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

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?

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 $$\xi=1$$.
• Thank you for your answer! I didn't realize that. Is this exactly the Euler's Method? So for instance if I define $\zeta = 2v_2u_2 = 0.002$, will the Feller condition follow from the continuous case? Sep 15 at 12:56