狠多，monte carlo其实是最慢的一个，最后一条路

我试过的有tree，finite difference，recursive integration，radial basis function，没有试过的有finite element，finite volume

[此贴子已经被作者于2007-11-8 8:52:26编辑过]

我试过一次三叉数，不错的。可以去掉zigzag效果。

Function Tri_put(S, X, r, d, sigma, Tmat, N, American_FLAG)

    Dim dt As Single
    Dim up_step As Single
    Dim down_step As Single
    Dim pup As Single
    Dim pmm As Single
    Dim pdown As Single
       
    ReDim opt(0 To 2 * N) As Single
    ReDim opt_new(0 To 2 * N) As Single
    ReDim S_array(0 To 2 * N) As Single

    dt = Tmat / N
    up_step = Exp(sigma * Sqr(3 * dt))
    down_step = Exp(-sigma * Sqr(3 * dt))
    pup = Sqr(dt / (12 * sigma ^ 2)) * (r - sigma ^ 2 / 2) + 1 / 6
    pmm = 2 / 3
    pdown = 1 - pup - pmm
   
    For i = 0 To 2 * N
        S_array(i) = S * (up_step ^ (i - N))
        opt(i) = max(X - S_array(i), 0)
    Next i
   
    For j = 1 To N
        For i = j To 2 * N - j
           opt_new(i) = Exp(-r * dt) * (pup * opt(i + 1) + pmm * opt(i) + pdown * opt(i - 1))
          
           If American_FLAG = 1 Then
               If (X - S_array(i)) > opt_new(i) Then
                    opt_new(i) = (X - S_array(i))
               End If
           End If
          
        Next i
        For i = j To 2 * N - j
            opt(i) = opt_new(i)
        Next i
    Next j
   
    Tri_put = opt(N)
   
End Function

兄台，没看懂你的评价。

这段东西，我是照着hull的书写的，好像hull的书里证明过pup pdown都是正的。

Sqr(dt / (12 * sigma ^ 2)) * (r - sigma ^ 2 / 2) 要小于1/6,你是这个意思？

需要加一个条件判断？