Stochastic differential equation Gompertz plotting a graph

3 vues (au cours des 30 derniers jours)
George Hendry
George Hendry le 26 Fév 2022
Commenté : Torsten le 26 Fév 2022
I have come stuck when entering the model which has a log(x(t) into my code and when looking at the Weiner process deciding what my value of dt would need to be. Because at the moment when producing the graph using this code it is blank and I don't understand why that is happening when I introduce the log(x(t)) as I get a graph when it isn't intrdocued.I have coded
th = 0.1;
mu = 0.3;
sig = 0.1;
dt = 1e-1 ;
t = 0:dt:50; % Time vector
x = zeros(1,length(t)); % Allocate output vector, set initial condition
rng(1); % Set random seed
for i = 1:length(t)-1
x(i+1) = x(i)+th*x(i)*dt-mu*x(i)*dt-mu*log(x(i))*dt+sig*sqrt(dt)*randn;
end
figure;
plot(t,x);

Réponse acceptée

Alan Stevens
Alan Stevens le 26 Fév 2022
You have x(1) = 0, so log(x(1)) is -Inf, which means that subsequent values of x will be NaNs.
  2 commentaires
George Hendry
George Hendry le 26 Fév 2022
Thanks I'm not sure what the value of the dt is this model as I have seen various differing values of dt due to the weiner process do you have any views on what it should be?
Torsten
Torsten le 26 Fév 2022
It's the "Wiener Process", not the "Weiner Process".

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