使用R的布朗运动模拟

Question

在时间[0,100]的倒数中布朗运动的模拟，并且通过模拟n = 1000点来绘制路径。我生成以下代码：

 n <- 1000
 t <- 100
 bm <- c(0, cumsum(rnorm(n,0,sqrt(t/n))))
 steps <- seq(0,t,length=n+1)
 plot(steps,bm,type="l")

我如何模拟标准布朗运动的50个样本路径，并以不同的颜色显示每条路径，例如一束轨迹？

我认为它将类似于replicate(50,bm)，但是当我这样做时，xy.coords中出现错误。感谢您的帮助！

[[1,1]上布朗桥的模拟，并且通过模拟n = 1000点来绘制路径。我生成以下代码

n <- 1000
t <- seq(0,1,length=n)
No.Ex<-10
bm <- c(0,cumsum(rnorm(n-1,0,1)))/sqrt(n)
B = replicate(No.Ex,{
  bb <- bm - t*bm[n]
})
matplot(B, type = "l", col = cols, lty = 1)

生成几何布朗运动的样本路径的代码

simGBM<- function(P0, mu, sigma, T, nSteps, nRepl){
  dt<- T/nSteps
  muT<- (mu-sigma^2/2)*dt
  sigmaT<- sqrt(dt)*sigma
  pathMatrix<- matrix(nrow = nRepl, ncol = nSteps+1)
  pathMatrix[,1]<- P0
  for(i in 1:nRepl){
    for(j in 2:(nSteps+1)){
      pathMatrix[i,j]<- pathMatrix[i,j-1]*exp(rnorm(1, muT, sigmaT))
    }
  }
  return(pathMatrix)
}

P0<- 1 #initial price
mu<- 0.1 #drift
sigma<- 0.5 #volatility
T<- 100/360 #100 days of a commercial year
nSteps<- 50 #No of steps
nRepl<- 100 #No of replications

paths<- simGBM(P0, mu, sigma, T, nSteps, nRepl)
yBounds<- c(min(paths),max(paths)) #bounds of simulated prices

plot(paths[1,], ylim = yBounds, type = 'l',col = 1, main = "Simulation of sample paths of GBM", xlab = "Time", ylab = "Price")
for(k in 2:numRepl) lines(paths[k,], col = k)

我正在尝试使用matplot函数，但无法生成相同的图

cols = rainbow(nSteps)
matplot(paths, ylim = yBounds, type = "l", col = cols, lty = 1, main = "Simulation of sample paths of GBM", xlab = "Time", ylab = "Price")

Answer 1

怎么样

n = 1000
t = 100
No.Ex = 10
steps = seq(0,t,length=n+1)
A = replicate(No.Ex, {
  bm <- c(0, cumsum(rnorm(n,0,sqrt(t/n))))
}) 

cols = rainbow(No.Ex)
matplot(A, type = "l", col = cols, lty = 1)

我修改了答案，并纳入了斯蒂芬·洛朗的matplot建议。这给出了下图。

编辑：

为了在评论中回答您的问题，我认为您应该保留bm的初始代码为bm <- c(0, cumsum(rnorm(n,0,sqrt(t/n))))。然后一切都很好！感谢您指出漂亮的matplot命令@Stephane Laurent。

编辑2：我刚刚意识到您对布朗桥提出了一个新的问题。您可以尝试此代码

n <- 1000
t <- seq(0,1,length=n)
No.Ex<-10
B = replicate(No.Ex,{
  bm <- c(0, cumsum(rnorm(n - 1,0,sqrt(t/n))))
  bb <- bm - t*rep(bm[length(bm)], length.out = length(bm))
})
matplot(B, type = "l", col = cols, lty = 1)

这产生

[另外，对于几何布朗摩天，尝试用较少的重复次数对代码进行此修改

simGBM<- function(P0, mu, sigma, T, nSteps, nRepl){
  dt<- T/nSteps
  muT<- (mu-sigma^2/2)*dt
  sigmaT<- sqrt(dt)*sigma
  pathMatrix<- matrix(nrow = nRepl, ncol = nSteps+1)
  pathMatrix[,1]<- P0
  for(i in 1:nRepl){
    for(j in 2:(nSteps+1)){
      pathMatrix[i,j]<- pathMatrix[i,j-1]*exp(rnorm(1, muT, sigmaT))
    }
  }
  return(pathMatrix)
}

P0<- 1 #initial price
mu<- 0.1 #drift
sigma<- 0.5 #volatility
T<- 100/360 #100 days of a commercial year
nSteps<- 50 #No of steps
nRepl<- 10 #No of replications

paths<- simGBM(P0, mu, sigma, T, nSteps, nRepl)
yBounds<- c(min(paths),max(paths)) #bounds of simulated prices

plot(paths[1,], ylim = yBounds, type = 'l',col = 1, main = "Simulation of sample paths of GBM", xlab = "Time", ylab = "Price")
for(k in 2:nRepl) lines(paths[k,], col = k)

cols = rainbow(nSteps)
matplot(paths, ylim = yBounds, type = "l", col = cols, lty = 1, main = "Simulation of sample paths of GBM", xlab = "Time", ylab = "Price")

在我的机器上，这产生