# Pricing in the Heston Model

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?

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 $$\xi=\epsilon\lambda$$.