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!:)
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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