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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?

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  • $\begingroup$ 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 Apr 15 '17 at 7:41
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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.

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  • $\begingroup$ yeah, you are right, it should be $dV - rVdt.$ $\endgroup$ – A.Oreo Apr 20 '17 at 1:27
  • $\begingroup$ nice explanation! $\endgroup$ – Richard Apr 20 '17 at 6:21

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