模拟随机资产价格路径的R代码

问题描述

考虑以下资产价格演变模型：

这就是我所做的（在 R 中）。我找不到随机输出 +1 或 -1 的函数，所以我决定调整内置的 rbinom 函数。

## This code is in R

rm(list = ls())

library(dplyr)
library(dint)
library(magrittr)
library(stats)

path  = 
  function(T,mu,sigma,p,x0) {
    x    = rep(NA,T)
    x[1] = x0
    
    for(i in 2:T){
    z    = if_else(rbinom(1,1,p) == 0,-1,1)
    x[i] = x[i-1] * exp(mu + sigma*z)
    }
    return(x)
  }

## Just some testing
x_sim = path(T = 4,mu = 0,sigma = 0.01,p = 0.5,x0 = 100)

## Actual answer
Np = 10000
mc = matrix(nrow = 17,ncol = Np)
for(j in 1:Np){
  mc[,j] = path(T = 17,x0 = 100)
}
test     = mc[2:nrow(mc),] >= 100
sum_test = colSums(test)
comp     = sum(sum_test >= 1)/length(sum_test)
prob     = 1 - comp

这有意义吗？任何帮助/提示/建议将不胜感激。谢谢！

解决方法

接近你的代码，我想出了这个。直觉上，如果你仔细想想，由于参数的原因，这个概率应该相当低，我得到的概率大约为 6.7%，这大概是我使用赋值中的参数运行你的代码时得到的概率。

simpath <- function(t,mu,sigma,p,x0,seed){
  # set seed
  if(!missing(seed)){
    set.seed(seed)  
  }
  # set up matrix for storing the results
  res <- matrix(c(1:t,rep(NA,t*2)),ncol = 3)
  colnames(res) <- c('t','z_t','x_t')
  res[,'z_t'] <- sample(c(1,-1),size = t,prob = c(p,1-p),replace = TRUE)
  res[1,3] <- x0
  for(i in 2:t){
    res[i,3] <- res[i-1,3] * exp(mu+sigma*res[i,2])
  }
  return(res)
}

x_sim <- simpath(t = 4,mu = 0,sigma = 0.01,p = 0.5,x0 = 100,seed = 123)
x_sim2 <- simpath(t = 36,sigma = 0.03,seed = 123)

## Actual answer
Np <- 100000
mc <- matrix(nrow = 36,ncol = Np)
for (j in 1:Np){
  mc[,j] <- simpath(t = 36,x0 = 100)[,3]
}
test <- mc > 100
sum_test <- colSums(test)
comp     = sum(sum_test == 0)/length(sum_test)
prob     = comp
> prob
[1] 0.06759