Heston stochastic volatility, Girsanov theorem

Question

How can we apply Girsanov's theorem to a stochastic volatility model? In Heston's model the dynamics are given by \begin{align*} dS_t &= \mu S_t dt + \sqrt{v_t}S_t d\widehat{W}^\mathbb{P}_{1,t}, \\ dv_t &= \kappa ( \theta - v_t) dt + \sigma \sqrt{v_t} \left( \rho d \widehat{W}_{1,t}^\mathbb{P} +\sqrt{1- \rho^2} d\widehat{W}^\mathbb{P}_{2,t} \right) \end{align*} where $\widehat{W}^\mathbb{P}_{1,t}$ and $\widehat{W}_{2,t}^\mathbb{P}$ are the independent standard Brownian motions, and $\rho \in (-1,1)$ the correlation coefficient. Using Girsanov's theorem we would have for the stock process $$\widehat{W}^\mathbb{Q}_{1,t} = \left(\widehat{W}^\mathbb{P}_{1,t} + \frac{\mu - r}{\sqrt{v_t}}t\right).$$ Can we also apply Girsanov's theorem for the variance SDE? More precisely, what happens with $d\widehat{W}^\mathbb{P}_{2,t}$. In some textbooks they write $d\widehat{W}^\mathbb{Q}_{2,t} = \left(d\widehat{W}^\mathbb{P}_{2,t} + \lambda dt\right). $ But then we have \begin{align*} dv_t &= \kappa ( \theta - v_t) dt + \sigma \sqrt{v_t} \left( \rho d \widehat{W}_{1,t}^\mathbb{P} +\sqrt{1- \rho^2} d\widehat{W}^\mathbb{P}_{2,t} \right) \\ &= \kappa ( \theta - v_t) dt + \sigma \sqrt{v_t} \left( \rho \left( d \widehat{W}_{1,t}^{\mathbb{Q}_\lambda} - \frac{\mu - r}{\sqrt{v_t}} dt \right) +\sqrt{1- \rho^2} \left( d \widehat{W}^\mathbb{Q}_{2,t} - \lambda d t) \right) \right) \\ &= \kappa \left( \theta - \frac{\rho}{\kappa} \sigma (\mu - r) - v_t - \frac{\sqrt{1- \rho^2 } }{\kappa} \lambda \sigma \sqrt{v_t} \right) dt + \sigma \sqrt{v_t} \left( \rho d\widehat{W}^\mathbb{Q}_{1,t} + \sqrt{1- \rho^2} d \widehat{W}^\mathbb{Q}_{2,t} \right) \end{align*} It's strange that I never saw it that way. Can anyone give me a clue?

Answer 1

Consider the Heston (1993) model under the real world measure ($\mathbb{P}$) \begin{align*} \mathrm{d}S_t&=\mu^\mathbb{P} S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{S,t}^\mathbb{P}, \\ \mathrm{d}v_t&=\kappa^\mathbb{P} (\bar{v}^\mathbb{P}-v_t)\mathrm{d}t+\sigma^\mathbb{P}\sqrt{v_t}\mathrm{d}B_{v,t}^\mathbb{P}, \end{align*} where $\mathrm{d}B_{S,t}^\mathbb{P}\mathrm{d}B_{v,t}^\mathbb{P}=\rho^\mathbb{P}\mathrm{d}t$.

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I define the market prices of risk (or ``Girsanov kernel'') to be \begin{align} \varphi_S &= \frac{(\mu^\mathbb{P}-r)S_t}{\sqrt{v_t}S_t}=\frac{\mu^\mathbb{P}-r}{\sqrt{v_t}}, \\ \varphi_v &= \frac{\lambda v_t}{\sigma^\mathbb{P}\sqrt{v_t}}=\frac{\lambda}{\sigma^\mathbb{P}}\sqrt{v_t}, \end{align} where $\lambda$ is a parameter.

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The two-dimensional Girsanov Theorem gives rise to \begin{align*} \frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}\bigg|_{\mathcal{F}_t}= \exp\bigg(&-\int_0^t \frac{\mu^\mathbb{P}-r}{\sqrt{v_s}}\mathrm{d}B_{S,s}^\mathbb{P} -\int_0^t \frac{\lambda\sqrt{v_s}}{\sigma^\mathbb{P}}\mathrm{d}B_{v,s}^\mathbb{P} \\ &+ \int_0^t \frac{ (\mu^\mathbb{P}-r)\lambda\rho^\mathbb{P}}{\sigma^\mathbb{P}}\mathrm{d}s-\frac{1}{2}\int_0^t \left(\frac{(\mu^\mathbb{P}-r)^2}{v_s}+\frac{\lambda^2v_s}{(\sigma^\mathbb{P})^2} \right)\mathrm{d}s\bigg), \end{align*} such that \begin{align*} \mathrm{d}W_{S,t}^\mathbb{Q} &= \mathrm{d}B_{S,t}^\mathbb{P} + \varphi_S\mathrm{d}t, \\ \mathrm{d}W_{v,t}^\mathbb{Q} &= \mathrm{d}B_{v,t}^\mathbb{P} + \varphi_v\mathrm{d}t, \end{align*} are increments of standard Brownian motions under $\mathbb{Q}$. Note that $\mathrm{d}W_{S,t}^\mathbb{Q}\mathrm{d}W_{v,t}^\mathbb{Q}=\rho^\mathbb{P}\mathrm{d}t$. Thus, when changing the measure, the correlation coefficient remains the same, $\rho^\mathbb{Q}=\rho^\mathbb{P}$.

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The new risk-neutral dynamics ($\mathbb{Q}$) are then \begin{align*} \mathrm{d}S_t&=r S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}W_{S,t}^\mathbb{Q}, \\ \mathrm{d}v_t&= \kappa^\mathbb{P} (\bar{v}^\mathbb{P}-v_t)\mathrm{d}t+\sigma^\mathbb{P}\sqrt{v_t}\mathrm{d}W_{v,t}^\mathbb{Q} -\lambda v_t\mathrm{d}t \\ &= \left(\kappa^\mathbb{P}\bar{v}^\mathbb{P}-(\kappa^\mathbb{P}+\lambda)v_t\right)\mathrm{d}t +\sigma^\mathbb{P}\sqrt{v_t}\mathrm{d}W_{v,t}^\mathbb{Q} \\ &= \kappa^\mathbb{Q} \left(\bar{v}^\mathbb{Q}-v_t\right)\mathrm{d}t+\sigma^\mathbb{Q}\sqrt{v_t}\mathrm{d}W_{v,t}^\mathbb{Q}, \end{align*} which is again a square-root diffusion, where the vol-of-vol has not changed, $\sigma^\mathbb{Q}=\sigma^\mathbb{P}$. However, as in Heston (1993, Equation (27)), the speed of mean reversion and the long-term mean are now \begin{align} \kappa^\mathbb{Q} &= \kappa^\mathbb{P}+\lambda, \\ \bar{v}^\mathbb{Q} &= \frac{\kappa^\mathbb{P}\bar{v}^\mathbb{P}}{\kappa^\mathbb{P}+\lambda}. \end{align}

Interestingly, $\kappa^\mathbb{P}\bar{v}^\mathbb{P}=\kappa^\mathbb{Q}\bar{v}^\mathbb{Q}$.

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What's happening economically?

• The difference between the $\mathbb{P}$ and $\mathbb{Q}$ drift of $S_t$ is $(\mu^\mathbb{P}-r)S_t$. The term $\mu^\mathbb{P}-r$ is the risk premium (= the return a risk averse agent demands for holding a unit exposure to the shocks driving the stock price).
• The difference between the $\mathbb{P}$ and $\mathbb{Q}$ drift of $v_t$ is $\lambda v_t$. We call $\lambda$ the variance risk premium (= the return a risk averse agents demands to hold a unit exposure to the variance innovations).
• The market prices of risk are Sharpe ratios. They divide the risk premium by the corresponding instantaneous volatilities (the $\text{d}B$-part of the SDEs).
• In equilibrium, $\lambda<0$ because rational agents do not like high volatility (deterioration of the investment opportunity set in an ICAPM sense, see Campbell et al. (2018, JFE)). Empirical evidence for this is given by Coval and Shumway (2001, JF) and Carr and Wu (2009, RFS).
• If $\lambda<0$, then $\bar{v}^\mathbb{Q}>\bar{v}^\mathbb{P}$ and $\kappa^\mathbb{Q}<\kappa^\mathbb{P}$, i.e. variance has higher mean levels but a slower rate of mean reversion. This means the risk-neutral distribution inflates the variance process. This is consistent with what the stochastic discount factor should do, see this answer. The stochastic discount factor, $M_t$, is simply $M_t=e^{-rt}\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}\bigg|_{\mathcal{F}_t}$.
• While the Radon Nikodym derivative seems to depend on the integrated variance, I'd reckon the SDF is still Markovian. The stock price is Markovian too and if you write down how $S_t$ looks like, it also seems to include the integrated variance. The characteristic function $\ln(S_t)$ reveals however that the probabilistic properties only depend on the current values of the state variables, $S_t$ and $v_t$.
• Because the market is incomplete, there exist infinitely many risk-neutral measures. I freely chose to define the market prices of risk to be of a particular form (such that $S_t$ and $v_t$ have the same distribution under both measures, just different parameters). Other parametrisations based on minimising the error of a Delta hedge in the Heston model are possible.

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More technical details on Girsanov's theorem. Suppose you want to work with independent Brownian motions. Set \begin{align*} \mathrm{d}S_t&=\mu^\mathbb{P} S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{1,t}^\mathbb{P}, \\ \mathrm{d}v_t&=\kappa^\mathbb{P} (\bar{v}^\mathbb{P}-v_t)\mathrm{d}t+\sigma^\mathbb{P}\sqrt{v_t}\left(\rho\mathrm{d}B_{1,t}^\mathbb{P}+\sqrt{1-\rho^2}\mathrm{d}B_{2,t}^\mathbb{P}\right), \end{align*} where $\mathrm{d}B_{2,t}^\mathbb{P}\mathrm{d}B_{1,t}^\mathbb{P}=0$. Set, as always, $\varphi_1=\frac{\mu^\mathbb{P}-r}{\sqrt{v_t}}$ and, importantly let the second market price of risk, $\varphi_2$, undetermined for now (we come back to this in in the end). Then, the Girsanov theorem is concerned with independent Brownian motions only and we have \begin{align*} \mathrm{d}W_{1,t}^\mathbb{Q} &= \mathrm{d}B_{1,t}^\mathbb{P} + \varphi_1\mathrm{d}t, \\ \mathrm{d}W_{2,t}^\mathbb{Q} &= \mathrm{d}B_{2,t}^\mathbb{P} + \varphi_2\mathrm{d}t. \end{align*} Thus, by construction, we get the usual risk-neutral stock price dynamics \begin{align*} \mathrm{d}S_t&=r S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}W_{1,t}^\mathbb{Q}. \end{align*} However, the variance process turns to \begin{align*} \mathrm{d}v_t&= \kappa^\mathbb{P} (\bar{v}^\mathbb{P}-v_t)\mathrm{d}t+\sigma^\mathbb{P}\sqrt{v_t}\left(\rho\mathrm{d}W_{1,t}^\mathbb{Q}+\sqrt{1-\rho^2}\mathrm{d}W_{2,t}^\mathbb{Q}\right)\\ &\;\;\;\;\;\;\underbrace{-\sigma^\mathbb{P}\rho(\mu^\mathbb{P}-r)\text{d}t-\sigma^\mathbb{P}\sqrt{v_t}\sqrt{1-\rho^2}\varphi_2\text{d}t}_{=\lambda(t,S_t,v_t)\text{d}t} \\ &= \left(\kappa^\mathbb{P}(\bar{v}^\mathbb{P}-v_t)+\lambda(t,S_t,v_t)\right)\mathrm{d}t +\sigma^\mathbb{P}\sqrt{v_t}\left(\rho\mathrm{d}W_{1,t}^\mathbb{Q}+\sqrt{1-\rho^2}\mathrm{d}W_{2,t}^\mathbb{Q}\right). \end{align*} If you now use the same parametrisation for the variance risk premium as before, $\lambda(t,S_t,v_t)=\lambda v_t$, you recover the same risk-neutral parameters as before. Note that we did not determined $\varphi_2$ directly. We only implicitly choose a price of risk for the orthogonal Brownian motion by choosing $\lambda(t,S_t,v_t)$. Again, the main reason for this parametrisation is to keep the distribution for $v_t$ under both measures the same (although Heston provides some intuition based on the CCAPM).

Answer 2

I don't think that you can apply Girsanov's theorem in this way, noting that I don't understand your (short) argument in the comments. This is how I would proceed, which makes sense mathematically, but the economic interpretation is then a bit strange.

Let us write the SDE's a bit different, noting that it still preservers the same correlation structure \begin{align*} \frac{dS_t}{S_t} &= \mu dt + \sqrt{v_t} d \left( \rho d \widehat{W}_{2,t}^\mathbb{P} +\sqrt{1- \rho^2} d\widehat{W}^\mathbb{P}_{1,t} \right) \\ dv_t &= \kappa (\theta - v_t )dt + \sigma \sqrt{v_t} d\widehat{W}^\mathbb{P}_2\\ \end{align*} where $\widehat{W}^\mathbb{P}_1$ and $\widehat{W}^\mathbb{P}_2$ are the independent standard Brownian motions, and $\rho \in (−1,1)$ the correlation coefficient. Assuming the same market price of (volatility) risk as @Kevin or Heston and denote it by $\lambda_2$, we have \begin{align*} \lambda_2= \frac{\lambda \sqrt{v_t} }{\sigma} \end{align*} Then, the relationships between the standard Brownian motions under the risk-neutral measure and the standard Brownian motions under the physical measure are given by \begin{align*} d \widehat{W}_1^{\mathbb{Q}_\lambda} = \frac{1}{\sqrt{1-\rho^2}} \left( \frac{\mu - r}{\sqrt{v}} - \frac{ \lambda \rho \sqrt{v_t}}{\sigma} \right)dt + d \widehat{W}^\mathbb{P}_1 \end{align*} and \begin{align*} d\widehat{W}^{ \mathbb{Q}_\lambda}_2 = \frac{\lambda \sqrt{v_t} }{\sigma}dt + d \widehat{W}_2^\mathbb{P}. \end{align*} The two-dim market price of risk process $(\lambda_1, \lambda_2)$ is given by \begin{align*} \lambda_1 =\frac{1}{\sqrt{1-\rho^2}} \left( \frac{\mu - r}{\sqrt{v}} - \frac{ \lambda \rho \sqrt{v_t}}{\sigma} \right) \end{align*} and \begin{align*} \lambda_2= \frac{\lambda \sqrt{v_t} }{\sigma} \end{align*} and not \begin{align*} \lambda_1 = \frac{\mu - r}{\sqrt{v_t}} \end{align*} Indeed, in the Black and Scholes model, the market price of risk is given by $(\mu^{BS}-r)/\sigma^{BS}$, but here I think we have to take both standard Brownians into account.

In particular, then the we have \begin{align*} \frac{d\mathbb{Q}}{d \mathbb{P}} \bigg\vert_{\mathcal{F}_T}= \exp \bigg( - \bigg( \int^T_0 \lambda_1(u) d \widehat{W}_{1,t}^\mathbb{P}(u) + \int^T_0 \lambda_2(u) d \widehat{W}_{2,t}^\mathbb{P}(u) \bigg) - \frac{1}{2} \bigg( \int^T_0 \lambda_1^2(u) du + \int^T_0 \lambda_2^2(u)du \bigg) \bigg) \end{align*}