Market Price of Volatility Risk in Heston Pricing

# Heston Model - PDE and Monte Carlo


# Heston Model - PDE and Monte Carlo












Why there is a "market price of volatility risk" variable in the PDE of Heston Model and no such variable in Monte Carlo Simulation?

Do we obtain the same price from both methods?

## Answer by ir7 (score 4, accepted)

https://quant.stackexchange.com/a/57045

One fixes the market price of volatility risk on the SDE first, then implies the pricing PDE. That way the SDE and PDE are consistent.

One starts with a Heston SDE: $$ dS/S = \mu dt + \sqrt{v} dW_1 $$ $$ dv = \kappa(\theta - v)dt + \eta \sqrt{v}dW_2$$ with $W =(W_1,W_2)^T$ correlated Brownian motion, $dW_1dW_2 = \rho dt$.

As we have two Brownian drivers but only one risky asset, the no-arbitrage drift conditions can only fix one of the components of the market price of risk process

$$ \lambda =(\lambda_1, \lambda_2)^T. $$

That is, we have $$ \lambda_1 = \frac{\mu-r}{\sqrt{v_t}}, $$

while $\lambda_2$ (market price of volatility risk) is unspecified.

This allows us to consider $\lambda_2$-dependent EMM's (equivalent martingale measure) under which process $W^\lambda =(W_1^\lambda, W_2^\lambda)^T$, defined by

$$ dW^\lambda = dW - \left(\frac{\mu-r}{\sqrt{v_t}},\lambda_2\right)^T dt, $$

is a Brownian motion.

The original Heston SDE transforms into:

$$ dS/S = r dt + \sqrt{v} dW_1^\lambda $$ $$ dv = (\kappa(\theta - v)-\eta \sqrt{v}\lambda_2) dt + \eta \sqrt{v}dW_2^\lambda$$

which is not of Heston type for all $\lambda_2$ choices.

We choose $\lambda_2$ such that $$\kappa(\theta - v)-\eta \sqrt{v}\lambda_2 $$ can be rewritten as

$$ \hat{\kappa}(\hat{\theta} - v) $$

for some $\hat{\kappa}$ and $\hat{\theta}$ (e.g., $\lambda_2=0$ or $\lambda_2 = \sqrt{v_t}$). This makes the variance a CIR dynamics again and the full SDE is again of Heston type.