Why Feedback Creates Covariance
In response-adaptive randomization, an estimated response parameter can influence the probability of assigning the next participant to each arm. The resulting allocation is therefore connected to the response estimates that helped determine it. Even when the randomization draw and the current response residual are conditionally independent, an estimate’s error can persist into later allocation decisions. This is a source of covariance between allocation and estimation fluctuations.
The purpose here is to isolate that covariance using a stable, linearized two-arm recursion. The recursion is a local approximation: its coefficients summarize how allocation deviations persist, how strongly an estimated response parameter changes later allocation, and how the estimation error evolves. It is not a universal formula for every response-adaptive design. In a particular design, the coefficients and innovation covariances must come from that design’s allocation rule and estimator.
A Linearized Two-Arm Recursion
Consider the local model
Here \(r\) describes persistence of allocation deviation in the absence of new inputs. The coefficient \(\beta\) is the local sensitivity of future allocation deviation to the current response-estimation error. For example, if a differentiable allocation rule assigns arm 1 with probability \(g(\hat\theta_m)\), then its local sensitivity at the target parameter \(\theta\) is \(g'(\theta)\); the coefficient in a chosen recursion can include additional scaling. The coefficient \(\kappa\) describes persistence of the estimation error on this local scale. The random terms \(\xi_{m+1}\) and \(\eta_{m+1}\) are new allocation and estimation innovations.
Assume the innovations have mean zero, are uncorrelated with the state \((U_m,V_m)\) before step \(m+1\), and have constant second moments
The innovation covariance matrix must be positive semidefinite; in particular, \(q_U\geq 0\), \(q_V\geq 0\), and \(q_{UV}^2\leq q_Uq_V\). Assume also \(|r|<1\) and \(|\kappa|<1\). These conditions make the triangular recursion stable and allow its second moments to settle to finite stationary limits. They are assumptions about this linearized model, not automatic facts about every adaptive design.
Deriving the Covariance Contribution
Write \(S_m=\operatorname{Var}(V_m)\), \(C_m=\operatorname{Cov}(U_m,V_m)\), and \(R_m=\operatorname{Var}(U_m)\). Expand the second recursion’s variance and use the fact that \(\eta_{m+1}\) is uncorrelated with the state at step \(m\):
For the cross-covariance, multiply out the two updates. The terms involving a current innovation and a previous state vanish by the innovation assumption, while the two same-step innovations contribute \(q_{UV}\):
At stationarity, set \(S_{m+1}=S_m=S\) and \(C_{m+1}=C_m=C\). Solving the first equation gives \(S=q_V/(1-\kappa^2)\). Substitution into the cross-covariance equation gives the central result:
This formula separates two contributions. The term \(\beta\kappa S/(1-r\kappa)\) is the covariance created by feedback and persistence: estimation error persists through \(\kappa\), influences later allocation through \(\beta\), and the allocation-estimation relationship persists through \(r\kappa\). The term \(q_{UV}/(1-r\kappa)\) propagates same-step innovation covariance. Either contribution can be positive or negative.
For completeness, the allocation variance obeys
The cross-term \(2r\beta C\) matters: adding the separate variances of the persistence term, feedback term, and innovation while omitting their covariance generally gives the wrong allocation variance.
Worked Example: Feedback With Independent New Innovations
Suppose a hypothetical local recursion has \(r=0.4\), \(\beta=0.6\), and \(\kappa=0.5\). Let the new estimation innovation have variance \(q_V=0.09\), and suppose the new allocation and estimation innovations are uncorrelated, so \(q_{UV}=0\). Assume the innovation covariance matrix is feasible, with \(q_U=0.16\). Both persistence coefficients have absolute value below 1, so the stated stability condition holds.
First calculate the stationary estimation-error variance:
Then calculate the allocation-estimation cross-covariance:
The covariance is positive despite zero same-step innovation covariance. In this model, a positive estimation error tends to be followed by a positive allocation deviation because \(\beta\) and \(\kappa\) are positive. The allocation persistence \(r\) further propagates that relationship.
The stationary allocation variance is
All variances and covariances here are on the hypothetical recursion’s chosen scales. They are not automatically variances in participant counts or response units; that interpretation depends on how \(U_m\) and \(V_m\) were scaled.
Worked Example: Shared Innovation Can Reverse the Sign
Now take \(r=0.5\), \(\beta=0.3\), \(\kappa=0.4\), \(q_V=0.25\), \(q_U=0.20\), and \(q_{UV}=-0.04\). These values satisfy the covariance feasibility condition because \((q_{UV})^2=0.0016\), whereas \(q_Uq_V=(0.20)(0.25)=0.05\). Both persistence coefficients again have absolute value below 1.
The stationary estimation-error variance is
The feedback part of the cross-covariance numerator is \((0.3)(0.4)(0.2976)\approx0.03571\). The shared-innovation part is \(-0.04\). Therefore,
The overall covariance is slightly negative. Positive feedback by itself would have produced a positive contribution, but the negative same-step innovation covariance is larger in magnitude in the numerator. This example shows why the sign cannot be inferred from \(\beta\) alone.
As a consistency check, the resulting allocation variance is
The negative cross-term reduces the allocation variance slightly; it does not make the variance negative.
Worked Example: Two Feedback Perturbations in an Allocation Innovation
Consider a deliberately simplified one-step linearization with target arm-1 allocation fraction \(v=0.25\). Let \(A\) be a Bernoulli allocation indicator with \(P(A=1)=v\). Suppose the allocation innovation includes the centered randomization draw and two independent response-estimation perturbations:
where \(h=0.01\) allocation-fraction units per response unit, and each \(Z_j\) is centered with variance \(\sigma^2/v\), using the hypothetical response variance \(\sigma^2=4\). Assume \(A,Z_1,Z_2\) are mutually independent for this illustration. The Bernoulli draw contributes \(v(1-v)=0.25(0.75)=0.1875\). Each feedback perturbation contributes
Thus the full allocation-innovation variance is
The covariance contribution from the first perturbation is
Its units are allocation-fraction units times response units. This covariance is not the variance contribution \(Q\): covariance is linear in \(h\), while the variance contribution is quadratic in \(h\). The calculation also depends on the simplifying independence assumptions; a real design’s innovation covariances must be derived from its own joint assignment-and-response mechanism.
Using the Calculation Responsibly
A practical derivation starts by writing down the state and its scale. A centered allocation count, an allocation proportion, and a square-root-scaled proportion are different quantities and generally have different recursion coefficients and innovation variances. Likewise, \(V_m\) must be the specific response-estimation error whose effect enters the allocation rule, not an unspecified “response noise.”
Differentiate the rule with respect to the estimated response parameter at the target value, and express the local effect in the scale used for \(U_m\) and \(V_m\). This identifies the feedback coefficient \(\beta\).
Calculate \(q_U\), \(q_V\), and \(q_{UV}\) for new innovations after accounting for what is already in the state. Do not assume \(q_{UV}=0\) merely because assignment occurs before the response is observed.
Verify \(|r|<1\), \(|\kappa|<1\), nonnegative innovation variances, and \(q_{UV}^2\leq q_Uq_V\). If these checks fail, the stationary formulas above do not apply as stated.
Calculate \(S\), then \(C\), keeping the feedback and direct innovation terms separate. State the scale and units of the resulting covariance.
Diagnostics should focus on the assumptions as well as the arithmetic. A fitted or derived recursion with \(|r|\) or \(|\kappa|\) near 1 can produce large stationary moments, so small coefficient errors can matter substantially. A negative calculated variance, or a covariance matrix violating \(C^2\leq RS\), signals an algebra error, an infeasible set of inputs, or a mismatch between the proposed recursion and the process being approximated. Persistent residual dependence on past states also contradicts the assumed innovation structure.
The recursion calculation is useful because it identifies which mechanism creates covariance and how that contribution propagates. It is not a substitute for deriving the design-specific joint limit law. Directly setting \(C=0\) discards feedback and any shared innovations; treating all fluctuations as independent can therefore give incorrect variance calculations. Simulation can check a derived approximation, but simulation alone does not establish the recursion’s assumptions or explain the source of a covariance.
| Approach | What it captures | Main caution |
|---|---|---|
| Stable linear recursion | Separates persistence, estimator-to-allocation feedback, and shared innovations | Requires correct local coefficients, scales, and innovation assumptions |
| Ignore cross-covariance | Simplifies calculations when independence is justified | Can omit both feedback and same-step covariance |
| Simulation | Shows finite-sample behavior under a specified design | Does not by itself justify a stationary or asymptotic approximation |
The main limitation is scope: this is a stable, time-homogeneous, linearized two-state calculation. If the allocation target is at a boundary, the rule is nonsmooth, the coefficients change with time, or the relevant state is not stable, these stationary equations may not describe the asymptotic covariance. In such cases the recursion itself and its limiting regime must be reconsidered rather than forcing the same formula onto the problem.
Check Your Understanding
Answer each question, then check the short solutions below.
- In the recursion, what does \(q_{UV}=0\) imply, and what does it not imply?
- If \(\beta=0\) but \(q_{UV}\neq0\), what is the stationary cross-covariance?
- For \(r=0.2\), \(\kappa=0.5\), \(\beta=0.4\), \(q_V=0.12\), and \(q_{UV}=0\), calculate \(S\) and \(C\).
- Why must \(q_{UV}^2\leq q_Uq_V\) hold?
- Name one reason the stationary formulas may fail to describe a particular adaptive design.
Solutions. (1) It removes same-step innovation covariance, but lagged feedback can still make \(C\) nonzero. (2) \(C=q_{UV}/(1-r\kappa)\), since the feedback numerator is zero. (3) \(S=0.12/[1-(0.5)^2]=0.16\), and \(C=(0.4)(0.5)(0.16)/[1-(0.2)(0.5)]=0.032/0.90\approx0.03556\). (4) The innovation covariance matrix must be positive semidefinite, which bounds the squared covariance by the product of the variances. (5) Examples include a nonstable recursion, time-varying coefficients, or a nonsmooth allocation rule at a boundary.