# Simulation Heston Model, markovianity

I am trying to simulate the instanteneous volatility of a Heston process.

My equations are the following :

wealth process: $$dX_t = r_t X_t + \theta \sqrt {V_t} u_t dt + u_t dW_{1t}$$

Volatility: $$dV_t = (\kappa \phi - \lambda V_t) dt + \sigma \sqrt {V_t} dB_t$$

With, I start my simulations with a 2D brownian motion : $$(W_1, W_2)$$ and another "corrolated" Brownian motion $$B_t = \rho d \tilde{W}_{1t} + \sqrt{1- \rho^2} dW_{2t}$$

My problem lies in the $$d \widetilde{W}_{1t}$$. Its definition is :

$$\widetilde{W}_{1t} = W_{1t} + 2 \theta \int_0^t \sqrt {V_s} ds$$.

So I know how to simulate the wealth process, it s a classical "flow".

The volatility follows the same pattern, iff the brownian motion $$dB_t$$ is a classical one. Here there is a drift movement which makes the whole simulation cyclic. I have no idea how to deal with it.

1. Is it possible to simulate that ? Is my problem markovian ?
2. How would one deal with that problem. I simply need a solution for $$\widetilde{W}_{1t}$$, I'll deal with the rest.

Thank you

• There are several undefined processes, e.g. $r_t$, $u_t$, and possibly also $\phi_t$. Could you specify which are parameters and otherwise the definitions of the other processes. – oliversm Jan 20 at 11:49
• @oliversm thanks for the comment. The parameters are constant, $\phi_t = \phi$. What for $r_t, u_t$, I could define them ( $r_t$ is the interest rate, can be considered constant, $u_t$ would take an equation in order to be defined), however, my main problem was with $\widetilde{W}_{1t}$ and those parameters do not intervene within the equation. Why do you need more information about it ? Knowing why you need that information would help me define and give you additional information that you'd need – Marine Galantin Jan 20 at 13:36

• so you mean that $$d \widetilde{W}_{1t} = d W_{1t} + 2 \theta \sqrt{V_t} dt$$ ? – Marine Galantin Jan 20 at 21:40
• you're right @user1987 but I should have thought about it myself ! Now that you say it, it is obvious that the answer you gave is the right one. I believe I thought that the expression is in fact : $$d\widetilde{W}_{1t} = d W_{1t} + 2 \theta \int_0^t \sqrt {V_s} ds$$ This is what I must have read in my head. But you're right, you can apply the transformation you did. Thank you so much. – Marine Galantin Jan 20 at 21:55