The non-negativity condition of a discretized mean-reverting Heston model with stochastic violatilities

Question

I happened to encounter the following discretized mean-reverting Heston model with stochastic volatilities in a paper $$ P(t) = P(t-1) + v_1(u_1-P(t-1))+\sqrt{\sigma(t)}\cdot \epsilon_1(t) \\ \sigma(t) = \sigma(t-1) + v_2(u_2-\sigma(t-1))+\sqrt{\sigma(t-1)}\cdot \epsilon_2(t) $$ where $v_1=v_2=0.1,u_1=100,u_2=0.01$ are pre-set parameters, and $\epsilon_1,\epsilon_2 \sim N(0,1)$ follow the normal distribution IID. Recall that in the original Heston Model formulation, there is a condition (known as the Feller condition) to make sure that the values under the square root is positive. See wiki for more info. But in this case, how can I ensure that the value of $\sigma$ to be positive?

Answer 1

The continuous version of your equation for $\sigma(t)$ reads $$ d\sigma(t)=v_2(u_2-\sigma(t))\,dt+\sqrt{\sigma(t)}\,dW^\sigma_t\,. $$ In this notation, the Feller condition ensuring $\sigma(t)>0$ is $2v_2u_2>1\,.$ This is not the case for the values $v_2=0.1,u_2=0.01$ you have chosen. Note that the Heston model also has a vol-of-vol parameter $\xi$: $$ d\sigma(t)=v_2(u_2-\sigma(t))\,dt+\xi\sqrt{\sigma(t)}\,dW^\sigma_t\, $$ and that the Feller conditon in full glory says $2v_2u_2>\xi^2\,.$ In other words, you should use that vol-of-vol $\xi$ and not make it as large as $\xi=1$.