fragile.

ApproachWhat it doesMain caution
Unadjusted studentized contrastUses inverse assignment probabilities and the observed outcomes, with \(g_i=0\).May carry avoidable baseline outcome variation into the contrast.
Prognostically adjusted studentized contrastCenters outcomes using predictable baseline values and studentizes with a design-based bound.Benefit is not guaranteed; valid calibration requires suitable asymptotic conditions.
Sharp-null randomization testEvaluates a statistic under the sharp null of no effect for every participant, using the assignment mechanism.Exactness for a sharp null does not by itself test only the weak average-effect null.

For a sharp-null randomization test, the statistic and any adjustment must be recomputed in each assignment generated according to the actual adaptive mechanism. A fixed-allocation permutation scheme is not a substitute when the design did not assign participants with fixed group counts. For the weak null, studentization supports asymptotic randomization-based inference; it does not turn the weak null into a sharp null or guarantee an exact finite-sample test.

The main limitations are practical as well as theoretical. Extreme assignment probabilities can cause unstable weights; a weakly prognostic or data-leaking predictor can undermine the intended adjustment; and small samples may not support the normal approximation. The conservative variance estimate may also be substantially larger than the actual variance because it omits the nonnegative squared-effect term from the exact conditional variance. Covariate adjustment can improve precision without changing assignment, but it cannot repair incorrect probabilities, interference, or a mismatch between the target estimand and the analysis.

Key takeaway: Use predictable baseline predictions to center the inverse-probability contrast, and studentize with a variance estimate that reflects the sequential assignment probabilities. Prognostic centering can improve weak-null inference, but the assignment rule stays intact and the asymptotic conditions still need to be credible.

Check Your Understanding

Work through each question, then compare your reasoning with the solutions below.

  1. For one participant with \(\pi_i=0.5\), \(Y_i(1)-g_i=1\), and \(Y_i(0)-g_i=1\), find the conditional variance of \(D_i(g)\) and the upper-bound contribution.
  2. Why must \(g_i\) be specified without using participant \(i\)'s post-assignment outcome?
  3. If a prognostic adjustment reduces the observed standard error, does that prove that the true design variance is lower for every possible assignment? Explain.
  4. What must be used to generate assignments for a sharp-null randomization test in a response-adaptive trial?

Solutions. (1) Here \(a_i=b_i=1\) and \(\tau_i=a_i-b_i=0\). The variance is \(1/0.5+1/0.5-0^2=4\), and the bound is also \(4\). (2) The prediction must be available before assignment so it is fixed with respect to the current assignment; using the realized outcome would make the centering depend on the treatment response and invalidate the stated conditional expectation argument. (3) No. The observed standard error is computed from one realization. A prediction can reduce the variance bound when it reduces weighted residual squares, but the gain is not guaranteed across assignments or for a poor prediction. (4) Use the trial's actual adaptive assignment mechanism, including its dependence on the observed history and baseline information, and recompute the statistic for each generated assignment.

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