# Heston Model: Quadratic exponential scheme

I am having trouble understanding the QE scheme of Andersen.

Leif Andersen: Efficient Simulation of the Heston Stochastic Volatility Model, 2006

Is it possible for the variance process to become zero? There is a switching rule. If $v=a(b+Z)^2$, then the probability is zero I think. (Only if b=Z since a>0). But for the other process, I don't know what will happen.

• In the exponential part of the scheme, the pdf of $v_{t+\Delta t} \vert v_t$ exhibits a probability mass at $0$ whose strength is given by $p$ (see equation (24)). This $p$ is calibrated by matching the first two theoretical moments of the conditional distribution of $v_{t+\Delta t}$ (those are known in close form since $v_t$ follows a CIR process). – Quantuple Aug 23 '17 at 15:16