We have pointed out that the random walk is closely similar in itsmathematics to the coin-tossing game we considered at the beginning of the chapter.If we imagine the direction of each step to be in correspondence with theappearance of heads or tails in a coin toss, then D is just NH−NT , the difference in the number of heads and tails. Since NH+NT=N, the total number of steps (and tosses), we have D=2NH−N. We have derived earlier an expression for the expected distributionof NH (also called k ) and obtained the result of Eq. (6.5).Since N is just a constant, we have the corresponding distribution for D. (Since for every head more than N/2 there is a tail “missing,” we have the factor of 2 between NH and D .) The graph of Fig. 6–2 representsthe distribution of distances we might get in 30 random steps (where k=15 is to be read D=0 ; k=16 , D=2 ; etc.).
在本章的开始,我们思考了抛硬币的数学,现在,我们指出,随机行走的数学与抛硬币的数学,非常近似。如果我们想象每一步的方向,都与抛硬币中的正面或反面相应,那么,D就正是NH−NT,正面次数与负面次数之差。由于NH+NT=N,即总步数(总抛掷数),我们就有D=2NH−N。对于被期待的NH(也被称为 k)的分布,我们导出过一个表达,并获得了结果,即方程(6.5)。由于N正是一个常数,所以对D,我们也有一个相应的分布。(由于对于每一个正面超过N/2的局,那么负面就“丢失了”,所以,NH和 D之间,我们有一个2的因子。)图6-2中的曲线,代表了在30次随机的步伐中,我们能得到的距离的分布(这里k=15,将被解读为 D=0 ; k=16 , D=2 ;等等。)。
在本章的开始,我们思考了抛硬币的数学,现在,我们指出,随机行走的数学与抛硬币的数学,非常近似。如果我们想象每一步的方向,都与抛硬币中的正面或反面相应,那么,D就正是NH−NT,正面次数与负面次数之差。由于NH+NT=N,即总步数(总抛掷数),我们就有D=2NH−N。对于被期待的NH(也被称为 k)的分布,我们导出过一个表达,并获得了结果,即方程(6.5)。由于N正是一个常数,所以对D,我们也有一个相应的分布。(由于对于每一个正面超过N/2的局,那么负面就“丢失了”,所以,NH和 D之间,我们有一个2的因子。)图6-2中的曲线,代表了在30次随机的步伐中,我们能得到的距离的分布(这里k=15,将被解读为 D=0 ; k=16 , D=2 ;等等。)。
















