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?

  • 2
    $\begingroup$ What Monte Carlo simulation? $\endgroup$
    – Bob Jansen
    Jul 31, 2020 at 19:23

1 Answer 1


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.

  • $\begingroup$ Thanks! So when people are using risk neutral Heston Model to price a derivative, which $\lambda_2$ do people use? 0 or $\sqrt{v_t}$? Or doesn't it really matter because when people calibrate the model, we directly estimate/calibrate the $\hat{k}$ and $\hat{\theta}$ from market? $\endgroup$
    – StupidMan
    Aug 1, 2020 at 8:03
  • $\begingroup$ As you said, $0$ should be ok. We want the variance process to be CIR and then calibrate its mean reversion speed, long run variance and vol of vol (with or without hats won't matter much) to liquidly traded derivatives (vanilla European options or even variance swaps, if respective market has them). $\endgroup$
    – ir7
    Aug 1, 2020 at 15:31
  • 1
    $\begingroup$ To continue this line of thought, change from the real world to risk neutral measure matters if you have a new model you want to prove is arbitrage free and somehow useful (or that some statistics carry over from the time series). But for risk-neutral pricing the general recipe is to assume some risk-neutral model up front and run with it. The market price of risk does not matter in the sense of getting a price consistent with your calibration instruments but only in the stability of the resulting hedge ratios (which is sadly usually of secondary concern) $\endgroup$
    – river_rat
    Aug 3, 2020 at 11:58
  • $\begingroup$ @river_rat Thank you for the insight. $\endgroup$
    – ir7
    Aug 3, 2020 at 16:22

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