next_allocation_rate_BAR {BAR}

Calculate the allocation probability for the next block of new patients using Bayesian adaptive randomization

Description

Calculate updated allocation probability for each arm based on the accumulative data with binary outcomes

Usage

next_allocation_rate_BAR(n, success_count, tot_num,
                         power_c = "n/2N",
                         lower_bound = .05,
                         control_arm = "",
                         seed = 100)

Arguments

n

the number of patients enrolled for each arm

success_count

the number of responders for each arm

tot_num

the total number of patients enrolled for the trial. If this number cannot be pre-planned, the user can choose argument "power_c" to be numeric instead of "n/2N". In this case, even if the "tot_num" is given a number, this number will not be used

power_c

the power correction of allocation probability. The default value is power_c = "n/2N" and can also be numeric, e.g., power_c = .5

lower_bound

the lower bound of the allocation probability. It must between 0 and \frac{1}{K}. The default value is lower_bound = .05; K indicates total number of arms (including control arm)

control_arm

if this argument is "fixed", then allocation probability of control arm (the first slot) will be fixed to \frac{1}{K}. The default of this argument will return unfixed results; K indicates total number of arms (including control arm)

seed

the seed. The default value is seed = 100

Details

We show how the updated allocation probabilities for each arm are calculated.

Treatments are denoted by k = 1,\ldots,K.\;N is the total sample size. If no burn-in(s), the BAR will be initiated start of a study, that is, for each enrolled patient, n = 1,\ldots,N, the BAR will be used to assign each patient. Denoting the true unknown response rates of K treatments by \pi_{1},\ldots,\pi_{K},\;we can compute K posterior probabilities: r_{k,n} = Pr(\pi_{k} = max\{\pi_{1},\ldots,\pi_{K}\}\;|\;Data_{n}), here, n refers to the n-th patient and k refers to the k-th arm. We calculate the updated probabilities of the BAR algorithm according to the following steps.
\;
Step 1: (Normalization) Normalize r_{k,n} as r_{k,n}^{(c)} = \frac{(r_{k,n})^{c}}{\sum_{j=1}^{K}(r_{j,n})^{c}}, here \;c = \frac{n}{2N}.

Step 2: (Restriction) To avoid the BAR sticking to very low/high probabilities, a restriction rule to the posterior probability r_{k,n}^{(c)} will be applied:

Lower\;Bound \le r_{k,n}^{(c)} \le 1 - (K - 1) \times Lower\;Bound,

0 \le Lower\;Bound \le \frac{1}{K}

After restriction, the posterior probability is denoted as r_{k,n}^{(c,re)}.

Step 3: (Re-normalization) Then, we can have the updated allocation probabilities by the BAR denoted as:

r_{k,n}^{(f)} = \frac{r_{k,n}^{(c,re)}\times(\frac{r_{k,n}^{(c,re)}}{\frac{n_{k}}{n}})^{2}}{\sum_{j=1}^{K}\{r_{j,n}^{(c,re)}\times(\frac{r_{j,n}^{(c,re)}}{\frac{n_{j}}{n}})^{2}\}}

where n_{k} is the number of patients enrolled on arm k up-to-now.

Step 4: (Re-restriction) Finally, restricts again by using

Lower\;Bound \le r_{k,n}^{(f)} \le 1 - (K-1) \times Lower\;Bound,

0 \le Lower\;Bound \le \frac{1}{K}

and denote r_{k,n}^{(ff)} as the allocation probability used in the BAR package.

Value

next_allocation_rate_BAR() returns the updated allocation probability for each arm

Author(s)

Chia-Wei Hsu, Haitao Pan

References

Wathen JK, Thall PF. A simulation study of outcome adaptive randomization in multi-arm clinical trials. Clin Trials. 2017 Oct; 14(5): 432-440. doi: 10.1177/1740774517692302.

Xiao, Y., Liu, Z. & Hu, F. Bayesian doubly adaptive randomization in clinical trials. Sci. China Math. 60, 2503-2514 (2017). doi: 10.1007/s11425-016-0056-1.

Hu F, Zhang L X. Asymptotic properties of doubly adaptive biased coin designs for multi-treatment clinical trials. Ann Statist, 2004, 30: 268–301.

Examples

## power_c = "n/2N"
next_allocation_rate_BAR(n = c(30, 30, 30),
                         success_count = c(5, 6, 12),
                         tot_num = 150)

## power_c = .5
next_allocation_rate_BAR(n = c(30, 30, 30),
                         success_count = c(5, 6, 12),
                         tot_num = 150, power_c = .5)