Converting Heston Parameters from Real-World to Risk-Neutral Dynamics

# Going from $\mathcal{P}$ to $\mathcal{Q}$


# Going from $\mathcal{P}$ to $\mathcal{Q}$












Under $\mathcal{P}$, we have the Heston Model given by:

$$ d S_{t}=\mu S_{t} d t+\sqrt{\nu_{t}} S_{t} d W_{t}^{S},\\ d \nu_{t}=\kappa\left(\theta-\nu_{t}\right) d t+\xi \sqrt{\nu_{t}} d W_{t}^{\nu}. $$

Assume I estimated these parameters appering in the Heston model $\underline{\text{using only stock returns}}$ with MCMC.

My question is; how can I price vanilla options under a risk neutral measure $\mathcal{Q}$ now? Since these parameters are estimated under the stocks $\mathcal{P}$-dynamics, are these parameters useless for option pricing?

I know there exists fourier-transform pricing methods. But can I use fourier-transform pricing methods with these estimated parameters?

## Answer by Kevin (score 2)

https://quant.stackexchange.com/a/51297

As you said, you estimated the $\mathcal{P}$ parameters but for option pricing, one needs the $\mathcal{Q}$ parameters. But there exists a transformation.

Under $\mathcal{P}$, Heston (1993) assumes \begin{align*} \mathrm{d}S_t &= \mu S_t\mathrm{d}t + \sqrt{v_t}S_t\mathrm{d}W_{1,t}^\mathcal{P}, \\ \mathrm{d}v_t &= \kappa(\theta-v_t)\mathrm{d}t + \xi\sqrt{v_t}\mathrm{d}W_{2,t}^\mathcal{P}, \end{align*} where $\mathbb{E}^\mathcal{P}[\mathrm{d}W_{1,t}^\mathcal{P}\mathrm{d}W_{2,t}^\mathcal{P}]=\rho\mathrm{d}t$.

These parameters include the market price of risk which equals zero in the risk-neutral world. Assume $\lambda(S_t,v_t,t)=\frac{\lambda}{\xi}\sqrt{v_t}$. Applying the two-dimensional Girsanov theorem, \begin{align*} \frac{\mathrm{d}\mathcal{Q}}{\mathrm{d}\mathcal{P}}\bigg|_{\mathcal{F}_t} =\exp\left(-\int_0^t\frac{\mu-r}{\sqrt{v_s}}\mathrm{d}W_{1,s}^\mathcal{P}-\int_0^t\frac{\lambda}{\xi}\sqrt{v_s}\mathrm{d}W_{2,s}^\mathcal{P}-\frac{1}{2}\int_0^t\frac{(\mu-r)^2}{v_s}+\frac{\lambda^2}{\xi^2}v_s\mathrm{d}s\right). \end{align*} This corresponds to \begin{align*} \mathrm{d}W_{1,t}^\mathcal{Q} &= \mathrm{d}W_{1,t}^\mathcal{P}+\frac{\mu-r}{\sqrt{v_t}}\mathrm{d}t \\ \mathrm{d}W_{2,t}^\mathcal{Q} &= \mathrm{d}W_{2,t}^\mathcal{P}+\frac{\lambda}{\xi}\sqrt{v_t}\mathrm{d}t \\ \end{align*} Now, similar to Black-Scholes, applying Ito's Lemma to $f(x)=\ln(x)$, we obtain under $\mathcal{Q}$ \begin{align*} \mathrm{d}\ln(S_t) &= \left(r-\frac{1}{2}v_t\right)\mathrm{d}t+\sqrt{v_t}\mathrm{d}W_{1,t}^\mathcal{Q}, \\ \mathrm{d}v_t &= \kappa^\mathcal{Q}(\theta^\mathcal{Q}-v_t)\mathrm{d}t + \xi\sqrt{v_t}\mathrm{d}W_{2,t}^\mathcal{Q}, \end{align*} where $\mathbb{E}^\mathcal{Q}[\mathrm{d}W_{1,t}^\mathcal{Q}\mathrm{d}W_{2,t}^\mathcal{Q}]=\rho\mathrm{d}t$ and $\kappa^\mathcal{Q}=\kappa+\lambda$ and $\theta^\mathcal{Q}=\frac{\kappa\theta}{\kappa+\lambda}$.

So, the good news is that you can convert the process (and its parameters) from the real-world into the risk-neutral world. It's also good that the vol-of-vol and correlation coefficient do not alter at all. The bad news is that the speed of mean-reversion and the long-term mean depend on the market price of volatility risk which requires further estimation.