Limiting Behavior at Gamma Zero and Large Gamma

Why the Endpoints of Gamma Matter

In a doubly adaptive biased coin design, the target allocation share and the probability of assigning the next participant to an arm are related, but they are not generally identical. The tuning parameter $\gamma$ controls how strongly the next assignment responds to the current difference between target and observed allocation shares. Examining $\gamma=0$ and the limit as $\gamma$ grows clarifies what the rule does at its two extremes.

This tutorial holds both the target and current allocation shares fixed while varying $\gamma$. That qualification matters: the assignment probability for a fixed state is a function of $\gamma$, whereas in an actual sequential trial the current shares can change after each assignment. The fixed-state limit therefore describes the rule at a given moment, not by itself the long-run behavior of a trial.

The Two-Arm Rule at a Fixed State

Let $\rho_1$ and $\rho_2$ be the target allocation shares for arms 1 and 2, and let $r_1$ and $r_2$ be their current allocation shares. The shares satisfy $\rho_1+\rho_2=1$ and $r_1+r_2=1$. Assume in this section that all four shares are strictly positive, so the ratios in the rule are defined. The tuning parameter satisfies $\gamma\geq 0$.

Definition: At a fixed state, the two-arm DBCD rule assigns the next participant to arm 1 with probability $$ \pi_1(\gamma)= \frac{\rho_1(\rho_1/r_1)^\gamma} {\rho_1(\rho_1/r_1)^\gamma+\rho_2(\rho_2/r_2)^\gamma}. $$ The probability of assignment to arm 2 is $\pi_2(\gamma)=1-\pi_1(\gamma)$. The target shares are $\rho_1,\rho_2$; $\pi_1(\gamma),\pi_2(\gamma)$ are next-participant probabilities.

The factor $\rho_i/r_i$ compares arm $i$'s target share with its current share. It exceeds 1 when the arm is underrepresented relative to its target and is below 1 when the arm is overrepresented. The target share $\rho_i$ also appears as a base weight in the rule; it is not removed when $\gamma$ changes.

For two arms, the rule is especially easy to study in odds form. Define the relative correction $R=(\rho_1/r_1)/(\rho_2/r_2)$. Dividing the arm 1 weight by the arm 2 weight gives

\frac{\pi_1(\gamma)}{1-\pi_1(\gamma)} = \frac{\rho_1}{\rho_2}R^\gamma, \qquad R=\frac{\rho_1/r_1}{\rho_2/r_2}.

Thus the odds start at the target odds $\rho_1/\rho_2$, then are multiplied by the relative correction raised to $\gamma$. This form makes both endpoint behaviors transparent.

Worked Example: Gamma Zero Gives the Target Share

Suppose the target shares are $\rho_1=0.70$ and $\rho_2=0.30$, while the current shares are $r_1=0.60$ and $r_2=0.40$. Find the assignment probability to arm 1 when $\gamma=0$.

Since each positive quantity raised to the zero power is 1, substitute directly into the rule:

\pi_1(0)= \frac{0.70(0.70/0.60)^0} {0.70(0.70/0.60)^0+0.30(0.30/0.40)^0} = \frac{0.70}{0.70+0.30} =0.70.

The next participant has a 0.70 probability of assignment to arm 1 and a 0.30 probability of assignment to arm 2. The current shares do not affect this calculation at $\gamma=0$. The result is a randomized assignment with probabilities equal to the target shares, not a guarantee that the next assignment or the eventual observed allocation will equal those shares.

Deriving the Two Endpoint Results

At $\gamma=0$, the correction factors disappear because $(\rho_i/r_i)^0=1$. More generally, the denominator becomes $\rho_1+\rho_2=1$, so

\pi_1(0)=\frac{\rho_1}{\rho_1+\rho_2}=\rho_1, \qquad \pi_2(0)=\rho_2.

This is the exact assignment rule at zero. It does not mean that the current shares are ignored by a design at every value of $\gamma$; it means that the current-share correction has exponent zero at this endpoint.

For the large-$\gamma$ limit, write $R$ as the ratio of the relative target-to-current shares. With positive shares, $R>0$. The odds identity gives three cases. If $R>1$, then $R^\gamma$ grows without bound, so the odds for arm 1 diverge and $\pi_1(\gamma)$ tends to 1. If $R<1$, then $R^\gamma$ tends to zero, so the odds tend to zero and $\pi_1(\gamma)$ tends to 0. If $R=1$, the correction leaves the odds unchanged at every $\gamma$.

Theorem: For fixed, strictly positive target and current shares in the two-arm DBCD rule, $$ \lim_{\gamma\to\infty}\pi_1(\gamma)= \begin{cases} 1, & \dfrac{\rho_1}{r_1}>\dfrac{\rho_2}{r_2},\\[6pt] \rho_1, & \dfrac{\rho_1}{r_1}=\dfrac{\rho_2}{r_2},\\[6pt] 0, & \dfrac{\rho_1}{r_1}<\dfrac{\rho_2}{r_2}. \end{cases} $$ Because both pairs of shares sum to 1, equality of the two ratios occurs exactly when $r_1=\rho_1$ and $r_2=\rho_2$.

To verify the equality statement, $\rho_1/r_1=\rho_2/r_2$ implies $\rho_1r_2=\rho_2r_1$. Using $r_2=1-r_1$ and $\rho_2=1-\rho_1$ gives $\rho_1(1-r_1)=(1-\rho_1)r_1$, hence $\rho_1=r_1$, and therefore $\rho_2=r_2$. At this state, $R=1$, so the odds remain $\rho_1/\rho_2$ and the assignment probability remains $\rho_1$ for every $\gamma$.

The limiting result follows directly from converting odds back to probability. If the odds $O(\gamma)$ tend to infinity, then $\pi_1=O/(1+O)$ tends to 1; if the odds tend to zero, $\pi_1$ tends to zero. If the odds remain $\rho_1/\rho_2$, the probability is $\rho_1$.

Worked Example: A Fixed State With Arm 1 Overrepresented

Let $\rho_1=0.70,\ \rho_2=0.30$, and suppose the current shares are $r_1=0.75,\ r_2=0.25$. Calculate the assignment probability for arm 1 at $\gamma=10$, then identify the limit as $\gamma$ grows.

The correction is

R=\frac{0.70/0.75}{0.30/0.25} =\frac{14/15}{6/5} =\frac{7}{9} \approx 0.7778.

Since $R<1$, the arm 1 odds decrease as $\gamma$ increases. At $\gamma=10$, use the odds formula, retaining the exact correction for the power:

\frac{\pi_1(10)}{1-\pi_1(10)} = \frac{0.70}{0.30}\left(\frac{7}{9}\right)^{10} \approx 2.3333(0.0810185) \approx 0.189043.

Convert odds to probability:

\pi_1(10)=\frac{0.189043}{1+0.189043} \approx 0.1590.

At this fixed state, the probability of assigning the next participant to arm 1 is about 0.1590 at $\gamma=10$, even though its target share is 0.70. This is because arm 1 is overrepresented relative to its target. As $\gamma$ tends to infinity, $R^{\gamma}$ tends to zero, so $\pi_1(\gamma)$ tends to 0. This is a limiting assignment probability at the stated fixed shares, not a claim that the observed trial allocation will be exactly 0% to arm 1.

How the Assignment Probability Changes With Gamma

The endpoint limits describe where the probability goes. A derivative gives a more precise statement about how it changes along the way. From the odds expression, define $L(\gamma)=\log\{\pi_1(\gamma)/(1-\pi_1(\gamma))\}$. Then $L(\gamma)=\log(\rho_1/\rho_2)+\gamma\log R$. The derivative of a probability expressed through its log odds is $\pi_1(1-\pi_1)$ times the derivative of those log odds, so

\frac{d\pi_1(\gamma)}{d\gamma} = \pi_1(\gamma)\{1-\pi_1(\gamma)\}\log R.

For interior shares, $0<\pi_1(\gamma)<1$, so the sign of this derivative is the sign of $\log R$. If $R>1$, the probability of arm 1 increases with $\gamma$; if $R<1$, it decreases; if $R=1$, it is constant. The derivative also shows why a tiny departure of $R$ from 1 can have a substantial effect when $\gamma$ is large: the log odds shift by $\gamma\log R$.

Worked Example: A Fixed State Where Arm 1 Is Underrepresented

Suppose the target is $(\rho_1,\rho_2)=(0.60,0.40)$, but the current allocation shares are $(r_1,r_2)=(0.40,0.60)$. Find $\pi_1$ for $\gamma=0,1,2$ and describe the direction of change.

First calculate the correction:

R=\frac{0.60/0.40}{0.40/0.60} =\frac{1.5}{0.6667} =2.25.

The target odds are $0.60/0.40=1.5$. At $\gamma=0$, the odds are 1.5 and $\pi_1(0)=1.5/(1+1.5)=0.60$. At $\gamma=1$,

\frac{\pi_1(1)}{1-\pi_1(1)}=1.5(2.25)=3.375, \qquad \pi_1(1)=\frac{3.375}{1+3.375}\approx0.7714.

At $\gamma=2$,

\frac{\pi_1(2)}{1-\pi_1(2)}=1.5(2.25)^2=7.59375, \qquad \pi_1(2)=\frac{7.59375}{1+7.59375}\approx0.8836.

Here arm 1 is underrepresented relative to its target, so $R>1$ and the assignment probability to arm 1 increases with $\gamma$. The limit is 1. The values at $\gamma=1$ and $\gamma=2$ remain probabilities for the next assignment, not allocations already achieved.

Assumptions, Diagnostics, and Limits

The derivations require a fixed pair of target shares, a fixed pair of current shares, the two-arm rule stated above, and positive shares so that each ratio is defined. The parameter $\gamma$ is nonnegative. At an exact state where an arm has current share zero, the displayed formula contains division by zero and cannot be applied as written. How a design handles such an initial or boundary state requires a separate specification; it should not be resolved by silently substituting zero into this formula.

Useful checks for a calculation are straightforward. The two assignment probabilities must sum to 1 and lie between 0 and 1. At $\gamma=0$, the result must equal the target share. If the current share of arm 1 is above its target, then $R<1$ and increasing $\gamma$ must decrease $\pi_1$; if it is below its target, $R>1$ and increasing $\gamma$ must increase $\pi_1$. Finally, if current and target shares agree, the correction is 1 and the assignment probability stays at the target for every $\gamma$. A result violating one of these checks usually indicates that the correction ratio or the odds-to-probability conversion has been reversed.

The endpoint comparison is also a comparison with simpler allocation choices. At $\gamma=0$, the rule reduces to fixed randomization at the target shares, regardless of the current shares. For positive $\gamma$, the rule adjusts next-patient probabilities according to the current departure from target. At very large $\gamma$, the fixed-state rule becomes increasingly concentrated on the arm with the larger relative target-to-current share. This is not the same as setting the target allocation itself to 0 or 1: the target remains fixed throughout.

A large-$\gamma$ probability near 0 or 1 can leave very few assignments to one arm for as long as the state remains similar. That can be undesirable for information or for maintaining assignment opportunities, even though the rule is moving in the direction that corrects the allocation imbalance. This endpoint analysis alone does not assess the full design’s ethical or inferential properties, nor does it establish how the shares evolve after repeated assignments. It isolates the immediate rule at a fixed state.

Common Mistake

A common mistake is to say, “At $\gamma=0$, the rule assigns according to the current shares,” or, “As $\gamma$ grows, the next-patient probability approaches the target share.” Both statements confuse current shares, target shares, and assignment probabilities. At zero, the probability equals the target share because the correction factors are raised to zero. At large $\gamma$, the probability tends to 1 or 0 according to which arm is relatively underrepresented, except at the state where current shares already equal target shares; there it remains at the target probability. Always inspect $R$ before interpreting the limit.

Key takeaway: At a fixed interior state, the two-arm DBCD odds equal the target odds multiplied by $R^\gamma$. Zero gamma recovers the target assignment probabilities; large gamma increasingly favors the arm with the larger target-to-current share ratio.

Check Your Understanding

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

For $(\rho_1,\rho_2)=(0.55,0.45)$, what is $\pi_1(0)$, regardless of the current shares?
If $\rho_1/r_1>\rho_2/r_2$, what happens to $\pi_1(\gamma)$ as $\gamma$ increases, and what is its limit?
Suppose the target and current shares are both $(0.30,0.70)$. What is $\pi_1(\gamma)$ for any nonnegative $\gamma$?
For target shares $(0.60,0.40)$ and current shares $(0.50,0.50)$, calculate $R$ and state whether $\pi_1$ increases or decreases with $\gamma$.
Why does the large-$\gamma$ limit at fixed shares not establish that an actual trial will assign every future participant to one arm?

Solutions. (1) $\pi_1(0)=\rho_1=0.55$. (2) Here $R>1$, so $d\pi_1/d\gamma>0$, and $\pi_1(\gamma)$ tends to 1. (3) The correction is $R=1$; the probability stays at the target, $\pi_1(\gamma)=0.30$. (4) $R=(0.60/0.50)/(0.40/0.50)=1.2/0.8=1.5$. Since $R>1$, $\pi_1$ increases with $\gamma$. (5) The limit holds with the shares fixed. In a trial, assignments can change the current shares, which changes $R$ and therefore changes later assignment probabilities.