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Distributions of patterns of two successes separated by a string of k-2 failures

机译:由一串k-2失败分隔的两次成功的模式分布

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摘要

Let Z 1, Z 2, . . . be a sequence of independent Bernoulli trials with constant success and failure probabilities p = Pr(Z t = 1) and q = Pr(Z t = 0) = 1 − p, respectively, t = 1, 2, . . . . For any given integer k ≥ 2 we consider the patterns ${mathcal{E}_{1}}$ : two successes are separated by at most k−2 failures, ${mathcal{E}_{2}}$ : two successes are separated by exactly k −2 failures, and ${mathcal{E}_{3}}$ : two successes are separated by at least k − 2 failures. Denote by ${ N_{n,k}^{(i)}}$ (respectively ${M_{n,k}^{(i)}}$ ) the number of occurrences of the pattern ${mathcal{E}_{i}}$ , i = 1, 2, 3, in Z 1, Z 2, . . . , Z n when the non-overlapping (respectively overlapping) counting scheme for runs and patterns is employed. Also, let ${T_{r,k}^{(i)}}$ (resp. ${W_{r,k}^{(i)})}$ be the waiting time for the r − th occurrence of the pattern ${mathcal{E}_{i}}$ , i = 1, 2, 3, in Z 1, Z 2, . . . according to the non-overlapping (resp. overlapping) counting scheme. In this article we conduct a systematic study of ${N_{n,k}^{(i)}}$ , ${M_{n,k}^{(i)}}$ , ${T_{r,k}^{(i)}}$ and ${W_{r,k}^{(i)}}$ (i = 1, 2, 3) obtaining exact formulae, explicit or recursive, for their probability generating functions, probability mass functions and moments. An application is given.
机译:设Z 1 ,Z 2 、。 。 。分别是具有恒定成功率和失败率的独立伯努利试验序列p = Pr(Z t = 1)和q = Pr(Z t = 0)= 1-p,t = 1 ,2, 。 。 。对于任何给定的k≥2的整数,我们考虑模式$ {mathcal {E} _ {1}} $:两个成功最多被k-2个失败分开,$ {mathcal {E} _ {2}} $:两个成功完全由k -2个失败和$ {mathcal {E} _ {3}} $分隔:两次成功至少由k-2个失败分隔。用$ {N_ {n,k} ^ {(i)}} $(分别是$ {M_ {n,k} ^ {{i)}} $)表示模式$ {mathcal {E}的出现次数_ {i}} $,i = 1、2、3,在Z 1 ,Z 2 ,中。 。 。 ,Z n ,当采用运行和模式的非重叠(分别重叠)计数方案时。同样,令$ {T_ {r,k} ^ {(i)}} $(分别为$ {W_ {r,k} ^ {{i)})} $是第r次出现的等待时间模式$ {mathcal {E} _ {i}} $,i = 1、2、3,在Z 1 ,Z 2 ,中。 。 。根据非重叠(重叠)计数方案。在本文中,我们对$ {N_ {n,k} ^ {(i)}} $,$ {M_ {n_k} ^ {(i)}} $,$ {T_ {r,k } ^ {(i)}} $和$ {W_ {r,k} ^ {(i)}} $(i = 1,2,3)获得精确的公式,无论是显式的还是递归的,因为它们的概率生成函数是质量功能和时刻。提出申请。

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