Local Sensitivity of Allocation Probabilities to Target Errors

Why Differentiate the Allocation Rule?

A response-adaptive design may use an estimated target share to determine the next participant’s assignment probability. If that estimated share changes slightly, how much does the assignment probability change? The derivative answers this local question. It describes the rule’s sensitivity to its target input at a specified allocation state; it does not, by itself, describe how uncertain the estimate is or how much the target estimate will change.

Here we study the two-arm doubly adaptive biased coin design (DBCD). Write $x$ for the current proportion assigned to arm 1, $\hat r$ for the estimated target share for arm 1, and $\gamma$ for the nonnegative biasing parameter. The analysis holds $x$ and $\gamma$ fixed while varying $\hat r$. This isolates the effect of a target error from other changes in the design state.

Definition: For $0<x<1$, $0<\hat r<1$, and $\gamma\geq 0$, the two-arm DBCD probability of assigning the next participant to arm 1 is $$ \pi_1(x,\hat r;\gamma) = \frac{\hat r\left(\frac{\hat r}{x}\right)^\gamma} {\hat r\left(\frac{\hat r}{x}\right)^\gamma+ (1-\hat r)\left(\frac{1-\hat r}{1-x}\right)^\gamma}. $$ The local sensitivity to the estimated target share is the partial derivative $$ \frac{\partial \pi_1(x,\hat r;\gamma)}{\partial \hat r}, $$ evaluated at a specified interior value of $\hat r$, with $x$ and $\gamma$ held fixed.

The numerator and denominator are the arm-specific DBCD weights. At alignment, $x=\hat r$, both adjustment factors are 1, so the weights are $\hat r$ and $1-\hat r$, proportional to the target shares. The assignment probability is then $\pi_1=\hat r$. Away from alignment, the adjustment factors tilt the probability toward the arm currently below its target.

Deriving the Sensitivity Formula

The probability is easier to differentiate after writing it in odds form. Let $w_1=\hat r^{\gamma+1}x^{-\gamma}$ and $w_2=(1-\hat r)^{\gamma+1}(1-x)^{-\gamma}$. Since $\pi_1=w_1/(w_1+w_2)$, its odds are $w_1/w_2$. Taking the logarithm gives

\log\left(\frac{\pi_1}{1-\pi_1}\right) =(\gamma+1)\log\left(\frac{\hat r}{1-\hat r}\right) -\gamma\log\left(\frac{x}{1-x}\right).

With $x$ and $\gamma$ fixed, differentiate both sides with respect to $\hat r$. The derivative of the left side is $\{1/[\pi_1(1-\pi_1)]\}(\partial\pi_1/\partial\hat r)$; the derivative of the right side is $(\gamma+1)/[\hat r(1-\hat r)]$. Equating them yields the sensitivity formula.

\boxed{ \frac{\partial \pi_1}{\partial \hat r} = \frac{(\gamma+1)\pi_1(1-\pi_1)} {\hat r(1-\hat r)} }

All quantities on the right are evaluated at the same $x,\hat r,\gamma$. For interior values, the derivative is positive: increasing the estimated target share for arm 1 increases its assignment probability, with the current share and $\gamma$ unchanged. The derivative is measured in probability change per unit change in target share. For example, a derivative of 2.95 means that a target increase of 0.01 corresponds locally to an assignment-probability increase of about 0.0295, or 2.95 percentage points.

Key result: At alignment $x=\hat r$, the probability equals $\hat r$, so $$ \left.\frac{\partial \pi_1}{\partial \hat r}\right|_{x=\hat r}=\gamma+1. $$ This is a local result: a small target-share change near alignment is amplified by approximately $\gamma+1$ in the assignment probability, provided the current share is held fixed.

Worked Calculations

Worked Example: Sensitivity Away from Alignment

Suppose the current arm 1 share is $x=0.30$, the estimated target is $\hat r=0.40$, and $\gamma=2$. Find the assignment probability and its local sensitivity to the target estimate.

The weights are $$ w_1=0.40\left(\frac{0.40}{0.30}\right)^2 =\frac{0.064}{0.09}=0.7111, \qquad w_2=0.60\left(\frac{0.60}{0.70}\right)^2 =\frac{0.216}{0.49}=0.4408. $$ Thus $$ \pi_1=\frac{0.7111}{0.7111+0.4408}\approx 0.6173. $$

Using the derivative formula, $$ \frac{\partial\pi_1}{\partial\hat r} = \frac{3(0.6173)(1-0.6173)}{(0.40)(0.60)} \approx 2.953. $$ A local target increase of 0.01 therefore predicts a probability increase of approximately $2.953(0.01)=0.02953$, or 2.953 percentage points. This is a first-order approximation, not an exact update. For comparison, substituting $\hat r=0.41$ into the original rule gives $w_1=0.41(0.41/0.30)^2\approx0.7658$, $w_2=0.59(0.59/0.70)^2\approx0.4191$, and $\pi_1\approx0.6463$. The exact increase, about 0.0290, is close to but not identical to the local prediction.

Worked Example: Sensitivity at Alignment

Let $x=\hat r=0.40$ and $\gamma=2$. At this state both adjustment factors equal 1, and the arm weights are 0.40 and 0.60. Thus $\pi_1=0.40$. The derivative at alignment is $$ \frac{\partial\pi_1}{\partial\hat r} = \frac{3(0.40)(0.60)}{(0.40)(0.60)} =3. $$

A target change of 0.01 predicts a probability change of 0.03, or 3 percentage points. To check against the nonlinear rule, keep the current share $x=0.40$ fixed and set $\hat r=0.41$. Then $$ w_1=0.41\left(\frac{0.41}{0.40}\right)^2=0.430756, \qquad w_2=0.59\left(\frac{0.59}{0.60}\right)^2=0.570497, $$ so $\pi_1=0.430756/(0.430756+0.570497)\approx0.4302$. The exact increase is about 0.0302, close to the derivative-based approximation of 0.0300. The small difference reflects curvature over this finite change.

Worked Example: Comparing Biasing Parameters

Suppose $x=0.25$ and $\hat r=0.40$. Compare local target sensitivity for $\gamma=1$ and $\gamma=3$. For $\gamma=1$, the weights are $0.40^2/0.25=0.64$ and $0.60^2/0.75=0.48$, giving $\pi_1=0.64/1.12=4/7\approx0.5714$. Therefore $$ \frac{\partial\pi_1}{\partial\hat r} = \frac{2(4/7)(3/7)}{(0.40)(0.60)} \approx2.041. $$

For $\gamma=3$, the weights are $0.40^4/0.25^3=1.6384$ and $0.60^4/0.75^3=0.3072$. Hence $\pi_1=1.6384/(1.6384+0.3072)=16/19\approx0.8421$, and $$ \frac{\partial\pi_1}{\partial\hat r} = \frac{4(16/19)(3/19)}{(0.40)(0.60)} \approx2.216. $$ At this particular state, the larger $\gamma$ gives a somewhat larger local sensitivity, but the comparison also reflects the different probabilities at which the derivative is evaluated. This example does not establish that sensitivity must increase with $\gamma$ for every state.

Interpreting the Derivative and Checking Its Use

A derivative is a local slope, not a guarantee about a change of any size. For a small perturbation $\delta$ of the target estimate, the first-order approximation is $$ \pi_1(x,\hat r+\delta;\gamma)-\pi_1(x,\hat r;\gamma) \approx \frac{\partial\pi_1}{\partial\hat r}\delta. $$ The closer $\hat r+\delta$ is to $\hat r$, the more appropriate the local approximation generally is. For a larger change, recompute the DBCD probability directly and compare it with the linear prediction.

The formula assumes an interior current share and target share, $0<x<1$ and $0<\hat r<1$, and a fixed $\gamma\geq0$. It also assumes that the probability is the stated, unclipped two-arm DBCD rule. If a design imposes an exploration floor, clips target estimates, or otherwise modifies the assignment rule, the derivative of that modified rule must be derived separately. At boundary values the displayed expression is not defined, so it should not be applied by direct substitution.

A practical check is to compute probabilities at $\hat r-\epsilon$, $\hat r$, and $\hat r+\epsilon$, for a small feasible $\epsilon$. The two finite-difference slopes, $(\pi_1(\hat r)-\pi_1(\hat r-\epsilon))/\epsilon$ and $(\pi_1(\hat r+\epsilon)-\pi_1(\hat r))/\epsilon$, should be close to the analytic derivative when $\epsilon$ is small. A sign disagreement or a large discrepancy can indicate an algebra or implementation error, an $\epsilon$ that is too large for a local check, or a modified rule that does not match the formula being differentiated.

Approach	What it answers	Main limitation
Analytic derivative	Instantaneous change in probability per unit target-share change at one specified state	Local; requires differentiability and the stated rule
Finite difference	Probability change over a chosen, finite target perturbation	Depends on perturbation size and can mix local slope with curvature
Direct recomputation	Exact probability under a specified alternative target value	Compares selected values rather than summarizing local sensitivity

Common Pitfall and Scope

A common mistake is to differentiate the assignment probability with respect to the current share $x$ when the question concerns error in the estimated target $\hat r$. These are different perturbations: changing $x$ changes the design state, whereas changing $\hat r$ changes the target input. The derivative derived here varies only $\hat r$. Another mistake is to treat the derivative as the probability itself or as a percentage change. It is an absolute probability slope; multiply it by a small target-share change to obtain the approximate absolute probability change.

This calculation isolates the rule’s response to a target perturbation. It does not quantify the uncertainty of the target estimate, the resulting variability over repeated assignments, or long-run operating characteristics. Those questions require additional analysis. Its value here is more focused: it provides a precise, state-specific way to compare how strongly an assignment probability reacts to small target errors.

Key takeaway: With current share $x$ and $\gamma$ fixed, the two-arm DBCD probability has target sensitivity $(\gamma+1)\pi_1(1-\pi_1)/[\hat r(1-\hat r)]$. Use it for small perturbations, verify it against direct probability calculations, and do not apply it to a modified rule or boundary state without a new derivation.

Check Your Understanding

Answer each question, then compare with the solutions below.

At $x=\hat r=0.30$ and $\gamma=1$, what are $\pi_1$ and the local sensitivity?
If a state has $\pi_1=0.50$, $\hat r=0.50$, and $\gamma=2$, calculate the derivative.
What does a derivative of 2.4 predict for a sufficiently small target increase of 0.005?
Why is a derivative calculation not a substitute for directly recalculating the probability after a large target change?

Solutions. (1) At alignment, $\pi_1=\hat r=0.30$, and the derivative is $\gamma+1=2$. (2) The derivative is $3(0.50)(0.50)/[(0.50)(0.50)]=3$. (3) The approximate probability increase is $2.4(0.005)=0.012$, or 1.2 percentage points. (4) A derivative describes the local slope; over a large change, curvature can make the linear approximation inaccurate, so the original probability rule should be evaluated at the new target value.