MATLAB changing variable inside a for loop

Question

I created the Matlab code below to implement the value of an European Put following the implementation inside this paper. I am trying to graph the value of M against the value of the European Put as M increases from 20 to 250 in time steps of 5.

In order to do this, I created a for loop to change the value of M,

for M = 20:5:250

I think that I need to create this for loop in order to change the value of M. Unit testing shows that I did something wrong. The for loop isn't working as intended. The graph produced by the code is referencing the original value of M (defined to be 200) instead of the changing values of M inside the for loop. I don't know why the code returns the original value of M instead of the values inside the for loop.

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

Unit testing shows that my code is wrong. Testing against M=20 returns a value less than one when the true value is 1.5076.

Did I write the for loop completely wrong? Why is it referencing the value of M=200 at every iteration instead of the increment specified in the for loop for M = 20:5:250?

As an example, running

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)

returns

Option value is
    1.5076

and running

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)

returns

Option value is
    1.4666

Although, I don't get these values in the graph of the for loop for M = 20:5:250. I must have made a mistake inside the for loop.

I added two unit tests. Please let me know if I should add more information. — Axion004
– Axion004, Commented Apr 21, 2018 at 16:38
Is there a way to write M without initializing it? If I delete the original value of M=200 then I cannot run the program. I think that I need to initialize M and then override it inside the for loop. — Axion004
– Axion004, Commented Apr 21, 2018 at 17:03

Sardar Usama · Accepted Answer · 2018-04-21 17:05:25Z

5

The problem is with M. You initialised M=200 before the start of the loop which is affecting all the calculations that you're doing before the loop. Whereas in the two unit tests that you have provided, you used M=20 and M=25 for all the calculations respectively.

So the fix is to simply move the calculations that are affected by M inside the loop. i.e

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

answered Apr 21, 2018 at 17:05

Sardar Usama

19.7k9 gold badges39 silver badges60 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

MATLAB changing variable inside a for loop

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related