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Bandit Formulations · Tutorial 292 of 501

Explore-Then-Commit Allocation

Build a balanced initial exploration stage, commit to the arm with the highest sample mean, and quantify the policy’s regret and limitations.

Advanced 12 min read

What You'll Learn

  • Define a finite-horizon explore-then-commit policy for multiple treatment arms.
  • State the reward and sampling assumptions needed for its analysis.
  • Calculate exploration regret and exact commitment regret in a two-arm example.
  • Derive a Hoeffding bound for the probability of committing to an inferior arm.
  • Recognize when a regret bound is loose and when exploration is inadequate.
  • Compare a one-time commitment policy with fixed allocation and continually adaptive approaches.

Why explore first and commit afterward?

A clinical trial may have a fixed enrollment horizon and a decision objective that rewards outcomes observed during the trial. Allocating every new patient to the arm with the best current sample mean can be risky: early estimates are noisy, and an initially lucky arm may stay ahead simply because it receives more patients. At the other extreme, continuing to divide patients evenly can spend many assignments on treatments that appear inferior. Explore-then-commit separates these competing aims into two stages: collect an initial balanced sample from every arm, then assign all remaining patients to the arm with the largest initial sample mean.

This policy is simple to specify and analyze, but it makes a consequential choice: once the exploration stage ends, new outcomes do not change the allocation. The number of initial observations per arm therefore controls how much the policy learns before it commits, while the remaining horizon controls the cost of choosing incorrectly.

Definition: Consider \(K\) treatment arms and a fixed horizon of \(T\) patients. Let \(m\) be a positive integer such that \(Km\leq T\). An explore-then-commit policy assigns \(m\) patients to each arm during exploration, calculates each arm’s sample mean, and assigns the remaining \(L=T-Km\) patients to an arm with the largest sample mean. A prespecified tie-breaking rule determines which arm is chosen if sample means tie.

Model and assumptions

For a standard analysis, represent the outcome on each patient as a reward \(X\) between 0 and 1, with larger values defined to be better. A binary success indicator is a common example. Arm \(i\) has an unknown mean reward \(\mu_i\), and \(\mu^*=\max_{1\leq i\leq K}\mu_i\) is the mean reward of a best arm. The sample mean from the \(m\) exploration outcomes on arm \(i\) is \(\bar{X}_i\). For Bernoulli rewards, it is the observed success proportion.

The usual finite-sample calculations assume that, for each arm, successive rewards are independent draws from a fixed distribution with mean \(\mu_i\), that rewards from different patients are independent, and that rewards lie in \([0,1]\). These conditions describe a stationary reward model: the distribution associated with an arm does not change over the trial or in response to which patients received it. The policy observes only the reward for the arm assigned to each patient, as in bandit feedback. In a clinical application, the reward must be defined before allocation, and its direction must be clear; an undesirable outcome cannot be treated as a benefit merely by calling it a reward.

These assumptions are substantive, not guaranteed by the allocation rule. If outcomes depend strongly on patient characteristics, calendar time, or previous treatment assignments, a single fixed mean \(\mu_i\) may not adequately describe an arm. The simple calculations below do not account for such changes or for the additional goals of estimating treatment effects and maintaining trial validity.

Constructing the policy

The exploration sample is balanced: each arm has exactly \(m\) outcomes before the policy compares arms. In a trial, the order of those assignments can be randomized while preserving the planned counts. After exploration, the policy selects an arm with maximum \(\bar{X}_i\), and every one of the \(L\) remaining patients is assigned to that arm. New observations in the commitment stage are not used to revise this choice.

1
Choose the exploration size.
Set \(m\) before observing outcomes, with \(m\geq1\) and \(Km\leq T\). A larger \(m\) gives more information per arm but leaves fewer patients for commitment.
2
Sample every arm.
Assign \(m\) patients to each arm and record the prespecified reward for each assigned patient.
3
Choose the empirical leader.
Calculate \(\bar{X}_i\) for every arm and apply the tie-breaking rule to select an arm with the largest value.
4
Commit for the remaining horizon.
Assign all \(L=T-Km\) remaining patients to the selected arm, without updating the decision.

Worked Example: Specifying a Three-Arm Policy

Suppose a hypothetical trial has \(T=60\) patients and \(K=3\) arms. The policy uses \(m=5\) patients per arm for exploration. Assume the exploration outcomes are binary successes, and suppose the observed success counts are 3, 4, and 3 for arms 1, 2, and 3, respectively.

The exploration sample means are \(\bar{X}_1=3/5=0.60\), \(\bar{X}_2=4/5=0.80\), and \(\bar{X}_3=3/5=0.60\). Thus arm 2 has the largest observed sample mean and is selected.

Exploration uses \(Km=3(5)=15\) patients, so the commitment stage has \(L=T-Km=60-15=45\) patients. The final allocation is 5 patients to arm 1, 50 to arm 2, and 5 to arm 3. These observed proportions determine the policy’s choice; they do not establish that arm 2 has the largest true mean.

Regret and the cost of commitment

The gap of arm \(i\) is \(\Delta_i=\mu^*-\mu_i\), which is nonnegative. The pseudo-regret of an allocation is the sum, over assigned patients, of the best arm’s mean reward minus the assigned arm’s mean reward. Expected pseudo-regret averages this quantity over the random outcomes and any random allocation. It measures expected reward foregone relative to assigning every patient to a best arm; it is not a measure of treatment-effect estimation accuracy.

Key result: For explore-then-commit with \(m\) exploration patients per arm, the expected exploration pseudo-regret is exactly \(m\sum_{i=1}^{K}\Delta_i\). If \(I\) denotes the arm selected for commitment, the expected commitment pseudo-regret is \(L\sum_{i=1}^{K}\Delta_i P(I=i)\). Arms with \(\Delta_i=0\) contribute zero regret, whether or not they are selected.

The exploration expression follows because each arm \(i\) is assigned exactly \(m\) times during exploration, and each such assignment has mean-reward gap \(\Delta_i\). For commitment, the policy assigns all \(L\) patients to the same selected arm \(I\). Conditional on selecting arm \(i\), the expected per-patient gap is \(\Delta_i\), so the conditional expected commitment regret is \(L\Delta_i\). Averaging over the possible selected arms gives the stated sum. Therefore, expected total pseudo-regret is the sum of these two contributions.

Worked Example: Exact Regret for Two Bernoulli Arms

Consider a hypothetical trial with two stationary Bernoulli arms, means \(\mu_1=0.60\) and \(\mu_2=0.40\), horizon \(T=20\), and \(m=4\). Suppose ties in the exploration sample means are always resolved in favor of arm 1. The arm gap is \(\Delta_2=0.20\), and the commitment stage has \(L=20-2(4)=12\) patients.

Write \(X_1\sim\operatorname{Binomial}(4,0.60)\) and \(X_2\sim\operatorname{Binomial}(4,0.40)\) for the exploration success counts. The policy selects the inferior arm 2 exactly when \(X_2>X_1\). By summing over possible values \(y\) of \(X_2\),

$$ P(I=2)=\sum_{y=1}^{4}P(X_2=y)P(X_1\leq y-1). $$

The four terms are \(0.3456(0.0256)=0.00884736\), \(0.3456(0.1792)=0.06193152\), \(0.1536(0.5248)=0.08060928\), and \(0.0256(0.8704)=0.02228224\). For example, \(P(X_2=2)=0.3456\), and \(P(X_1\leq1)=0.0256+0.1536=0.1792\). Adding the terms gives \(P(I=2)=0.1736704\).

Exploration regret is \(m\Delta_2=4(0.20)=0.80\) expected successes. Commitment regret is \(L\Delta_2P(I=2)=12(0.20)(0.1736704)=0.41680896\) expected successes. Total expected pseudo-regret is \(0.80+0.41680896=1.21680896\), or about 1.217 expected successes. This calculation uses the stated Bernoulli model and tie rule; changing the tie rule could change the probability of selecting arm 2.

A probability bound for selecting an inferior arm

Exact selection probabilities can be tedious to calculate. A useful general bound follows from Hoeffding’s inequality: if \(m\) independent observations lie in \([0,1]\) and have mean \(\mu\), then for \(a>0\), each one-sided deviation probability satisfies \(P(\bar{X}-\mu\geq a)\leq\exp(-2ma^2)\) and \(P(\mu-\bar{X}\geq a)\leq\exp(-2ma^2)\). This result requires independent, bounded observations with a common mean for the arm’s exploration sample.

Fix an optimal arm \(j\), so \(\mu_j=\mu^*\), and consider a suboptimal arm \(i\) with gap \(\Delta_i>0\). If arm \(i\) is selected, its sample mean must be at least as large as the optimal arm’s sample mean. That event requires at least one of two deviations: arm \(i\)’s sample mean is at least \(\Delta_i/2\) above its mean, or arm \(j\)’s is at least \(\Delta_i/2\) below its mean. Applying the one-sided inequality to each deviation and adding their probabilities yields

$$ P(I=i) \leq \min\left\{1,\;2\exp\left(-\frac{m\Delta_i^2}{2}\right)\right\}. $$

The minimum with 1 is included because a probability cannot exceed 1. Combining this bound with the regret decomposition gives the valid, though potentially loose, bound

$$ E[\text{pseudo-regret}] \leq m\sum_{i=1}^{K}\Delta_i +L\sum_{i:\Delta_i>0}\Delta_i \min\left\{1,\;2\exp\left(-\frac{m\Delta_i^2}{2}\right)\right\}. $$

The first term is exact; the second bounds commitment regret. The exponential term decreases as \(m\) or the gap increases, reflecting that more exploration or more clearly separated means makes an inferior selection less likely under the model. The bound does not say that a particular trial has a high probability of choosing correctly unless its numerical value supports that conclusion.

Worked Example: A Conservative Regret Bound

Suppose a hypothetical two-arm trial has \(T=50\), \(m=18\), and Bernoulli means \(\mu_1=0.65\) and \(\mu_2=0.35\). Thus the gap is \(\Delta_2=0.30\), there are \(L=50-2(18)=14\) commitment patients, and exploration regret is \(18(0.30)=5.40\) expected successes.

The Hoeffding bound for selecting arm 2 is \(P(I=2)\leq\min\{1,2\exp[-18(0.30)^2/2]\}=\min\{1,2\exp(-0.81)\}\). Using the unrounded exponential, \(2\exp(-0.81)=0.8897161324\ldots\), so this probability is bounded above by that value.

The commitment contribution is at most \(14(0.30)(0.8897161324\ldots)=3.7368078\ldots\) expected successes. Hence total expected pseudo-regret is at most \(5.40+3.7368078\ldots=9.1368078\ldots\), approximately 9.13681 expected successes (or conservatively, at most 9.1369 to four decimal places). The bound is valid but weak: it does not imply that the actual selection error probability is close to 0.8897.

Choosing \(m\), diagnosing trouble, and alternatives

The exploration size involves a finite-horizon trade-off. Increasing \(m\) raises exploration regret because more patients are assigned to arms below the best mean. At the same time, it reduces \(L\) and can lower the chance of committing to an inferior arm. Since the unknown gaps determine how much information is needed to distinguish arms, there is no universally best \(m\) without further assumptions or design objectives. A choice of \(m\) should be made with the horizon, plausible reward differences, and the consequences of allocation errors in view.

The probability bound can also reveal when a nominal guarantee is uninformative. For example, with \(m=25\) and gap \(\Delta_i=0.20\), the bound is \(\min\{1,2\exp[-25(0.20)^2/2]\}=\min\{1,2\exp(-0.50)\}=1\), since \(2\exp(-0.50)\approx1.2131\). It supplies no nontrivial upper bound on the error probability. Omitting the factor 2 and reporting about 0.6065 would be an arithmetic and inequality error: 0.6065 is approximately \(\exp(-0.50)\), not \(2\exp(-0.50)\).

Practical warning signs include an exploration sample so small that one or two outcomes determine the selected arm, nearly tied sample means, or a commitment period that is long compared with exploration. In those situations, a selection based on noisy early data can have substantial consequences. Conversely, if \(Km\) is close to \(T\), the policy has little or no commitment period and may offer little allocation benefit from selecting a leader. The Hoeffding bound may be especially loose when the gap is small; it is a guarantee, not a diagnostic test of the actual trial’s selection error.

Explore-then-commit differs from a fixed balanced allocation, which continues sampling all arms and therefore does not make a single irreversible allocation decision, but may allocate many patients to arms with lower mean reward. It also differs from a continually adaptive policy, such as Thompson sampling, that can use outcomes throughout enrollment to revise assignments. Those alternatives have different allocation and inferential properties; explore-then-commit’s distinctive feature is the one-time switch from balanced exploration to commitment. Compared with simply selecting the arm with the largest early mean, its deliberate initial sampling protects against leaving an arm nearly unobserved.

The policy’s limitations matter in clinical trials. Maximizing a prespecified reward during enrollment is not the same as maximizing patient welfare across all outcomes, estimating treatment effects precisely, or preserving a conventional randomized comparison. A long commitment to one arm may leave limited concurrent information about alternatives. If the reward distribution changes over time, if patient groups differ in ways relevant to response, or if safety and allocation constraints apply, the stationary bounded-reward calculation may not describe the actual design. The policy should therefore be judged against the trial’s scientific and ethical objectives, not regret alone.

Common mistake

  • Wrong reasoning: “Arm 2 had the largest exploration sample mean, so it is the best treatment.” Correction: The selected arm is only the empirical leader in a finite sample. Its true mean may be lower, and commitment makes the cost of that selection error depend on the remaining horizon.
  • Wrong reasoning: “The error bound is 0.8897, so the error probability equals 0.8897.” Correction: It is an upper bound, not an exact probability; the exact two-arm example above calculated the probability directly.
  • Wrong reasoning: “More exploration is always better.” Correction: More exploration can improve selection but also increases assignments made before commitment and reduces the number of patients remaining for it.
Key takeaway: Explore-then-commit is a finite-horizon policy with a clear allocation rule and a transparent regret decomposition. Its performance depends on both the exploration sample size and how reliably that sample distinguishes the best arm; commitment makes an early selection error costly when many patients remain.

Check Your Understanding

Try each problem, then check the concise solutions below.

  1. A trial has \(K=4\) arms, \(T=48\), and \(m=6\). How many patients are used for exploration and how many for commitment?
  2. Two arms have means 0.55 and 0.40. If \(m=8\) patients are assigned to each during exploration, what is the expected exploration pseudo-regret?
  3. With \(K=2\), \(T=30\), and \(m=5\), suppose the probability of committing to the inferior arm is 0.12 and the gap is 0.10. Calculate expected commitment regret.
  4. For a bounded-reward arm with \(m=10\) and gap \(\Delta=0.40\), calculate the stated Hoeffding upper bound on selecting that arm over an optimal arm. Is it an exact error probability?
  5. Why can a policy that commits after exploration be vulnerable even when the arm with the largest sample mean is selected correctly according to its rule?

Solutions. (1) Exploration uses \(Km=4(6)=24\) patients; commitment uses \(48-24=24\). (2) The suboptimal-arm gap is \(0.55-0.40=0.15\), so exploration regret is \(m\Delta=8(0.15)=1.20\) expected rewards. (3) Commitment regret is \(L\Delta P(I=\text{inferior})\), where \(L=30-2(5)=20\); hence \(20(0.10)(0.12)=0.24\) expected rewards. (4) The bound is \(\min\{1,2\exp[-10(0.40)^2/2]\}=\min\{1,2\exp(-0.80)\}\approx0.8987\). It is an upper bound, not an exact error probability. (5) The empirical leader can still be inferior in true mean because the exploration estimates are random; commitment then assigns every remaining patient to that arm without using later outcomes to correct the choice.

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