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Asymptotic Theory for Adaptive Allocation · Tutorial 197 of 501

ODE Limits for Allocation Proportions

Learn how allocation recursions become deterministic ODEs on a logarithmic time scale, and which conclusions require more than trajectory tracking.

Advanced 13 min read

What You'll Learn

  • Write the allocation-proportion recursion as drift plus martingale noise and a vanishing perturbation.
  • Derive the ODE and its logarithmic time scale from the sample-size update.
  • Check conditions for the stochastic interpolation to track ODE trajectories.
  • Compute fluid trajectories for fixed and state-dependent allocation rules.
  • Distinguish an ODE equilibrium from a proven limit of the adaptive allocation process.
  • Diagnose failures caused by an incorrect drift, persistent perturbations, or boundary behavior.

Why an ODE describes allocation proportions

An adaptive allocation rule changes the probability of assigning the next participant as data accumulate. The resulting allocation proportions are random, but each new assignment changes a proportion by only about the reciprocal of the current sample size. On the right time scale, these small updates can be approximated by a deterministic differential equation. That approximation is useful because it turns a complicated random path into a trajectory whose direction and possible stationary points can be examined.

The ODE is not automatically a proof that the observed proportions converge to a particular target. Its role is more precise: under suitable regularity and noise conditions, the stochastic path follows ODE trajectories over bounded intervals of an algorithmic time scale. Conclusions about a specific limit require additional arguments.

Definition: For \(K\) treatment arms, let \(N_{n,i}\) be the number assigned to arm \(i\) among the first \(n\) participants, and let \(x_n=(N_{n,1}/n,\ldots,N_{n,K}/n)\) be the allocation-proportion vector. It lies in the probability simplex \(\Delta_K=\{x\in[0,1]^K:\sum_i x_i=1\}\). If the conditional probability of assigning participant \(n+1\) to arm \(i\) is approximately \(q_i(x_n)\), define the allocation drift \(h(x)=q(x)-x\). The fluid ODE is \(\dot{x}(t)=h(x(t))\), with initial state \(x(0)\in\Delta_K\).

Here \(q(x)\) is the vector of next-assignment probabilities, so its entries are nonnegative and sum to one. It is not the same object as the current allocation proportions \(x\). The difference \(q(x)-x\) is the mean direction in which those proportions change. The simplex is a natural state space: a valid trajectory should retain nonnegative components summing to one.

From the discrete update to fluid time

Let \(I_{n+1}\) be the \(K\)-vector that has a one in the arm receiving participant \(n+1\) and zeros elsewhere. The proportions satisfy the exact identity

$$ x_{n+1}=x_n+\frac{1}{n+1}(I_{n+1}-x_n). $$

If the conditional assignment probabilities are \(q_n\), then \(E(I_{n+1}\mid\mathcal{F}_n)=q_n\), where \(\mathcal{F}_n\) records the information available before the next assignment. Write \(q_n=q(x_n)+r_{n+1}\), where \(r_{n+1}\) represents the discrepancy between the actual probabilities and the limiting probability map. Define the martingale difference \(M_{n+1}=I_{n+1}-q_n\), which has conditional mean zero. Substitution gives

$$ x_{n+1}=x_n+a_{n+1}\{h(x_n)+M_{n+1}+r_{n+1}\}, \qquad a_{n+1}=\frac{1}{n+1}, \qquad h(x)=q(x)-x. $$

The increments \(a_{n+1}\) shrink, but their cumulative sum diverges. The appropriate continuous time is therefore \(t_n=\sum_{j=1}^{n}1/j\), which grows approximately like \(\log n\), rather than ordinary sample size \(n\). An interval of fixed length on this time scale spans a multiplicative change in sample size: increasing \(t\) by about one corresponds roughly to multiplying \(n\) by \(e\).

Tracking principle: Suppose the iterates remain in a compact state space, \(h\) is Lipschitz there, the assignment noise is a martingale difference with uniformly bounded conditional second moments, and the perturbations \(r_n\) vanish (or their weighted sums vanish over every bounded interval of fluid time). With \(a_n=1/n\), for which \(\sum_n a_n=\infty\) and \(\sum_n a_n^2<\infty\), the linearly interpolated allocation path tracks solutions of \(\dot{x}=h(x)\) over bounded fluid-time intervals, with approximation error tending to zero. This is a trajectory approximation, not by itself a convergence theorem.

The compactness condition is automatic for allocation proportions because they lie in the simplex. The Lipschitz condition ensures that nearby starting points generate nearby ODE solutions over finite intervals. Bounded assignment indicators give bounded noise; more generally, the stated conditional second-moment bound suffices for the weighted martingale noise to become negligible over late bounded fluid-time windows. Vanishing perturbations are important when assignment probabilities depend on estimated response quantities: the limiting map \(q(x)\) must accurately describe the conditional probabilities in the long run.

Why the tracking approximation is justified

The justification has two parts. First, over a bounded fluid-time interval, the accumulated random noise becomes small. Second, the deterministic part of the recursion is an Euler approximation to the ODE. To see the second point, freeze the state near some \(x\) for a short time increment \(a_{n+1}\). The ODE changes by approximately \(a_{n+1}h(x)\), matching the recursion’s drift term. Lipschitz continuity controls the error from the state changing during that increment.

For the noise, the weighted partial sums are sums of \(a_{n+1}M_{n+1}\). The martingale-difference property removes cross terms in their second moments, and bounded conditional second moments give a tail variance bounded by a constant times \(\sum a_{n+1}^2\). Since that series converges, the late noise accumulated over a bounded fluid-time window vanishes. If \(r_n\to0\), its weighted sum over such a window also vanishes: the total step size in the window is bounded, while the largest perturbation in a sufficiently late window is small. These facts, together with the Lipschitz bound, yield the finite-window tracking statement.

One immediate implication is that if the allocation proportions themselves converge to a point \(x^*\), then under these conditions \(h(x^*)=0\). If some coordinate \(h_i(x^*)\) were nonzero, the recursion would keep changing that coordinate in the same direction over a late fluid-time interval, by an amount approaching the interval length times \(h_i(x^*)\). That is incompatible with convergence to \(x^*\). Thus a convergent allocation proportion must be an equilibrium, meaning \(q(x^*)=x^*\). This necessary condition does not show that a particular equilibrium is reached.

Worked Example: A Fixed Assignment Probability

Consider two arms. Suppose the probability of assigning each new participant to arm 1 is always \(0.20\), regardless of the current allocation proportion. Let \(p_n\) be the proportion assigned to arm 1. The conditional assignment probability is constant, so the model has no state-dependent estimation perturbation; assignment indicators are bounded and their centered versions are martingale differences. The scalar drift is \(h(p)=0.20-p\), a Lipschitz function on \([0,1]\).

The fluid equation is \(\dot{p}(t)=0.20-p(t)\). With initial proportion \(p(0)=0.50\), solve the linear equation by multiplying the deviation from \(0.20\) by \(e^t\):

$$ p(t)=0.20+\{p(0)-0.20\}e^{-t} =0.20+0.30e^{-t}. $$

At fluid time \(t=1\),

$$ p(1)=0.20+0.30e^{-1} \approx 0.3104. $$

The ODE predicts movement from the initial share of \(0.50\) toward the assignment probability \(0.20\). It is a fluid-time prediction: \(t=1\) corresponds approximately to multiplying the sample size by \(e\), not to adding one participant. The trajectory approaches \(0.20\), which is the equilibrium of this fixed-probability rule.

Worked Example: A State-Dependent Assignment Rule

Suppose a two-arm design assigns the next participant to arm 1 with probability \(q(p)=0.20+0.60p\), where \(p\) is the current arm-1 proportion. For \(p\in[0,1]\), this probability ranges from \(0.20\) to \(0.80\), so it is valid; arm 2 receives the complementary probability. The resulting drift \(h(p)=q(p)-p=0.20-0.40p\) is Lipschitz, and the rule keeps the proportions in their valid state space. Assume assignments are drawn using these probabilities and no additional persistent estimation error is present.

The equilibrium solves \(0.20-0.40p^*=0\), giving \(p^*=0.50\). The ODE is

$$ \dot{p}(t)=0.20-0.40p(t). $$

Starting from \(p(0)=0.80\), the solution is

$$ p(t)=0.50+(0.80-0.50)e^{-0.40t} =0.50+0.30e^{-0.40t}. $$

At \(t=2\), \(e^{-0.8}\approx0.4493\), so \(p(2)\approx0.50+0.30(0.4493)=0.6348\). The drift is negative whenever \(p>0.50\) and positive whenever \(p<0.50\), as also follows from \(h(p)=0.20-0.40p\). This calculation describes the ODE path; the tracking conditions are what connect that path to the random allocation process.

Worked Example: Several Arms With a Fixed Target Map

Consider three arms with fixed assignment probabilities \(q=(0.20,0.30,0.50)\), and initial proportions \(x(0)=(0.50,0.30,0.20)\). The probabilities are nonnegative and sum to one; the assignment indicators are bounded, and the drift \(h(x)=q-x\) is Lipschitz. Thus the fluid ODE applies under the stated fixed-rule model.

For each arm, \(\dot{x}_i=q_i-x_i\), so the solution is \(x_i(t)=q_i+\{x_i(0)-q_i\}e^{-t}\). At \(t=\log 2\), \(e^{-t}=1/2\), giving

$$ x(\log 2) = (0.20,0.30,0.50) +\frac{1}{2}\{(0.50,0.30,0.20)-(0.20,0.30,0.50)\} =(0.35,0.30,0.35). $$

The components remain nonnegative and sum to \(1.00\), and the path moves halfway from its initial vector toward the fixed assignment-probability vector over this fluid-time interval. The equilibrium is \(x=q\).

Reading the ODE without overclaiming

The main alternatives answer different questions. Direct simulation of the adaptive design gives finite-sample paths and can reveal variability, but does not by itself establish a limiting behavior. Solving the equilibrium equation \(q(x)=x\) identifies stationary points, but does not describe how the system moves between them. The ODE provides that directional, large-sample-scale description. It still does not prove that the stochastic process converges to a particular equilibrium; tracking only says that late segments resemble ODE segments.

Several checks help determine whether the approximation is being used appropriately:

  • Check the drift. The assignment-probability vector is \(q(x)\); the proportion drift is \(q(x)-x\). Omitting the \(-x\) term models assignment probabilities, not changes in allocation proportions.
  • Check the state space. The map \(q\) should be a valid probability vector, and the ODE should preserve the simplex. If a proposed approximation leaves the simplex, revisit the drift or boundary assumptions.
  • Check perturbations. If response estimates or target estimates do not settle into the proposed map, \(r_n\) may not vanish. A deterministic ODE with the wrong drift may then give a misleading path.
  • Check time units. The ODE time is approximately logarithmic in sample size. Treating \(t=1\) as one participant is a scale error.

A common mistake is to find \(q(x^*)=x^*\) and conclude that the adaptive design must converge to \(x^*\). The equation only says that \(x^*\) is stationary for the ODE. Trajectory tracking does not rule out other equilibria or more complicated limiting behavior, nor does the equilibrium equation establish that the process enters the relevant region. Showing that a particular target attracts the stochastic process requires a separate stability argument.

The fluid approximation also has limits. It is a large-sample description, not a finite-sample confidence statement, and it does not quantify the random fluctuations around a trajectory. It can be unreliable when the probability map is discontinuous, the noise is not controlled, or the estimation perturbation persists. Near boundaries, one must verify that the rule remains valid and that the limiting ODE respects the simplex; a formal algebraic solution outside the simplex is not a possible allocation proportion.

Key takeaway: Under regular drift, controlled martingale noise, and negligible perturbations, allocation proportions track the ODE \(\dot{x}=q(x)-x\) on logarithmic fluid time. A convergent proportion must be an ODE equilibrium, but tracking alone does not identify which equilibrium, if any, is reached.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. For a two-arm design with constant probability \(q\) of assigning to arm 1, write the fluid ODE for its proportion \(p(t)\).
  2. Why is the relevant time scale approximately \(\log n\) rather than \(n\) for the proportion recursion?
  3. A proposed allocation rule has \(q(p)=0.10+0.70p\). Find its equilibrium proportion for arm 1.
  4. State one reason that solving \(q(x)=x\) is not enough to prove convergence of the allocation proportions.
  5. If an estimated target creates a discrepancy \(r_n\) that does not vanish, what part of the ODE tracking justification is at risk?

Solutions. (1) The drift is \(q-p\), so \(\dot{p}=q-p\). (2) Each update is scaled by \(1/(n+1)\); the accumulated step sizes are harmonic sums, \(\sum_{j=1}^n1/j\approx\log n\). (3) Solve \(0.10+0.70p=p\), giving \(p^*=1/3\approx0.3333\). The probability map is valid on \([0,1]\), since it ranges from \(0.10\) to \(0.80\). (4) An equilibrium is only a stationary point; the process might not approach it, and trajectory tracking alone does not establish convergence to a specific equilibrium. (5) The perturbation’s weighted contribution may remain substantial over bounded fluid-time windows, so the recursion need not track the proposed limiting ODE.

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