Why joint covariance kernels matter
A Gaussian limit for an allocation process and a Gaussian limit for a response process do not, by themselves, describe how the two processes vary together. That dependence is encoded by their cross-covariance kernel: the covariance between the allocation fluctuation at one time and the response fluctuation at another. Under adaptive allocation, this covariance can arise directly because the limiting martingale increments share outcome-driven noise. It can also arise indirectly when one process feeds into the dynamics of the other.
The calculation is useful whenever an inference concerns both allocation and response over time. Separate marginal variance formulas cannot recover the cross-covariance, and omitting it can give an incorrect variance for a sum, difference, or other joint quantity. The earlier material on Gaussian process limits supplies the setting: here the focus is on calculating the kernels once the joint limiting dynamics and their noise covariance are specified.
Joint dynamics and the covariance-kernel formula
Consider a \(d\)-dimensional limiting state \(X(t)\), for \(0\leq t\leq T\), whose first coordinates represent allocation fluctuations and whose remaining coordinates represent response fluctuations. Assume \(X(0)=0\) and that its limit has the linear stochastic differential representation
Here \(B(t)\) describes how current fluctuations affect later dynamics, \(L(t)\) maps the noise into the state, and \(M(t)\) is a mean-zero Gaussian martingale with independent increments and covariance rate \(Q(t)\). In particular,
The matrix \(Q(t)\) must be symmetric and positive semidefinite for each \(t\). Its off-diagonal blocks describe covariance between different sources of martingale noise at the same time. In an adaptive trial, these sources may include allocation and outcome contributions. The representation assumes that the stated limiting dynamics hold, that the coefficients are sufficiently regular for the integrals below, and that the joint process limit is Gaussian. It does not assert that every adaptive design has this form; the design’s own limiting argument must justify the representation.
Let \(\Phi(t,s)\) be the transition matrix for the deterministic dynamics: it solves \(\partial\Phi(t,s)/\partial t=B(t)\Phi(t,s)\), with \(\Phi(s,s)=I\). Solving the linear system gives
Because martingale increments from disjoint time intervals have zero covariance, only noise accumulated up to both \(t\) and \(u\) contributes to their covariance. Thus the kernel is
The allocation-response kernel is the corresponding block of this matrix. This formula accounts for both direct shared noise, through off-diagonal entries of \(Q(s)\), and propagated dependence, through the transition matrices and the way \(L(s)\) injects noise into the state.
Justification and a useful special case
To justify the formula, substitute the integral representation for \(X(t)\) and \(X(u)\). The covariance of two stochastic integrals with respect to the same Gaussian martingale is the integral of their integrands on the time interval shared by both integrals, weighted by the covariance rate \(Q(s)\). Their shared interval is \(0\leq s\leq\min(t,u)\), which yields the displayed expression. Noise after either observation time cannot affect that observation.
When there is no drift propagation and \(L(t)=I\), this reduces to \(K_X(t,u)=\int_0^{\min(t,u)}Q(s)\,ds\). For a constant covariance rate, the kernel is simply \(\min(t,u)Q\). When the process is a time-normalized cumulative process, its covariance must also be divided by the relevant time factors; that normalization is not part of the martingale covariance rate itself.
Worked Example: A Direct Shared-Noise Kernel
Suppose a hypothetical two-coordinate Gaussian limit has \(X(t)=(U(t),R(t))^\mathsf{T}=M(t)\) on \(0\leq t\leq1\), with constant covariance rate
The first coordinate is an allocation fluctuation and the second is a response fluctuation. The matrix is a valid covariance rate: its diagonal entries are positive and its determinant is \(0.25(0.16)-0.06^2=0.0364>0\). Therefore it is positive definite.
Since the dynamics are pure cumulative martingales, \(K_X(t,u)=\min(t,u)Q\). The cross-kernel is consequently
At \(t=0.4\) and \(u=0.8\), \(K_{UR}(0.4,0.8)=0.06(0.4)=0.024\). In these hypothetical units, the covariance between the allocation fluctuation at time 0.4 and the response fluctuation at time 0.8 is 0.024. The factor is the earlier time, not the later one, because only increments up to time 0.4 are shared.
Feedback can create covariance after the innovations
A cross-covariance kernel need not come only from a nonzero off-diagonal block in the innovation covariance. Even if allocation and response innovations are uncorrelated at each instant, a response process that accumulates past allocation fluctuations can be correlated with the allocation process. This is represented by off-diagonal terms in \(B(t)\), or equivalently by the transition matrices in the kernel formula.
Worked Example: Covariance Induced by Feedback
Consider a hypothetical limiting system with independent standard Brownian motions \(W_A\) and \(W_R\):
The allocation fluctuation \(U\) mean-reverts, and the response state \(R\) accumulates a contribution from \(U\). The Brownian noises are independent, so there is no direct cross-covariance between their increments. Nevertheless,
The independent \(W_R\) term has zero covariance with \(U(1)\), so \(\operatorname{Cov}\{U(1),R(1)\}=0.5\int_0^1\operatorname{Cov}\{U(1),U(r)\}\,dr\). The solution \(U(t)=\int_0^t e^{-(t-s)}dW_A(s)\) implies, for \(0\leq r\leq1\),
Therefore
The positive covariance is generated by the feedback term \(0.5U(t)\,dt\), despite independent innovation sources. Setting the cross-covariance to zero merely because \(W_A\) and \(W_R\) are independent would miss this contribution.
Normalizing a cumulative response process
Sequential response estimates often have a time normalization. A cumulative response fluctuation may be a martingale \(S(t)\), while the corresponding estimate fluctuation is proportional to \(S(t)/t\). This changes the kernel and makes the process undefined at \(t=0\) without a separate extension or rescaling. For positive \(t,u\), if \(U(t)\) and \(S(t)\) are cumulative martingales with constant joint covariance rate \(c\), then
Worked Example: Kernel for a Time-Normalized Response
Use the hypothetical covariance rate from the first example, so the allocation and cumulative response increments have cross-covariance rate \(c=0.06\). Let the response fluctuation at time \(u>0\) be \(S(u)/u\). At \(t=0.4\), \(u=0.8\), the cross-kernel is
The value 0.0300 is the covariance for the normalized response fluctuation, not for the cumulative response \(S(0.8)\). Dividing the response coordinate by \(u\) divides its covariance with any other coordinate by \(u\); dividing both coordinates would instead divide by the product of their two time factors.
Calculation checks, alternatives, and limitations
A reliable calculation starts by identifying which state coordinates are allocation fluctuations and which are response fluctuations, then deriving or specifying the joint noise covariance rate. Before integrating, check that each \(Q(t)\) is symmetric positive semidefinite. For a two-coordinate rate with variances \(q_A,q_R\) and cross-rate \(c\), this requires \(q_A\geq0\), \(q_R\geq0\), and \(c^2\leq q_Aq_R\). A cross-rate outside these bounds cannot be part of a valid covariance matrix.
- Check the time overlap. For cumulative martingales, the upper integration limit is \(\min(t,u)\), not \(\max(t,u)\).
- Check the coordinate order. The reverse cross-kernel satisfies \(K_{RU}(u,t)=K_{UR}(t,u)^\mathsf{T}\). A scalar cross-kernel need not be symmetric in \(t\) and \(u\) after time normalization or feedback.
- Check both sources of dependence. A zero direct cross-block in \(Q\) does not imply a zero process cross-kernel if feedback propagates one state into another.
- Check units and scaling. If a response coordinate is divided by time, its covariance is divided by the same time factor. The kernel’s units are the product of the units of its two coordinates.
The main alternative is to calculate the allocation and response marginal covariance kernels separately. That can be sufficient for a question about one process alone, but it does not determine the cross-kernel needed for joint uncertainty. Another approach is to derive the cross-kernel directly from a joint stochastic-integral representation; that can be shorter for simple cumulative processes, but is less convenient when feedback creates nontrivial transition matrices. The matrix formula handles both cases in one calculation.
The formula is conditional on having the correct joint Gaussian limit and the correct limiting dynamics. It does not establish those facts for a particular design. In particular, an adaptive rule may alter the drift, noise covariance, or scaling, and a proposed kernel must be reconsidered if the assumed limit fails. Near time zero, normalized response processes may also require special care because division by \(t\) can make their variance large or the process undefined.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- For cumulative martingales with constant cross-covariance rate \(c=0.08\), calculate the cross-kernel at \(t=0.3\), \(u=0.7\).
- If the response coordinate is \(S(u)/u\), what is its cross-kernel with \(U(t)\) when the cumulative cross-kernel is \(c\min(t,u)\)?
- Why can feedback create a nonzero allocation-response cross-kernel when the two innovation sources are independent?
- A proposed two-coordinate covariance rate has \(q_A=0.20\), \(q_R=0.10\), and \(c=0.16\). Is it a valid covariance rate?
Solutions. (1) The kernel is \(0.08\min(0.3,0.7)=0.08(0.3)=0.024\). (2) Divide the cumulative response coordinate by \(u\), giving \(c\min(t,u)/u\) for \(u>0\). (3) Feedback carries past fluctuations in one process into the other, so their states can be correlated even if their direct noise increments are not. (4) No. Positive semidefiniteness requires \(c^2\leq q_Aq_R\), but \(0.16^2=0.0256>0.20(0.10)=0.0200\).