Strong Laws for Adaptive Allocation Counts

Why Allocation Counts Need a Strong Law

In a response-adaptive trial, the chance of assigning the next participant to an arm can depend on the trial history. Consequently, assignment indicators are generally not independent and need not have the same distribution. A familiar law of large numbers for independent, identically distributed assignments is therefore not the right general tool. The useful structure is conditional: at each assignment, the allocation probability is known from the history available just before that assignment.

The key result separates the cumulative number assigned to an arm into two parts. One is the sum of its conditional allocation probabilities, which is determined by the evolving design and history. The other is a martingale sum representing the random deviation of assignments from those probabilities. Under bounded assignment increments, that second part divided by the number of assignments converges to zero almost surely. Thus, if the average conditional allocation probability has a limit, the realized allocation share has the same limit.

Predictable Allocation and Assignment Noise

Consider a sequential trial with \(K\) treatment arms. Let \(\mathcal{F}_{k-1}\) denote the information available immediately before assignment \(k\), including the past assignments and responses. For arm \(j\), let \(A_{j,k}=1\) if participant \(k\) is assigned to arm \(j\), and let \(A_{j,k}=0\) otherwise. Define the conditional allocation probability \(\pi_{j,k}=P(A_{j,k}=1\mid\mathcal{F}_{k-1})\). Since assignment \(k\) goes to exactly one arm, the indicators satisfy \(\sum_{j=1}^K A_{j,k}=1\) and the probabilities satisfy \(\sum_{j=1}^K\pi_{j,k}=1\).

The conditional-mean property follows directly from the definition of \(\pi_{j,k}\): \(E(D_{j,k}\mid\mathcal{F}_{k-1})=E(A_{j,k}\mid\mathcal{F}_{k-1})-\pi_{j,k}=0\). Thus \(D_{j,k}\) is a martingale difference with respect to the trial history. Summing the definition of \(D_{j,k}\) gives the exact decomposition

The first term may be random, because the history that determines future probabilities is random. It is nevertheless predictable one step at a time. The second term is the accumulated difference between actual assignments and their conditional probabilities.

Strong-Law Result and Justification

The crucial fact is that the martingale noise is negligible on the scale \(n\). Here is a direct justification for this bounded-increment setting. Because \(D_{j,k}\) is a martingale difference and \(|D_{j,k}|\leq1\), its second moment is at most 1. The martingale series with increments \(D_{j,k}/k\) has total expected squared increments bounded by

Martingale differences at distinct times are orthogonal in expectation: for \(r<k\), \(D_{j,r}\) is measurable with respect to \(\mathcal{F}_{k-1}\), so \(E(D_{j,r}D_{j,k})=E[D_{j,r}E(D_{j,k}\mid\mathcal{F}_{k-1})]=0\). Therefore the partial sums of \(\sum_k D_{j,k}/k\) form an \(L^2\)-bounded martingale. The martingale convergence theorem implies that this series converges almost surely.

To see why this implies that the unweighted noise average vanishes, set \(s_n=\sum_{k=1}^nD_{j,k}/k\) and \(s_0=0\). Since \(D_{j,k}=k(s_k-s_{k-1})\), summation gives

The limit is zero because a convergent sequence and its arithmetic averages have the same limit. Dividing the count decomposition by \(n\) now proves the theorem: its predictable term tends to \(q_j\), and its martingale term tends to zero. This proof uses no independence assumption and allows the allocation probabilities to change with the observed history.

Applying the Result in Two-Arm Trials

For a two-arm trial, label the treatment arm \(T\) and the control arm \(C\). Write \(A_k=A_{T,k}\), \(p_k=P(A_k=1\mid\mathcal{F}_{k-1})\), and \(N_T(n)=\sum_{k=1}^nA_k\). Then \(N_C(n)=n-N_T(n)\), and the theorem yields

If the average \(n^{-1}\sum_{k=1}^np_k\) converges almost surely to \(q\), then the treatment share converges almost surely to \(q\), and the control share converges to \(1-q\). This conclusion is conditional on the predictable average having a limit; the strong law for assignment noise does not itself ensure that the design’s probabilities converge or have a convergent average.

How to Use and Check the Conclusion

In a theoretical analysis, the conditional probability should match the actual randomization mechanism, given the full history on which that mechanism depends. If a recorded or simplified probability omits relevant history, the claimed conditional-mean-zero property may fail. In a simulation, plotting \(N_T(n)/n\) against \(n^{-1}\sum_{k=1}^np_k\) can help diagnose discrepancies or implementation errors, but a finite plot is not a proof of almost-sure convergence.

A common mistake is to argue that the treatment count is close to its expected value and conclude that its realized share has a particular limit. An expectation alone does not establish an almost-sure statement. The correction is to use the conditional decomposition: prove the centered assignment sum is negligible almost surely, and separately establish the limit of the average conditional probabilities. If that average does not converge, this theorem gives no fixed limiting share, even though the assignment noise remains negligible.

The result also clarifies how this argument differs from the independent, identically distributed law of large numbers. Independent, identically distributed allocation with a fixed probability is a special case, but adaptive assignments need not be independent. The martingale decomposition instead uses conditional probabilities and bounded increments, so it applies to history-dependent randomization. Its scope is limited: it identifies the allocation-count limit when the predictable average has one; it does not show that a particular adaptive design makes that average converge.