# Heston stochastic volatility, Girsanov theorem

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?

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 $$\mathrm{d}B_{S,t}^\mathbb{P}\mathrm{d}B_{v,t}^\mathbb{P}=\rho^\mathbb{P}\mathrm{d}t$$.

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.

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, $$\rho^\mathbb{Q}=\rho^\mathbb{P}$$.

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

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.

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).

• Nice answer, your final sentence does make one wonder about these alternative parameterisations. Do they have practical use? If so, which? Is there enough to say about this to create a new question? – Bob Jansen Mar 23 at 18:42
• @BobJansen It's a great question. The parametrisation I chose is motivated by Heston (1993) using the CCAPM. There exists (at least) one other popular choice which chooses the variance risk premium to minimise the delta hedging error in the Heston model. That's clearly very important for traders and practical purposes (even if it's lacks some general equilibrium beauty). But I can't for the life of me remember who came up with this other variance risk premium parametrisation. Gatheral? Bergomi? Guyon? I can’t remember! :( – Kevin Mar 23 at 18:54
• Thank you for your nice answer. In fact, I still have a question about this. To obtain the risk-neutral measure, we can modify each SDE separately by applying Girsanov's theorem, i.e. the multi-dim Girsanov, right? What mainly confuses me is actually the correlation. I thought the multidimensional Girsanov theorem applies to uncorrelated processes. So do I understand correctly that we can apply the Girsanov theorem to correlated standard BMs and therefore do not need to write the process in terms of uncorrelated BMs? – quantmath Mar 23 at 20:11
• @NN2 it's akin to why $\mu$ is so difficult to estimate. Option prices only reveal risk-neutral parameters that merge’’ $\mathbb{P}$ parameters and risk premiums. You can try to disentangle them using time series econometrics but that's difficult. Another way is to estimate the expected return of assets that only depend on volatility (e.g. var swaps), see Carr and Wu (2009, RFS). But estimating mean returns with high statistical accuracy is difficult too. Getting information about the $\mathbb{P}$ and $\mathbb{Q}$ distribution simultaneously remains an open question in quant finance – Kevin Mar 23 at 21:16
• Very very nice answer, IMHO, especially the last point: I think it is sometimes not stated sufficiently explicitly that (any) market-price-of-risk ansatz is usually rooted in some notion of equilibrium or in the 'wish' to stay within the distribution class. – Kermittfrog Mar 24 at 8:19

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*}

• I added some more details to my answer to show how to do the calculations with independent Brownian motions. I hope that helps? The trick is to parametrise the variance risk premium at the right step during the derivation – Kevin Mar 27 at 20:47
• It’s normally always defined equivalently. It’s more like that the notion changes. Normally, you can reconcile seemingly different approaches (see my edit for independent Brownian motions). I’m not quite sure what still confuses you? – Kevin Mar 28 at 8:37
• @Kevin thank you for the execution. I have read both "your" and "my" explanations of the "market price of risk" in a number of well-known books, and this market price of risk has always been defined differently, which is exactly what confuses me. Can someone perhaps address this point? – quantmath Mar 28 at 8:41
• Could you please share which books define the market price of risk to be different to $\frac{\lambda v_t}{\sigma\sqrt{v_t}}=\frac{\lambda}{\sigma}\sqrt{v_t}$. Always make sure to differentiate between the “market price of risk” (a Sharpe ratio) that appears in Girsanov's theorem and the “risk premium” (difference in expected returns). These are different concepts and often confused by authors. – Kevin Mar 28 at 8:58
• See the following article emis.de/journals/HOA/JAMSA/Volume2006/18130.pdf But I agree with you that many people confuse the "market price of risk" with the "risk premium", as in the above article. – quantmath Jun 29 at 13:40