In a 2006 paper Zhang and Zhu propose a model for VIX and VIX Futures based on Heston.

I am struggling in understanding how they get equation 6 and 8 (where they define the parameters).

Can anyone of you help me?enter image description here


1 Answer 1


Heston - Change of measure

Consider the following Heston dynamics written under the real world measure $\Bbb{P}$ \begin{gather} \frac{dS_t}{S_t} = \mu_t dt + \sqrt{v_t} dW_S^{\Bbb{P}}(t),\ S(0) = S_0 \\ dv_t = \kappa(\theta-v_t)dt + \xi \sqrt{v_t} dW_v^{\Bbb{P}}(t),\ v(0) = v_0 \\ d\langle W_S^\Bbb{P}, W_v^\Bbb{P} \rangle_t = \rho dt \end{gather} In order to be able to use that model to price financial instruments, arbitrage-free pricing theory (APT) tells us that we need to move to an equivalent measure $\Bbb{Q}$ under which discounted asset prices are martingales (or more generally: the value of any self-financing portfolio, when expressed in the risk-free money market account numéraire, should emerge as a $\Bbb{Q}$ martingale).

Because the Heston model is incomplete, there exists infinitely many such measures. Mathematically, these will differ by the drift attributed to the instantaneous variance process i.e. $$ dv_t = \kappa(\theta-v_t)dt - \lambda(t,S_t,v_t) dt + \xi \sqrt{v_t} dW_v^\Bbb{Q}(t) $$ where the term $\lambda(t,S_t,v_t)$ is often referred to as the market price of volatility risk.

In his original 93 paper, Heston makes a particular assumption regarding the market price of volatility risk which he considers being proportional to $v_t$ relying on some economic arguments $$ \lambda(t,S_t,v_t) = \lambda v_t$$ In that particular case, the dynamics under (Heston's) $\Bbb{Q}$ may be rewritten \begin{gather} \frac{dS_t}{S_t} = (r_t - q_t) dt + \sqrt{v_t} dW_S^{\Bbb{Q}}(t),\ S(0) = S_0 > 0 \\ dv_t = \kappa^*(\theta^*-v_t)dt + \xi \sqrt{v_t} dW_v^{\Bbb{Q}}(t),\ v(0) = v_0 \\ d\langle W_S^\Bbb{Q}, W_v^\Bbb{Q} \rangle_t = \rho dt \end{gather} with \begin{align} \kappa^* = \kappa + \lambda \\ \theta^* = \theta \frac{\kappa}{\kappa + \lambda} \tag{1} \end{align} and $r_t$ (resp. $q_t$) figures the risk-free rate (resp. equity dividend yield).

Heston - Variance swaps

For a pure diffusion model, the fair variance strike $\hat{\sigma}^2_T(0)$ of a fresh-start variance swap of maturity $T$ calculated at $t=0$ is defined as $$ \hat{\sigma}_T(0)^2 = \frac{1}{T} \Bbb{E}_0^\Bbb{Q} \left[ \int_0^T d\langle \ln S \rangle_t \right] $$ In the particular case of the Heston model we can further write \begin{align} \hat{\sigma}_T(0)^2 &= \frac{1}{T} \Bbb{E}_0^\Bbb{Q} \left[ \int_0^T v_t dt \right] \tag{2} \\ &= \theta^* + (v_0 - \theta^*) \frac{1-e^{-\kappa^* T}}{\kappa^* T} \end{align} where the second equality can be obtained either by:

  • Permuting integral and expectation operators in $(2)$ (Fubini), noting that $v_t$ is CIR so that its conditional expectation is known in closed form for any time $t$, integrating the result;
  • Integrating the SDE verified by $v_t$, taking the expectation, solving the resulting ODE for $\Bbb{E}_0[v_t]$ and again integrating the result.


Because the VIX squared is by definition the fair variance strike of an (idealised) variance swap of maturity $T=\tau_0$ equal 30 days we then have, under Heston \begin{align} VIX^2(0) &= \theta^* + (v_0 - \theta^*) \frac{ 1-e^{-\kappa^* \tau_0}}{\kappa^* \tau_0} \\ &= \underbrace{\theta^* \left( 1 - \frac{1-e^{-\kappa^* \tau_0}}{\kappa^* \tau_0} \right)}_{A} + v_0 \underbrace{\frac{1-e^{-\kappa^* \tau_0}}{\kappa^* \tau_0}}_{B} \tag{3} \end{align} which is exactly the equation you mention, with the risk-neutral parameters of the Heston dynamics $(\kappa^*, \theta^*)$ related to the parameters under the real world measure $(\kappa, \theta)$ through $(1)$

Let the Heston dynamics under the $\Bbb{P}$ measure be given by \begin{align} \frac{dS_t}{S_t} &= \mu_t dt + \sqrt{v_t} dW_S^{\Bbb{P}}(t) \\ dv_t &= \kappa(\theta-v_t)dt + \xi \sqrt{v_t} dW_v^{\Bbb{P}}(t) \\ d\langle W_S^{\Bbb{P}},W_v^{\Bbb{P}}\rangle_t &= \rho dt \end{align}

Define the Radon-Nikodym derivative of $\Bbb{Q}$ with respect to $\Bbb{P}$ as \begin{align} \frac{d\Bbb{Q}}{d\Bbb{P}} &= \mathcal{E}\left( -\lambda_S W_S^{\Bbb{P}}(t) - \lambda_{S,\bot} W_{S,\bot}^{\Bbb{P}}(t) \right) \\ &:= \mathcal{E}(X_t) \end{align} where $\lambda_S$ is the market price of equity risk $$ \lambda_S = \frac{\mu_t-r_t}{\sqrt{v_t}} $$ $\lambda_{S,\bot}$ another risk-premium (yet to be defined) and $\mathcal{E}[X_t]$ the Doélans-Dade exponential of process $X_t$. Assuming that $\lambda_S$ and $\lambda_{S,\bot}$ verify the Novikov condition , Girsanov theorem then stipulates that \begin{align} W_S^{\Bbb{Q}}(t) &= W_S^{\Bbb{P}}(t) - \left\langle W_S^{\Bbb{P}}, X \right\rangle_t \\ W_v^{\Bbb{Q}}(t) &= W_v^{\Bbb{P}}(t) - \left\langle W_v^{\Bbb{P}}, X \right\rangle_t \end{align} are 2 standard $\Bbb{Q}$-Brownian motions verifying $d\langle W_S^{\Bbb{Q}},W_v^{\Bbb{Q}} \rangle_t = \rho dt$.

From the bilinearity property of quadratic variation, the first of the above equation yields \begin{align} W_S^{\Bbb{Q}}(t) = W_S^{\Bbb{P}}(t) + \lambda_S t \end{align}

Using a Cholesky decomposition to re-express $W_v^{\Bbb{P}}(t)$ as $$ W_v^{\Bbb{P}}(t) = \rho W_S^{\Bbb{P}}(t) + \sqrt{1-\rho^2} W_{S,\bot}^{\Bbb{P}}(t) $$ the second equation in turn gives $$ W_v^{\Bbb{Q}}(t) = W_v^{\Bbb{P}}(t) + \lambda_S \rho t + \lambda_{S,\bot} \sqrt{1-\rho^2} t $$ which allows us to rewrite the dynamics under $\Bbb{Q}$ as \begin{align} \frac{dS_t}{S_t} &= r_t dt + \sqrt{v_t} dW_S^{\Bbb{Q}}(t) \\ dv_t &= \kappa(\theta-v_t)dt - \xi \sqrt{v_t} (\lambda_S \rho + \lambda_{S,\bot} \sqrt{1-\rho^2} ) dt + \xi \sqrt{v_t} dW_v^{\Bbb{Q}}(t) \\ d\langle W_S^{\Bbb{Q}},W_v^{\Bbb{Q}} \rangle_t &= \rho dt \end{align}

Observe that the change of measure we introduced makes discounted asset prices emerge as $\Bbb{Q}$-martingales. Also observe how, the dynamics under $\Bbb{Q}$ is yet merely defined up to a constant $\lambda_{S,\bot}$. In other words, it is not unequivocally determined until we make an assumption regarding the market price of volatility risk. Let's define $$ \lambda(t,S_t,v_t) = \xi \sqrt{v_t} (\lambda_S \rho + \lambda_{S,\bot} \sqrt{1-\rho^2} ) $$ so that \begin{align} \frac{dS_t}{S_t} &= r_t dt + \sqrt{v_t} dW_S^{\Bbb{Q}}(t) \\ dv_t &= \kappa(\theta-v_t)dt - \lambda(t,S_t,v_t) dt + \xi \sqrt{v_t} dW_v^{\Bbb{Q}}(t) \\ d\langle W_S^{\Bbb{Q}},W_v^{\Bbb{Q}} \rangle_t &= \rho dt \end{align}

Some common choices:

Choice 1 Assume $$ \lambda(t,S_t,v_t) = \alpha v_t $$ Then we get \begin{align} dv_t &= \kappa(\theta-v_t)dt - \alpha v_t dt + \xi \sqrt{v_t} dW_v^{\Bbb{Q}} \\ &= (\kappa \theta - (\kappa+\alpha)v_t) dt + \xi \sqrt{v_t} dW_v^{\Bbb{Q}} \\ &= (\kappa+\alpha) (\frac{\kappa \theta}{\kappa+\alpha} - v_t) dt + \xi \sqrt{v_t} dW_v^{\Bbb{Q}} \end{align} so that \begin{align} \kappa^* &= \kappa + \alpha \\ \theta^* &= \frac{\kappa \theta}{\kappa+\alpha} \end{align} This is the choice formulated in my answer above, and also the one in Heston's original paper [http://web.math.ku.dk/~rolf/Heston93.pdf] (end of p.329 and p.335).

Choice 2 Assume $$ \lambda(t,S_t,v_t) = \lambda_S \sqrt{v_t} \rho \xi = (\mu_t - r_t)\rho\xi := \alpha $$ \begin{align} dv_t &= \kappa(\theta-v_t)dt - \alpha dt + \xi \sqrt{v_t} dW_v^{\Bbb{Q}} \\ &= \kappa((\theta-\alpha/\kappa)-v_t) dt + \xi \sqrt{v_t} dW_v^{\Bbb{Q}} \end{align} so that \begin{align} \theta^* &= \theta-\alpha/\kappa \\ \end{align} This corresponds to a replication minimising the Delta-hedging standard error (minimum variance $\Delta$ à la Bergomi).

  • 2
    $\begingroup$ @Davide don't hestitate to mark this answer as accepted if it helped you. $\endgroup$
    – Quantuple
    Mar 19, 2018 at 16:02
  • $\begingroup$ Great answer (+1)!! I struggle with the change of measure though. From Girsanov, $dW_S^\mathbb Q=dW_S^\mathbb P+\frac{\mu-r}{\sqrt{v_t}}dt$ and $dW_v^\mathbb Q=dW_v^\mathbb P+\lambda(t,S_t,v_t)dt$. Then, $$dv_t=\kappa(\theta-v_t)dt+\xi\sqrt{v_t}dW_v^\mathbb P=\kappa(\theta-v_t)dt+\xi\sqrt{v_t}dW_v^\mathbb Q-\xi\sqrt{v_t}\lambda(t,S_t,v_t)dt.$$ I now have to assume $\lambda(t,S_t,v_t)=\frac{\lambda}{\xi}\sqrt{v_t}$ to obtain $dv_t=\kappa^*(\theta^*-v_t)dt+\xi dW_v^\mathbb Q$. So, I assume the variance risk premium is proportional to $\sqrt{v_t}$ and not $v_t$. Can you see where am I wrong? $\endgroup$
    – Alex
    Aug 28, 2020 at 11:54
  • 2
    $\begingroup$ Notice that "my" $\lambda(t,S_t,v_t)$ is your $\xi \sqrt{v_t} \lambda(t,S_t,v_t)$ so indeed we agree. But note that this is only one way of doing things (which Heston justifies in his original paper using economic argument). But you could also pick $\lambda(\mu-r-q)\rho\xi$ for instance (like Bergomi does in his papers) to also land with another risk-neutral dynamics (where $\theta^* = f(\theta)$ but $\kappa^*=\kappa$. At the end of the day, both are related to hedging strategies in which the error will be zero in expectation... but the standard deviation will differ! $\endgroup$
    – Quantuple
    Aug 28, 2020 at 13:05
  • 2
    $\begingroup$ This comes from the fact that the model is incomplete, see this question: quant.stackexchange.com/questions/18305/… $\endgroup$
    – Quantuple
    Aug 28, 2020 at 13:06
  • 3
    $\begingroup$ @Alex, I hope I made the developments clearer in my edits (and there are no typos, I have to run, no time to check). The key is that yes the Girsanov theorem includes the market price of risk, but there are two sources of risk which complicates things slightly. $\endgroup$
    – Quantuple
    Aug 28, 2020 at 15:03

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