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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?

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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 $\xi=1$.

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  • $\begingroup$ 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? $\endgroup$
    – Tab1e
    Sep 15 at 12:56
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    $\begingroup$ Your discretization method is Euler's method (fully explicit). I think that it is hard to figure out a Feller condition for the discrete case. Better than getting theoretical headaches is to test what your discretization is producing. $\endgroup$
    – Kurt G.
    Sep 15 at 13:05
  • $\begingroup$ Agree. Now I have some ideas to test and to modify the model. Thank you! $\endgroup$
    – Tab1e
    Sep 15 at 13:18

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