從回歸中繪制擬合值

Question

嘿，我在 R 中有以下代碼

S0 = 40
r = log(1 + 0.07)
sigma = 0.3
K = 45
n_steps_per_year = 4
dt = 1 / n_steps_per_year
T = 3
n_steps = n_steps_per_year * T
R = n_paths
Q = 70
P = 72
n_paths = P * Q
d = exp(-r * dt)

N = matrix(rnorm(n_paths * n_steps, mean = 0, sd = 1), n_paths, n_steps)
  
paths_S = matrix(nrow = n_paths, ncol = n_steps + 1, S0)  
  
for(i in 1:n_paths){
  for(j in 1:n_steps){
    paths_S[i, j + 1] = paths_S[i, j] * exp((r - 0.5 * sigma ^ 2) * dt + sigma * sqrt(dt) * N[i, j])
  }
}

I = apply(K - paths_S, c(1,2), max, 0)
V = matrix(nrow = n_paths, ncol = n_steps + 1)
V[, n_steps + 1] = I[, n_steps + 1]
dV = d * V[, n_steps + 1]
model = lm(dV ~ poly(paths_S[, n_steps], 10))

pred = predict(model, data.frame(x = paths_S[, n_steps]))
plot(paths_S[, n_steps], d * V[, n_steps + 1])
lines(paths_S[, n_steps], pred)

但是當我運行最后兩行時，我會變得非常奇怪 plot （多行而不是一行）。 到底是怎么回事？

Answer 1

您沒有提供n_paths ，讓我們假設：

n_paths = 7
set.seed(111)

然后運行你的代碼，在你 plot 之前，你需要在繪圖之前訂購你的 x 值：

o = order(paths_S[,12])
plot(paths_S[o, n_steps], d * V[o, n_steps + 1],cex=0.2,pch=20)
lines(paths_S[o, n_steps], pred[o],col="blue")