在R中生成晶格路径

Rob*_*sch 4 r matrix mathematical-lattices

例如,如果我有一个如下所示的晶格:

          133.1
         /
       121
      /  \
    110   108.9
   /  \  / 
 100   99
   \  /  \
    90    89.1
      \  /
       81
         \
          72.9
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格子从100开始,并且要么以1.1的系数上升,要么以0.9的系数下降.这个晶格有3个周期,它上升或下降.很明显,这个矩阵可以填充更长的时间.

矩阵形式的晶格如下所示:

     [,1] [,2] [,3]  [,4]
[1,]  100  110  121 133.1
[2,]   NA   90   99 108.9
[3,]   NA   NA   81  89.1
[4,]   NA   NA   NA  72.9
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我在R.工作.生成晶格矩阵的代码如下:

#Parameters
S0 <- 100 #price at t0
u <- 1.1 #up factor
d <- 0.9 #down factor
n <- 3 #number of periods

#Matrix for the prices
prices <- matrix(data=NA, nrow=(n+1), ncol=(n+1))
prices[1,1] <- S0

#Fill the matrix
for(column in 2:(n+1)){

  for(row in 1:(column-1)){

    prices[row,column] <- u*prices[row,column-1];
  }

  prices[column,column] <- d*prices[column-1,column-1];
}
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我想创建一个代码,生成一个矩阵,其中包含通过晶格的所有可能路径.对于此示例,它看起来像这样:

     [,1] [,2] [,3]  [,4]
[1,]  100  110  121 133.1
[2,]  100  110  121 108.9
[3,]  100  110   99 108.9
[4,]  100  110   99  89.1
[5,]  100   90   99 108.9
[6,]  100   90   99  89.1
[7,]  100   90   81  89.1
[8,]  100   90   81  72.9
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我几个小时以来一直在努力使用这段代码,所以任何帮助都会非常感激!提前致谢!:)

Vin*_*ynd 5

每条长度路径n对应一系列向上和向下移动:您只需枚举所有这些序列.如果已经有长度序列n-1,作为矩阵u,长度序列n可以获得为

rbind( 
  cbind( u,  .9 ), 
  cbind( u, 1.1 ) 
)
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你可以把它放在一个函数中,然后调用它n.

n <- 4
up   <- 1.1
down <- .9
m <- Reduce( 
  function(u,v) rbind( cbind( u, up ), cbind( u, down ) ), 
  rep(NA,n), 
  100
)
t(apply(m, 1, cumprod))
#  [1,] 100 110 121 133.1 146.41
#  [2,] 100  90  99 108.9 119.79
#  [3,] 100 110  99 108.9 119.79
#  [4,] 100  90  81  89.1  98.01
#  [5,] 100 110 121 108.9 119.79
#  [6,] 100  90  99  89.1  98.01
#  [7,] 100 110  99  89.1  98.01
#  [8,] 100  90  81  72.9  80.19
#  [9,] 100 110 121 133.1 119.79
# [10,] 100  90  99 108.9  98.01
# [11,] 100 110  99 108.9  98.01
# [12,] 100  90  81  89.1  80.19
# [13,] 100 110 121 108.9  98.01
# [14,] 100  90  99  89.1  80.19
# [15,] 100 110  99  89.1  80.19
# [16,] 100  90  81  72.9  65.61
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