Why Removal Rules Need a Feasibility Check
When every update only adds balls, a valid urn composition stays nonnegative automatically. That guarantee disappears as soon as a rule can remove balls. An update that is sensible in a large urn may demand more balls of a color than are currently present. The rule must therefore be checked against the states where it can be applied, not just described in general terms.
This matters for response-adaptive randomization because the urn determines which treatment can be assigned next. A negative count is not a rare or inconvenient result of the arithmetic: it is not a possible urn state. A rule that produces one has not defined the intended randomization process at that state.
The word net is important. The increment describes the difference between the composition just before the draw and the composition after the replacement operation is complete. Thus, if the drawn ball is returned and two balls of another color are added, that is a net change of zero in the drawn color and plus two in the other color. If the drawn ball is not returned, its removal is part of the net change. This convention prevents a common ambiguity about whether the drawn ball is included in a stated replacement vector.
For a fixed composition and draw, feasibility is checked for each outcome that can occur. If an outcome has positive conditional probability, its update must be feasible. For a rule that depends on the current composition, the check must be repeated at every state the design can reach. Counts are integers because they represent balls, so noninteger changes also fail to define an ordinary urn.
The Coordinate-Wise Feasibility Condition
Let \(\Delta_j(\mathbf{x},i,y)\) denote the increment for color \(j\). The exact check is
for every positive-probability draw-and-outcome combination. Equivalently, if a rule removes \(r_j\) balls of color \(j\) and adds \(a_j\), with \(r_j,a_j\geq 0\), then its net increment is \(a_j-r_j\), and the check is \(x_j-r_j+a_j\geq 0\). It is not enough that the total number added across colors is large: an excess of one color cannot compensate for a negative count in another.
A nonnegative composition may still be unable to support another assignment. If the updated urn has no balls, then its counts are nonnegative but the next draw cannot be made. A design intended to continue randomizing must therefore also ensure a positive total count at each state from which another draw is required. Whether that condition holds can be checked by summing the updated counts; detailed analysis of total-size changes is a separate question.
Worked Examples
Worked Example: Checking Two Outcomes from One Draw
Suppose a hypothetical two-color urn has composition \(\mathbf{x}=(4,2)\), with colors A and B. A ball is drawn, and if it is A, the response is success with probability \(0.7\) and failure with probability \(0.3\). On success, the rule removes one A and adds two B, so \(\boldsymbol{\Delta}=(-1,2)\). On failure, it adds one A and removes one B, so \(\boldsymbol{\Delta}=(1,-1)\). Check the updates and calculate the probabilities of the two next compositions.
There are \(4+2=6\) balls, so the probability of drawing A is \(4/6=2/3\). For success, the updated composition is
Both counts are nonnegative integers. For failure,
which is feasible as well. The probability of the A-draw followed by success is \((2/3)(0.7)=0.4667\). The probability of the A-draw followed by failure is \((2/3)(0.3)=0.2000\). These are probabilities of reaching the stated next compositions through these particular branches; other branches, beginning with a B draw, are not included.
Worked Example: Detecting an Invalid Boundary Update
Consider a hypothetical state \(\mathbf{x}=(1,3)\). If B is drawn and the response is failure, suppose a proposed rule removes two A balls and adds one B. Failure has conditional probability \(0.25\) after this draw. Is the proposed update valid?
The net increment is \((-2,1)\), so the proposed composition is
The A count is negative. The rule is therefore invalid at this state. This is not repaired by the fact that the total count would be \(3\): one cannot have minus one A ball. Because the failure outcome has positive probability, the invalid branch is part of the stated design and cannot be ignored. The probability of drawing B in this state is \(3/4\), so the problematic branch has probability \((3/4)(0.25)=0.1875\), not zero.
One possible redesign is to remove two A balls only when at least two are present, and specify a different feasible update otherwise. That is a change to the rule and should be made explicitly. Simply replacing the negative count by zero after the calculation would not implement the originally stated rule.
Worked Example: A State-Dependent Rule at a Boundary
A hypothetical two-color urn is in state \((1,0)\). Only A can be drawn. Suppose success has probability \(0.6\) and failure has probability \(0.4\). A proposed boundary-aware rule says: on success, add one A and remove one B if \(x_B\geq 1\); otherwise add one A and remove no B. On failure, remove one A and add one B if \(x_A\geq 2\); otherwise remove no A and add one B. Find the possible next states and check feasibility.
At \((1,0)\), the success condition \(x_B\geq 1\) is false, so the success increment is \((1,0)\). The resulting state is
For failure, \(x_A\geq 2\) is also false, so the failure increment is \((0,1)\), giving
Both next states are feasible, and both have positive total counts. Because the only available draw is A, the next-state probabilities are \(0.6\) for \((2,0)\) and \(0.4\) for \((1,1)\). The boundary clauses matter: without the success clause, the instruction to remove one B would produce \((2,-1)\). A state-dependent rule is valid only when its branches are fully specified, including what happens when a removal condition is not met.
Why the Feasibility Condition Is Necessary
The check follows directly from what a composition represents. Before the update, there are \(x_j\) balls of color \(j\). The net change for that color is \(\Delta_j\), so after the update the count must be \(x_j+\Delta_j\). Since a count cannot be negative, feasibility requires \(x_j+\Delta_j\geq 0\) for each color. Conversely, if every updated coordinate is a nonnegative integer, the vector describes a valid composition. If continued sampling is required, at least one ball must remain, adding the separate requirement that the sum of the updated coordinates be positive.
This argument also explains why checking only the expected increment is insufficient. A favorable average can conceal an impossible outcome. For example, two outcomes might have increments \((-2,1)\) and \((1,0)\); their average could be nonnegative in a particular state even though the first branch creates a negative count. Feasibility concerns every possible realized update, not an average over them.
When the probability of drawing color \(i\) is \(x_i/\sum_j x_j\), and the conditional probability of response \(y\) is \(q(y\mid i,\mathbf{x})\), a particular branch has probability
Thus, an invalid branch can be excluded from the check only if its probability is exactly zero under the stated design. A small positive probability is still positive, as in the second example.
Diagnostics, Alternatives, and Limitations
A practical audit starts by listing the reachable compositions and the possible draws and outcomes at each one. For every branch, calculate the updated vector explicitly. Pay particular attention to colors with small counts: a removal that is harmless in a large state may fail when that color has zero or one ball. For a state-dependent rule, record the condition selecting each branch and verify that all relevant conditions are mutually intelligible and cover the states where the rule is supposed to apply.
Common warning signs include a negative coordinate, a fractional count, an unspecified fallback when a removal is impossible, or an update that leaves an empty urn when another assignment is required. A further warning sign is an implementation that silently clips negative values to zero. Clipping makes the actual update different from the stated update and can change which treatments are drawn later.
| Approach | Feasibility at a boundary | Main trade-off |
|---|---|---|
| Add-only increments | Nonnegative starting counts cannot become negative through the update. | Simple to check, but does not implement a rule that requires removal. |
| State-dependent guard or fallback | Can prevent removal when the required balls are unavailable. | Must define and check each branch; the fallback changes the rule at the boundary. |
| Stop randomization at a specified state | Avoids applying an invalid update if stopping occurs first. | Changes the planned continuation and must be specified in advance. |
No remedy is universally best. A guarded rule can preserve the design's intended behavior in ordinary states while defining a feasible boundary action, but the guard itself affects the process. Stopping may be appropriate for a design that explicitly allows termination, but it is not a way to claim that an otherwise invalid update is valid. Add-only updates are easier to audit, although they may not express the desired reinforcement or removal behavior.
The central limitation of a feasibility check is that it establishes only whether the urn operation can be carried out. It does not show that the rule has desirable allocation properties, that it improves patient outcomes, or that later statistical analyses are valid. Those are distinct design questions. Here the essential requirement is more basic: every update that the stated randomization can actually select must correspond to a possible urn composition.
Common Mistake
Check Your Understanding
Answer each question by applying the coordinate-wise check. Solutions are included after the questions.
- From state \((2,1)\), an outcome has increment \((-2,1)\). Is the resulting state feasible?
- From state \((3,1)\), an outcome has increment \((-2,1)\). Is it feasible, and can another draw be made?
- At state \((1,2)\), a draw has probability \(1/3\), and a response after that draw has probability \(0.2\). That response leads to an invalid update. What is the probability of that invalid branch?
- Why does clipping a negative count to zero fail to implement the stated replacement rule?
Solutions. (1) \((2,1)+(-2,1)=(0,2)\), so the counts are nonnegative integers and the update is feasible. (2) \((3,1)+(-2,1)=(1,2)\), which is feasible and has total \(3\), so another draw can be made. (3) The branch probability is \((1/3)(0.2)=0.06667\); because it is positive, the invalid update cannot be dismissed. (4) Clipping changes the prescribed net increment and therefore changes the urn process; a feasible fallback must be specified as part of the rule rather than applied silently.