I have a code in which the variable changes in every loop iteration for a fixed time point t=5, by assuming that x is changing between 0 and 5, and t=0 to 30. how can I plot the changes ?

r = 0.02;                  
mu = 0.08;                  
sigma = 0.2;                  
T = 30;                    
rho = 0.02;                   

for x = 0: 1: 5
    t = 5;
    for i = 1:30
    gamma (i) = 5+ (0.07 * t (i));
    end
    f (t) = (exp (-rho * t) / gamma (i)). * exp (((1-gamma (i)) / gamma (i)). * (r + (mu - r) ^ 2 / (2 * sigma ^ 2. * Gamma (i))). * (T - t));
    v1 (t, x) = (f (t). ^ gamma (i)). * (x. ^ (- gamma (i)));
    v2 (t, x) = -gamma (i). * (f (t). ^ gamma (i)). * (x. ^ (1-gamma (i)));
    c1 (t, x) = (exp (-rho * t) / gamma (i)) * x * f (t) ^ (- 1);
   
end
 plot (x, c1)

Answer 1

[darova]

I found some mistakes in your code. Look

[Fae ba]

In this case I want to show that gamma is time dependent and for x between 0 and 5,by ignoring i from gamma (i) = 5+ (0.07 * t (i)) does it mean the same?

[darova]

What is this?

Why do you use t(i) if it fixed

[Fae ba]

yes you right, therefore if I want to develope the code to three cases as below, I mean there should be specific f,v1,v2 and c1 with the same formula for gammaB and gammaC then how can I creat the loop such that plot (x, c1,x,c2,x,c3) wrt to three cases of gamma for a fixed point eg t = 5?

gammaA (i) = 4;

gammaB (i) = 5+ (0.07 * t (i));

gammaC (i) = 5- (0.07 * t (i));

[darova]

Why are you talking about 3 cases? I see 30

for i = 1:30
    gamma (i) = 5+ (0.07 * t (i));
end

[Fae ba]

what I wrote firstly is the primary code but for developing the code to three cases of gamma then its a bit complicated. Then you mean the loop for i = 1:30 should be removed for considering three cases?

[darova]

If t then gamma is fixed also. So yes, for loop can be removed

gamma = 5+0.07*t;