Consider the stochastic vol: $$dS = \mu Sdt + \sigma SdW_1$$ $$d\sigma = p(\sigma,S,t)dt + q(\sigma,S,t)dW_2$$ $$dW_1dW_2 = \rho dt$$ We want to obtain the price of option $V(\sigma,S,t),$ we use the underlying asset $S$ and another option $V_1(\sigma,S,t)$ to build the hedging portfolio: $$\Pi = V -\Delta S - \Delta_1 V_1$$ then make $$d \Pi = r\Pi dt$$ eliminate the risk terms we have $$\dfrac{\partial V}{\partial t} + \dfrac{1}{2}\sigma^2S^2\dfrac{\partial^2 V}{\partial S^2} + \rho\sigma Sq\dfrac{\partial^2 V}{\partial S\partial \sigma} + \dfrac{1}{2}\sigma^2q^2\dfrac{\partial^2 V}{\partial \sigma^2} + rS\dfrac{\partial V}{\partial S} -rV = -(p-\lambda q)\dfrac{\partial V}{\partial \sigma}.$$ Here $\lambda$ is called market price of risk, since we can understand $\lambda$ as $$d V -rV dt = q\dfrac{\partial V}{\partial S}(\lambda d t + d W_2) = q\Delta(\lambda d t + d W_2)$$ this is the unit of extra return.

And we have another way to price $V,$ the discounted value of $V$ is martingale, namely $dt$ term of $d(e^{-rt} V)$ is zero, then we find that, the PDE of $V$ is exactly $$\lambda = 0$$ in above PDE. So does that mean, the discounted value of $V$ is martingale is equivalent to the market price of risk is zero?

  • $\begingroup$ Discounted value of $V$ is martingale under the risk neutral measure, and a risk neutral investor would not price market risk. So I guess your statement is reasonable. But what is really going on is a change of measure (Girsanov's theorem), where downward increment in return is assigned more weight. $\endgroup$
    – Michael
    Commented Apr 15, 2017 at 7:41

1 Answer 1


I think you misunderstood the underlying idea of the risk-neutrality and the market price of risk.

The basic idea is to price the option with a portfolio consisting of the underlying asset $S$ and another option. In order to make this portfolio risk-free and because of no-arbitrage arguments, the change in the portfolio should correspond to the change of the risk-free portfolio, i.e.

$d\Pi= r\Pi dt$

This delivers (I didn't check calculations in details) the partial differential equation with $V$ you obtained.

Your mistake is then coming when you write:

$d\Pi- r\Pi dt\ $=$\ q \Delta( \lambda dt+d W_2) $.

Of course, this is impossible because, given the risk-free portfolio, the left hand side is zero but the right hand side is always different from zero whatever the value of $\lambda$.

Probably that this formula is for the option to price:

$dV- rV dt\ $=$\ q \Delta( \lambda dt+d W_2) $

which expresses the idea of risk compensation by $\lambda$.

The right hand side contains a deterministic term in $dt$ and a stochastic term in $dW_2$. The term in $dW_2$ shows that the portfolio is a risky portfolio and $\lambda$ can then be interpreted as the excess return (on top of the risk-free return $r$) for accepting a certain level of risk. On the one hand, you have a risk but on the other hand, you have excess return via $\lambda$ (which explains the name "market price of risk").

By the Feynman–Kac formula, we can also express this PDE as a conditional expectation, which justifies the martingale approach. But nowhere, in the reasoning, we need to assume that $\lambda=0$. Otherwise, it would mean that the risk-neutral measure and the real-world measure coincide, which does not really make sense.

  • $\begingroup$ yeah, you are right, it should be $dV - rVdt.$ $\endgroup$
    – A.Oreo
    Commented Apr 20, 2017 at 1:27
  • 1
    $\begingroup$ nice explanation! $\endgroup$
    – Richi Wa
    Commented Apr 20, 2017 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.