本論文主要是發掘變異數在Two-armed Bandit問題中的影響。在文中我們假設兩種治療法的成功率分別是θ1和θ2，且以π1~Beta(cα,cβ)和π2~Beta(α,β)為其驗前機率分配。此外，我們假設所有病人數(N)已知。 我們證明了當N=2、3，變異因子(c)>1時，最佳的策略是k1*=0，也就是說，我們不應在成功率的變異數較小的治療法上做試驗。這個結果和One-armed Bandit問題(c=∞)的結論是一樣的。但是，當N=10、12的例子中，我們發現k1*=0就並非是最佳的策略。 當α=β時，我們證明了效用函數是c的遞減函數。也就是說，其中一個治療法的變異越小，效用亦越小。當α=β=c=1時，最佳的策略是k^*=k_2^*≈√(1+N)-1。此外，我們也證明了效用函數是c的連續函數。 / The focus of the report is to find the influence of variance in Two-armed Bandit problems. In this report, we consider the case when the success probabilities of the two treatmentsθ1,θ2 haveπ1~Beta(cα,cβ) andπ2~Beta(α,β) as their priors, and the total number of patients, N is known. We showed that for N=2 and 3 the optimal strategy is k1*=0 if variance factor, c>1. That is, we should not make trials on the treatment which variance is smaller. But when N=10 and 12, we showed that k1*=0 is not optimal. When α=β we showed that the utility function is a decreasing function of the c. That is, the smaller variance of a treatment is the smaller utility will be. We have found that k^*=k_2^*≈√(1+N)-1 whenα=β=c=1. Besides, we also have the continuity of utility function in c.

No description available.