Interrelations between random walks on diagrams (graphs) with and without cycles.
Interrelations between random walks on diagrams (graphs) with and without cycles.
复制标题
带循环和不带循环的图（图）上的随机游走之间的相互关系。
DOI：
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发表时间：
1988
影响因子：
11.1
通讯作者：
Terrell L. Hill
中科院分区：
文献类型：
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作者：
Terrell L. Hill
讨论了三个主题。具有一个或多个吸收状态的离散状态、连续时间随机游走可以通过一种可能的新方法来研究：可以从修改后的图（图）中找到一些平均属性，包括平均吸收时间，其中每个吸收状态都被返回到起始状态的单向循环所取代。第二个问题是在一个有圈的图（图）上的随机行走。行走在第一个循环完成时终止。这种走线可以用一种等价的走线来代替，这种走线是在一个修改过的图上进行的。这个吸收图又可以用另一个修改过的图来代替，它可以像第一个问题那样，单向循环回到起始状态。第三个问题，在生物物理学中很重要，涉及到一个长时间的连续行走的图表与周期。这个图可以被转换（分两步）为修改过的、更详细的、只有单向循环的图。因此，可以从修改后的图的状态概率中找到原始图的单向循环通量。这些概率本身可以通过简单的矩阵求逆得到（概率由线性代数稳态方程确定）。因此，现在有一种简单的方法可以精确地找到单向循环通量（以前需要蒙特卡罗模拟来找到这些通量，以及伴随的波动，对于任何复杂的图表）。上述过程的附带好处是，它提供了单向循环通量关系Jn +/- = IIn +/- sigma n/sigma的简单证明，其中n是原始图的任何循环。
Three topics are discussed. A discrete-state, continuous-time random walk with one or more absorption states can be studied by a presumably new method: some mean properties, including the mean time to absorption, can be found from a modified diagram (graph) in which each absorption state is replaced by a one-way cycle back to the starting state. The second problem is a random walk on a diagram (graph) with cycles. The walk terminates on completion of the first cycle. This walk can be replaced by an equivalent walk on a modified diagram with absorption. This absorption diagram can in turn be replaced by another modified diagram with one-way cycles back to the starting state, just as in the first problem. The third problem, important in biophysics, relates to a long-time continuous walk on a diagram with cycles. This diagram can be transformed (in two steps) to a modified, more-detailed, diagram with one-way cycles only. Thus, the one-way cycle fluxes of the original diagram can be found from the state probabilities of the modified diagram. These probabilities can themselves be obtained by simple matrix inversion (the probabilities are determined by linear algebraic steady-state equations). Thus, a simple method is now available to find one-way cycle fluxes exactly (previously Monte Carlo simulation was required to find these fluxes, with attendant fluctuations, for diagrams of any complexity). An incidental benefit of the above procedure is that it provides a simple proof of the one-way cycle flux relation Jn +/- = IIn +/- sigma n/sigma, where n is any cycle of the original diagram.