Why a Zero Allocation Target Needs More Than a Limit
An adaptive design can favor one arm so strongly that another arm’s share of assignments tends to zero. That limit alone does not say whether the less-favored arm stops receiving assignments. Its allocation fraction can vanish even while its cumulative number of assignments grows without bound. The distinction matters: a growing but sparse sample and a sample that eventually stops growing provide very different amounts of information.
To analyze this boundary behavior, focus on the probability of assignment at each step, not only on the fraction assigned by the end. The cumulative sum of those conditional probabilities determines whether continued sampling is possible and, under suitable conditions, gives the rate of growth of the arm’s count.
The probabilities \(\pi_t\) may depend on past assignments and observed responses, so they need not be deterministic. In a design that sets the current randomization probability equal to a time-varying target, \(\pi_t\) is that target. More generally, a target allocation fraction and the actual conditional probability used at each step need not be identical; rate conclusions require information about the latter.
The Cumulative Probability Determines Whether Sampling Continues
The process \(M_n=N_n-A_n=\sum_{t=1}^n(I_t-\pi_t)\) is a martingale: each increment has conditional mean zero given the information available before the assignment. Its conditional variance at step \(t\) is \(\pi_t(1-\pi_t)\), which is at most \(\pi_t\). These facts link the random count to its cumulative conditional allocation.
The count approximation follows from a martingale strong law. The key bound is that the conditional variance of each count increment is no greater than its conditional mean. Consequently, when \(A_n\) diverges, the martingale fluctuation is negligible relative to \(A_n\), so \(M_n/A_n\to0\) almost surely and \(N_n/A_n=1+M_n/A_n\to1\).
The finite-versus-infinite sampling statement is the conditional Borel–Cantelli criterion: for events whose conditional probabilities given the past are \(\pi_t\), infinitely many events occur almost surely when their conditional probabilities have a divergent sum; a finite sum yields only finitely many events almost surely. These results concern the count relative to cumulative assignment probability. They do not say that every adaptive design has the same rate; that rate depends on how \(\pi_t\) changes.
Worked Example: A Power-Law Boundary Target
Suppose, hypothetically, that at step \(t\) the probability of assigning a patient to a less-favored arm is \(\pi_t=0.4t^{-1/2}\). These probabilities are valid, since \(0<\pi_t\leq0.4\), and assignments are made using the stated conditional probabilities. Find the order of the count and allocation fraction.
The cumulative conditional allocation is \(A_n=0.4\sum_{t=1}^n t^{-1/2}\). Since \(\sum_{t=1}^n t^{-1/2}\sim2\sqrt{n}\),
Because \(A_n\to\infty\), the theorem gives \(N_n/A_n\to1\) almost surely. Thus \(N_n\sim0.8\sqrt{n}\) almost surely, while
For example, at \(n=10{,}000\), the leading-term approximation to the cumulative allocation is \(0.8\sqrt{10{,}000}=80\), and the corresponding allocation fraction is about \(80/10{,}000=0.008\), or 0.8%. These are asymptotic approximations, not an exact prediction of the realized count. The arm’s share is shrinking, but the count continues to grow.
Power-Law Rates and Their Boundary
Suppose, for a constant \(c>0\) and exponent \(\gamma>0\), that the conditional probability has the asymptotic form \(\pi_t\sim c t^{-\gamma}\). If this relation holds almost surely, then summing the probabilities gives the following regimes. The count conclusions use the theorem when the cumulative sum diverges; when it converges, the conditional Borel–Cantelli criterion applies.
| Decay exponent | Cumulative conditional allocation \(A_n\) | Count and fraction behavior |
|---|---|---|
| \(0<\gamma<1\) | \(A_n\sim\frac{c}{1-\gamma}n^{1-\gamma}\) | \(N_n\sim A_n\); the count diverges and \(R_n\to0\) |
| \(\gamma=1\) | \(A_n\sim c\log n\) | \(N_n\sim c\log n\); the count diverges slowly and \(R_n\to0\) |
| \(\gamma>1\) | \(A_n\) converges to a finite value | Only finitely many assignments occur almost surely; \(R_n\to0\) |
For \(0<\gamma<1\), the sum grows at the same leading order as the integral \(c\int_1^n x^{-\gamma}\,dx\), which is \(c(n^{1-\gamma}-1)/(1-\gamma)\). At \(\gamma=1\), that integral is \(c\log n\). For \(\gamma>1\), the corresponding integral to infinity is finite. The series test therefore separates persistent, though increasingly sparse, sampling from a finite total expected allocation.
Worked Example: Harmonic Decay Means Slow Continued Sampling
Consider a hypothetical design with \(\pi_t=0.3/t\). The probabilities are positive and no greater than 0.3, so they are valid conditional assignment probabilities. Since \(A_n=0.3H_n\), where \(H_n=\sum_{t=1}^n1/t\), and \(H_n\sim\log n\), the cumulative allocation diverges, but only logarithmically.
At \(n=1{,}000{,}000\), \(H_n\) is approximately 14.3927, so \(A_n=0.3H_n\approx4.318\). The expected cumulative allocation at this horizon is therefore only about 4.318 assignments; the realized count is random. Nevertheless, because the infinite sum \(\sum_t0.3/t\) diverges, the arm is assigned infinitely often almost surely over an indefinitely long sequence. Its fraction tends to zero, at the approximate rate \(0.3\log(n)/n\).
Worked Example: Faster Decay Can Stop Sampling
Now consider the hypothetical probabilities \(\pi_t=0.2t^{-1.2}\). They are valid because each is positive and at most 0.2. Since \(x^{-1.2}\) is decreasing,
The sum of conditional probabilities is finite. Thus the arm is assigned only finitely often almost surely, even though each step has a positive assignment probability. The expected total number of assignments is at most 1.2, since \(E[N_n]=E[A_n]\) for every \(n\) and the bound carries to the infinite count. In particular, for any positive integer \(k\), Markov’s inequality gives \(P(N_\infty\geq k)\leq1.2/k\), where \(N_\infty\) is the total number of assignments over all steps.
The arm’s allocation fraction also tends to zero, but here the count itself eventually stops increasing. A zero limiting fraction alone would not distinguish this case from either of the preceding examples.
What a Boundary Target Does—and Does Not—Tell You
If \(\pi_t\to0\), the average conditional allocation \(A_n/n\) tends to zero: it is the average of a sequence tending to zero. Also, \(N_n/n\to0\) almost surely, since the martingale fluctuations in \(N_n-A_n\), whose increments are bounded, are negligible relative to \(n\). This establishes a vanishing allocation fraction under those conditions. It does not determine whether \(A_n\) diverges. That requires studying the sum of the probabilities themselves.
A target allocation fraction of zero is even less informative if the design’s actual conditional probabilities do not match that target at every step. For rate analysis, identify the assignment probabilities the algorithm actually uses and examine their cumulative sum. A limiting target without a rate, a positive exploration floor, or other design details cannot by itself determine the number of assignments to the arm.
Diagnostics, Alternatives, and Limitations
- Track the count and fraction separately. Plotting \(R_n\) alone can hide whether \(N_n\) is growing. Report both, along with the cumulative conditional allocation \(A_n\) when it can be calculated.
- Check the decay of the actual probabilities. A fitted or specified power law can suggest a regime, but finite-sample behavior may differ. Compare the cumulative sum of observed conditional probabilities with the theoretical rate rather than inferring a rate from a single final fraction.
- Do not infer an infinite-horizon guarantee from a finite run. A slowly divergent sum, such as the harmonic case, can yield very few assignments at a practical sample size. Conversely, a finite-sample count cannot establish that the infinite series of probabilities converges.
- Consider an exploration floor when continued access to both arms is required. If the design enforces \(\pi_t\geq\varepsilon\) for a fixed \(\varepsilon>0\), then \(A_n\geq\varepsilon n\), so the arm cannot have a vanishing allocation fraction under that rule. This protects sampling but changes the allocation policy.
An interior target, bounded away from zero, leads to a fundamentally different allocation regime from a boundary target. The earlier material on Gaussian fluctuations around interior targets concerns fluctuations around a positive limiting allocation fraction; those conclusions should not be transferred automatically to an arm with a vanishing fraction. When counts grow sublinearly, the arm may have far less information than an arm sampled at a positive fraction of the trial. If its count stays finite, its response mean cannot be consistently estimated from an indefinitely growing number of observations on that arm. Even when its count diverges, the rate of growth must be checked before applying an inference result that assumes a particular sample-size rate.
A common mistake is to reason, “The allocation fraction tends to zero, so the arm eventually receives no assignments.” The correction is to distinguish the fraction \(N_n/n\) from the count \(N_n\): the fraction can vanish while the count diverges, as in the \(t^{-1/2}\) and \(t^{-1}\) examples. To decide between continued and finite sampling, examine whether \(\sum_t\pi_t\) diverges or converges.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- If \(\pi_t=0.6t^{-0.4}\), what are the leading orders of \(A_n\), \(N_n\), and \(R_n\)?
- If \(\pi_t=0.1/t\), does the arm receive infinitely many assignments almost surely? What is the leading order of its count?
- If \(\pi_t=0.05t^{-1.5}\), does the arm receive infinitely many assignments almost surely?
- Why does observing \(R_n=0.002\) at a particular sample size not establish that the arm will receive no further assignments?
Solutions. (1) Since \(0<0.4<1\), \(A_n\sim[0.6/(1-0.4)]n^{0.6}=n^{0.6}\). The cumulative allocation diverges, so \(N_n\sim n^{0.6}\) almost surely, while \(R_n\sim n^{-0.4}\to0\). (2) The harmonic sum diverges, so the arm is assigned infinitely often almost surely; \(A_n\sim0.1\log n\) and \(N_n\sim0.1\log n\). (3) The exponent 1.5 is greater than 1, so \(\sum_t0.05t^{-1.5}\) converges. The arm is assigned only finitely often almost surely. (4) The observed fraction is a ratio of a cumulative count to the total sample size. It gives neither the future conditional probabilities nor their cumulative sum, so it cannot distinguish a count that will keep growing slowly from one that will eventually stop.