I'm having difficulty understanding the well known property of the CIR model that it can't go below zero. Wikipedia says that this is because the random shock on the rate will grow very small as r moves closer to zero, but won't the drift term also do so? Especially if the volatility term is high, isn't it possible for the random shock to be of a greater negative value than the drift is positive even as r goes to zero?
I'm trying to implement the method in matlab currently but it happens to me that r becomes negative if I increase the volatility. Could it be a problem with the discretization as well perhaps? The code snippet is below if it's of interest.
theta=0.5; %Long run mean sigma=14; %Volatility of drawdowns k = 7.326; %Mean reversion constant n = 100; %number of time steps, t. dt = T/n; %time step M=10^3; %Number of realizations d0 = theta; d=ones(M,1).*d0;%Starting value for d for i = (j-1:n) dW = sqrt(dt)*randn(M,1); % Wiener increments d(:,i+1) = d(:,i) + k.*(theta-d(:,i)).*dt + sigma.*sqrt(d(:,i)).*dW; %drawdown rate end ```