是否有工具箱或可用的MATLAB函数可以解决随机积分的近似逼近,其中z是Wiener过程:
答案 0 :(得分:0)
%Lets say n is 100 and dt is 1/252 and k = .1
n = 100;
dt = 1/252;
k = 0.1;
dz = randn(n,1); %get random increments: normal
%dz = 2*(randi(2,n,1)-1.5); % or plus/minus ones : bernoulli
fnt = exp(-k*(n*dt - [0:n-1]*dt))*sqrt(dt)*dz;