Worked Example: Comparing Gamma Values at the Same State

Comparing Gamma at a Fixed Allocation State

In a doubly adaptive biased coin design (DBCD), the chance of assigning the next participant to an arm depends on both the target allocation and the allocation already observed. The tuning parameter $\gamma$ controls how strongly the rule responds to a mismatch between those two quantities. To see that effect clearly, compare different values of $\gamma$ while holding the target shares and current allocation proportions fixed. If the state changes between calculations, the difference in assignment probabilities cannot be attributed to $\gamma$ alone.

This comparison is conditional on a specified state and fixed target. In an ongoing trial, the target may itself be calculated from response estimates, and the current allocation proportions change as participants are randomized. Here, treat the target and current proportions as given inputs, and calculate the next assignment probability at each chosen $\gamma$.

Definition: In a two-arm DBCD, let $\rho_1,\rho_2$ be the target allocation shares, with $\rho_1+\rho_2=1$, and let $\pi_1,\pi_2$ be the current allocation proportions, with $\pi_1+\pi_2=1$. If both target shares and both current proportions are positive, the conditional probability of assigning the next participant to arm $k$ is $$ \phi_k(\gamma)= \frac{\rho_k(\rho_k/\pi_k)^\gamma} {\rho_1(\rho_1/\pi_1)^\gamma+\rho_2(\rho_2/\pi_2)^\gamma}, \qquad k=1,2,\quad \gamma\geq 0. $$ The unnormalized quantity $\rho_k(\rho_k/\pi_k)^\gamma$ is the allocation weight for arm $k$; dividing by the sum of the two weights makes the assignment probabilities add to one.

What Is Held Fixed?

If $n_k$ participants have so far been assigned to arm $k$, out of $n=n_1+n_2$ participants, the current allocation proportion is $\pi_k=n_k/n$. The target shares $\rho_1,\rho_2$ are the desired allocation proportions at the state being considered. To compare gamma values at the same state, use the same $\rho_k$ and $\pi_k$ in every calculation. Only $\gamma$ changes.

At $\gamma=0$, each factor $(\rho_k/\pi_k)^\gamma$ equals one, so the assignment probability is the target share: $\phi_k(0)=\rho_k$. For positive $\gamma$, the ratio $\rho_k/\pi_k$ adjusts the arm's weight. A ratio greater than one indicates that the arm's current share is below its target share; a ratio less than one indicates that its current share is above target. The adjustment is then scaled relative to the other arm and normalized.

The calculations below assume two arms, positive target shares and current proportions, a fixed state, and a nonnegative gamma. These conditions make each ratio and weight well-defined. They describe the DBCD rule without an additional allocation floor or other modification.

Worked Example: A Target Share of 0.60 and a Current Share of 0.40

Suppose the target shares are $\rho_1=0.60$ and $\rho_2=0.40$, while the current proportions are $\pi_1=0.40$ and $\pi_2=0.60$. Both sets of shares sum to one and are positive. Calculate the probability of assigning the next participant to arm 1 for $\gamma=0,\ 0.5,\ 1,$ and $2$.

The two adjustment ratios are $\rho_1/\pi_1=0.60/0.40=1.5$ and $\rho_2/\pi_2=0.40/0.60=0.6667$. For $\gamma=0$, the weights are $0.60$ and $0.40$, so $\phi_1(0)=0.60/(0.60+0.40)=0.6000$.

For $\gamma=0.5$, the weights and normalized probability are $$ w_1=0.60(1.5)^{0.5}=0.7348,\qquad w_2=0.40(0.6667)^{0.5}=0.3266, $$ $$ \phi_1(0.5)=\frac{0.7348}{0.7348+0.3266}=0.6923. $$

For $\gamma=1$, the weights are $w_1=0.60(1.5)=0.9000$ and $w_2=0.40(0.6667)=0.2667$. Thus $$ \phi_1(1)=\frac{0.9000}{0.9000+0.2667}=0.7714. $$

For $\gamma=2$, the weights are $w_1=0.60(1.5)^2=1.350$ and $w_2=0.40(0.6667)^2=0.1778$. Therefore $$ \phi_1(2)=\frac{1.350}{1.350+0.1778}=0.8836. $$

At this fixed state, arm 1 is below its target share, and its assignment probability rises as $\gamma$ increases through these values. The probability is not the current proportion, nor is it automatically the target share once $\gamma$ is positive.

Why the Probability Moves in That Direction

The two-arm rule can be understood through the odds of assigning to arm 1 rather than arm 2. Taking the ratio of the two probabilities cancels their common normalizing denominator:

\frac{\phi_1(\gamma)}{\phi_2(\gamma)} = \frac{\rho_1}{\rho_2} \left[ \frac{\rho_1/\pi_1}{\rho_2/\pi_2} \right]^\gamma.

The first factor is the target odds. The bracketed factor determines how the current allocation state changes those odds. If $\rho_1>\pi_1$, then $\rho_1/\pi_1>\rho_2/\pi_2$, so the bracketed factor exceeds one. Increasing $\gamma$ therefore increases the odds of assigning to arm 1. If $\rho_1<\pi_1$, the bracketed factor is below one, and increasing $\gamma$ decreases those odds.

A derivative makes the monotonicity explicit. Write $L=\log[(\rho_1/\pi_1)/(\rho_2/\pi_2)]$. The odds expression implies that $\phi_1(\gamma)=1/(1+(\rho_2/\rho_1)e^{-\gamma L})$. Differentiating gives

\frac{d\phi_1(\gamma)}{d\gamma} = \phi_1(\gamma)\bigl(1-\phi_1(\gamma)\bigr)L.

Because both assignment probabilities are between zero and one, the product $\phi_1(\gamma)(1-\phi_1(\gamma))$ is positive. Consequently, the sign of the derivative is the sign of $L$: positive when $\rho_1>\pi_1$, negative when $\rho_1<\pi_1$, and zero when $\rho_1=\pi_1$. This establishes the direction of change for every finite $\gamma\geq0$, not just for selected values in a table.

This result is conditional on the fixed state. It does not claim that the observed allocation proportion instantly becomes the target, or that a particular assignment is guaranteed. The rule changes the probability of the next assignment; random assignment and later changes in state remain part of the design.

Worked Example: Arm 1 Is Above Its Target Share

Suppose $\rho_1=0.40,\ \rho_2=0.60$, while $\pi_1=0.60,\ \pi_2=0.40$. The target and current shares are positive and each pair sums to one. Compare $\gamma=0,\ 1,$ and $2$.

The adjustment ratios are $0.40/0.60=0.6667$ for arm 1 and $0.60/0.40=1.5$ for arm 2. At $\gamma=0$, $\phi_1(0)=0.40$.

At $\gamma=1$, the weights are $w_1=0.40(0.6667)=0.2667$ and $w_2=0.60(1.5)=0.9000$. Thus $$ \phi_1(1)=\frac{0.2667}{0.2667+0.9000}=0.2286. $$ At $\gamma=2$, the weights are $w_1=0.40(0.6667)^2=0.1778$ and $w_2=0.60(1.5)^2=1.350$, giving $$ \phi_1(2)=\frac{0.1778}{0.1778+1.350}=0.1164. $$

Arm 1 is above its target share, so the rule reduces its next-assignment probability below the target share. The larger gamma value makes that reduction stronger at this fixed state.

Reading and Checking the Calculations

A useful hand-calculation routine is to write down both adjustment ratios first, calculate both weights using the same gamma, and then normalize. The probability for arm 2 is $1-\phi_1(\gamma)$, which provides a quick check. The resulting probabilities must be between zero and one and sum to one; a weight itself need not be a probability and may exceed one.

The comparison is most informative when the target and current shares differ. If $\rho_1=\pi_1$, then both adjustment ratios equal one, so the weights equal the target shares for every gamma and $\phi_1(\gamma)=\rho_1$. In that case, changing gamma has no effect at that state.

Worked Example: A Balanced Target With an Underallocated Arm

Suppose both target shares are $0.50$, while the current proportions are $\pi_1=0.20$ and $\pi_2=0.80$. The target and current shares are positive and sum to one. Compare $\gamma=0,\ 0.5,$ and $1$.

The ratios are $0.50/0.20=2.5$ and $0.50/0.80=0.625$. At $\gamma=0$, $\phi_1(0)=0.50$. At $\gamma=0.5$, the weights are $$ w_1=0.50\sqrt{2.5}=0.7906,\qquad w_2=0.50\sqrt{0.625}=0.3953, $$ so $\phi_1(0.5)=0.7906/(0.7906+0.3953)=0.6667$. At $\gamma=1$, the weights are $w_1=0.50(2.5)=1.250$ and $w_2=0.50(0.625)=0.3125$, giving $\phi_1(1)=1.250/(1.250+0.3125)=0.8000$.

Although the target is balanced, arm 1 is currently underallocated, so its assignment probability exceeds $0.50$ for positive gamma. This example shows why the target alone is not enough to determine the probability when $\gamma>0$: the current allocation proportions matter too.

Comparisons, Diagnostics, and Limitations

The direct comparison is with $\gamma=0$, which assigns according to the target shares regardless of the current allocation proportions. Positive gamma modifies the target odds in the direction indicated by the current imbalance; a larger gamma strengthens that odds adjustment at a fixed state. These are different conditional probabilities, not different target allocations.

Several simple checks help identify a calculation or interpretation error:

Check the state: confirm that the same target and current shares are used for every gamma being compared.
Check the ratios: an arm below target has $\rho_k/\pi_k>1$; an arm above target has $\rho_k/\pi_k<1$.
Check normalization: divide each weight by the sum of both weights, and verify that the two probabilities sum to one.
Check the direction: when arm 1 is below target, increasing gamma should increase $\phi_1$; when it is above target, increasing gamma should decrease $\phi_1$.

A common mistake is to increase only the weight for an underallocated arm and leave the other arm's weight unchanged, or to report the adjusted weight as the assignment probability. The DBCD formula adjusts both arm weights and then normalizes them. Another mistake is to compare results calculated at different current proportions and attribute the entire difference to gamma. Recalculate each gamma from the identical state before drawing that conclusion.

The formula as written requires positive current proportions and positive target shares. If an arm has no participants yet, its current proportion is zero and the ratio $\rho_k/\pi_k$ is undefined; a separate initialization or boundary-handling rule is needed rather than substituting zero into this formula. Likewise, if a target share is zero, the positive-share calculation and its odds interpretation do not apply as written. Additional restrictions, such as minimum assignment probabilities, modify the probabilities and must be included explicitly before comparing them.

Finally, increasing gamma is not a guarantee that a finite trial will achieve its target at each step. The calculation gives a conditional randomization probability for one assignment at a specified state. It does not remove the randomness of the assignment or account for the state changing after that assignment.

Key takeaway: At a fixed, positive two-arm state, $\gamma=0$ gives the target probability. Positive gamma changes the target odds according to the relative target-to-current ratios: the probability rises for an underallocated arm and falls for an overallocated arm as gamma increases.

Check Your Understanding

Answer each question using the two-arm rule, keeping the target and current shares fixed whenever gamma values are compared. Solutions follow the questions.

If $\rho_1=0.70,\ \rho_2=0.30,\ \pi_1=0.50,$ and $\pi_2=0.50$, what is $\phi_1(0)$? Is arm 1 above or below target?
For the state in question 1, calculate $\phi_1(1)$ by finding both weights and normalizing.
If the current shares equal the target shares, how does $\phi_1(\gamma)$ depend on gamma?
For $\rho_1=0.30,\ \rho_2=0.70,\ \pi_1=0.50,$ and $\pi_2=0.50$, should $\phi_1$ increase or decrease as gamma increases from zero? Explain.
Why is an unnormalized arm weight not itself an assignment probability?

Solutions. (1) At $\gamma=0$, $\phi_1(0)=\rho_1=0.7000$. Since $\pi_1=0.50<\rho_1=0.70$, arm 1 is below target. (2) At $\gamma=1$, the ratios are $0.70/0.50=1.4$ and $0.30/0.50=0.6$. The weights are $w_1=0.70(1.4)=0.9800$ and $w_2=0.30(0.6)=0.1800$, so $\phi_1(1)=0.9800/(0.9800+0.1800)=0.8448$. (3) If $\rho_k=\pi_k$ for both arms, both adjustment ratios are one. The weights are the target shares for every gamma, so $\phi_1(\gamma)=\rho_1$. (4) Arm 1 is above target because $0.30<0.50$, so its assignment probability decreases as gamma increases. (5) The two weights need not sum to one; dividing each by their sum produces probabilities that sum to one.