MATLAB在for循环内更改变量

Question

在this paper 内部实现之后，我创建了下面的 Matlab 代码来实现欧式看跌期权的价值。我试图将 M 的值与欧洲看跌期权的值作图，因为 M 在 5 的时间步长内从 20 增加到 250。

为了做到这一点，我创建了一个for循环来改变M的值，

for M = 20:5:250

我认为我需要创建这个 for 循环来更改 M 的值。单元测试表明我做错了。 for 循环没有按预期工作。代码生成的图形引用了 M 的原始值（定义为 200），而不是 for 循环内 M 的变化值。我不知道为什么代码返回 M 的原始值而不是 for 循环内的值。

clear all;
close all;
% EURO9  Binomial method for a European put.
%
% Uses explicit solution based on binomial expansion.
% Vectorized, based on logs to avoid overflow,
% and avoids computing with zeros.
%%%%%%%%%% Problem and method parameters %%%%%%%%%%%%%
S = 9 ;E = 10 ;T = 3 ;r = 0.06 ;sigma = 0.3 ; M = 200;
dt = T/M ; A = 0.5*(exp(-r*dt)+exp((r+sigma^2)*dt)) ;
u= A + sqrt(A^2-1) ; d = 1/u ; p = (exp(r*dt)-d)/(u-d) ;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% cut-off index

for M = 20:5:250
    z = max(1,min(M+1,floor( log((E*u)/(S*d^(M+1)))/(log(u/d)) )));
    % Option values at time T
    W=E-S*d.^([M:-1:M-z+1]').*u.^([0:z-1]');
    % log/cumsum version using cut-off index z
    tmp1 = cumsum(log([1;[M:-1:M-z+2]'])) - cumsum(log([1;[1:z-1]']));
    tmp2 = tmp1 + log(p)*([0:z-1]') + log(1-p)*([M:-1:M-z+1]');
    value = exp(-r*T)*sum(exp(tmp2).*W);
    disp('M is'), disp(M)
    disp('Option value is'), disp(value)
    hold on;
    xlabel('M') % x-axis label
    ylabel('European Put') % y-axis label
    plot(M,value,'r*')
end

单元测试表明我的代码是错误的。当真实值为 1.5076 时，针对 M=20 进行测试会返回小于 1 的值。

我写的 for 循环是不是完全错了？为什么每次迭代都引用 M=200 的值，而不是 for 循环中为 M = 20:5:250 指定的增量？

例如，运行

clear all;
close all;
% EURO9  Binomial method for a European put.
%
% Uses explicit solution based on binomial expansion.
% Vectorized, based on logs to avoid overflow,
% and avoids computing with zeros.
%%%%%%%%%% Problem and method parameters %%%%%%%%%%%%%
S = 9 ;E = 10 ;T = 3 ;r = 0.06 ;sigma = 0.3 ; M = 20;
dt = T/M ; A = 0.5*(exp(-r*dt)+exp((r+sigma^2)*dt)) ;
u= A + sqrt(A^2-1) ; d = 1/u ; p = (exp(r*dt)-d)/(u-d) ;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% cut-off index


z = max(1,min(M+1,floor( log((E*u)/(S*d^(M+1)))/(log(u/d)) )));
% Option values at time T
W=E-S*d.^([M:-1:M-z+1]').*u.^([0:z-1]');
% log/cumsum version using cut-off index z
tmp1 = cumsum(log([1;[M:-1:M-z+2]'])) - cumsum(log([1;[1:z-1]']));
tmp2 = tmp1 + log(p)*([0:z-1]') + log(1-p)*([M:-1:M-z+1]');
value = exp(-r*T)*sum(exp(tmp2).*W);
disp('M is'), disp(M)
disp('Option value is'), disp(value)

返回

Option value is
    1.5076

并运行

clear all;
close all;
% EURO9  Binomial method for a European put.
%
% Uses explicit solution based on binomial expansion.
% Vectorized, based on logs to avoid overflow,
% and avoids computing with zeros.
%%%%%%%%%% Problem and method parameters %%%%%%%%%%%%%
S = 9 ;E = 10 ;T = 3 ;r = 0.06 ;sigma = 0.3 ; M = 25;
dt = T/M ; A = 0.5*(exp(-r*dt)+exp((r+sigma^2)*dt)) ;
u= A + sqrt(A^2-1) ; d = 1/u ; p = (exp(r*dt)-d)/(u-d) ;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% cut-off index


z = max(1,min(M+1,floor( log((E*u)/(S*d^(M+1)))/(log(u/d)) )));
% Option values at time T
W=E-S*d.^([M:-1:M-z+1]').*u.^([0:z-1]');
% log/cumsum version using cut-off index z
tmp1 = cumsum(log([1;[M:-1:M-z+2]'])) - cumsum(log([1;[1:z-1]']));
tmp2 = tmp1 + log(p)*([0:z-1]') + log(1-p)*([M:-1:M-z+1]');
value = exp(-r*T)*sum(exp(tmp2).*W);
disp('M is'), disp(M)
disp('Option value is'), disp(value)

返回

Option value is
    1.4666

虽然，我没有在 M = 20:5:250 的 for 循环图中得到这些值。我一定是在 for 循环中犯了一个错误。

Answer 1

问题在于M。您在循环开始之前初始化了M=200，这会影响您在循环之前进行的所有计算。而在您提供的两个单元测试中，您分别使用M=20 和M=25 进行所有计算。

所以解决方法是简单地将受M 影响的计算移到循环内。即

S=9;   E=10;   T=3;   r=0.06;  sigma=0.3; 

for M = 20:5:250
    dt = T/M; 
    A = 0.5*(exp(-r*dt)+exp((r+sigma^2)*dt)) ;
    u = A + sqrt(A^2-1); 
    d = 1/u; 
    p = (exp(r*dt)-d)/(u-d);

    %And here goes what you already have in your loop
    %....
    %....
end