Pricing in the Heston Model

Question

The dynamics of the Heston Model is

\begin{align*} \frac{dS}{S} & = \lambda \sqrt{\nu} d W^S \\[0.5em] d \nu & = k (1- \nu )dt + \epsilon \sqrt{\nu} dW^\sigma \end{align*}

where $\lambda$ is the instantaneous volatility. Let $\nu_0 = 1$. The Brownian motions are correlated with $\rho dt$.

Now I want to use this online pricer: https://kluge.in-chemnitz.de/tools/pricer/heston_price.php to determine the prices.

How would I go about this without $\lambda$ being specified in the model of the pricer?

Say I want to find the price for $S_0=100$, $K=90$, $\epsilon = 0.3$, $\kappa = 0.05$, $\rho = 0.5$ and $\lambda = 0.2$.

I know that by using Ito on the spot I get

$$ d \log S_t = \lambda \sqrt{\nu_t} d W_t^S - \frac{1}{2} \lambda^2 \nu_t dt $$

Do I somehow need to use the relationship between $\lambda^2 \nu_t$? If so, how?

Answer 1

This is not the typical Heston stochastic differential equation (SDE). In the original Heston paper, the SDE is defined without $\lambda$, that is $\lambda=1$ and $v(0)=v_0$ not necessarily 1.

In your case you have to do the change of variable $y= \lambda^2 v$ which leads to $$dS/S = \sqrt{y}dW_S$$ $$dy = k(\lambda^2 - y) + \epsilon\lambda\sqrt{y} dW_y$$ and $y(0)=\lambda^2$.

the initial vol is $\sqrt{y(0)}$ and the vol of vol (actually really a vol of var) is $\xi=\epsilon\lambda$.