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Girsanov Change of Measure in the Heston Model

Article Quant Q&A · Author: Mr Frog
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Summary

The document explains how the Radon–Nikodym derivative for a Heston model differs from the single-Brownian version used in Black–Scholes. Heston has correlated Brownian drivers for the asset and its variance, so the market price of risk has components associated with both sources of uncertainty. The density process is formed from stochastic integrals against those drivers and includes a correction reflecting their correlation.

The answer gives a specific kernel for the stated physical-measure dynamics and writes the corresponding exponential density, then relates it to the stochastic discount factor. It also recalls the one-dimensional Black–Scholes expression for comparison. These formulas are model-specific: the variance-risk premium parameterization and regularity conditions matter, and the document does not discuss when the density is guaranteed to be a true martingale. The derivative itself is a scalar random variable, even though it is driven by a two-component Brownian process.

Key ideas

  • Heston dynamics involve correlated Brownian drivers for the asset and its variance.
  • The change of measure uses a market price of risk with components for both sources of uncertainty.
  • The density includes a correction term for the correlation between the Brownian drivers.
  • The Radon–Nikodym derivative is scalar-valued, although its driving Brownian motion is multidimensional.
  • The stated formula depends on the chosen variance-risk premium and model assumptions.

Tags

Derivatives pricingVolatilityStatisticsRisk management
Full text
# What is the Radon-Nikodym derivative in the Heston model?


# What is the Radon-Nikodym derivative in the Heston model?












It is clear to me that $$ \frac{dQ}{dP} = e^{-\lambda W_T-\frac{\lambda^2}{2}T}$$ is the Radon-Nikodym derivative that defines the change of measure in the framework described by Black and Sholes. But what is its counterpart in the framework of Heston?

I was thinking that it should have the same shape, with the exception that $\lambda$ and $W_T$ are now bi-dimensional processes. Am I right? In this case, would $\frac{dQ}{dP}$ be two or one dimensional?

## Answer by Kevin (score 7, accepted)

https://quant.stackexchange.com/a/63268

Let \begin{align*} \mathrm{d}S_t&=\mu S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{S,t}, \\ \mathrm{d}v_t&=\kappa(\bar{v}-v_t)\mathrm{d}t+\xi\sqrt{v_t}\mathrm{d}B_{v,t}, \end{align*} where $\mathrm{d}B_{S,t}\mathrm{d}B_{v,t}=\rho\mathrm{d}t$.

The market price of risk (or Girsanov kernel or Sharpe ratio) is ${\varphi}_t=\left(\frac{\mu-r}{\sqrt{v_t}},\frac{\lambda \sqrt{v_t}}{\xi}\right)$. Then, Girsanov Theorem suggests \begin{align*} A_t=\frac{\mathrm{d}\mathbb Q}{\mathrm{d}\mathbb P}= \exp\bigg(&-\int_0^t \frac{\mu-r}{\sqrt{v_s}}\mathrm{d}B_{S,s} -\int_0^t \frac{\lambda\sqrt{v_s}}{\xi}\mathrm{d}B_{v,s} + \int_0^t \frac{ (\mu-r)\lambda\rho}{\xi}\mathrm{d}s\\ &-\frac{1}{2}\int_0^t \left(\frac{(\mu-r)^2}{v_s}+\frac{\lambda^2v_s}{\xi^2} \right)\mathrm{d}s\bigg). \end{align*} This process $A_t$ is a martingale and solves $\text{d}A_t=-\varphi_tA_t\text{d}\mathbf{B}_t$, where $\mathbf{B}_t=\left(B_{S,t},B_{v,t}\right)$.

The corresponding stochastic discount factor is $M_t=e^{-rt}A_t$.

In the one-dimensional case (Black-Scholes model), you have $\varphi_t=\frac{\mu-r}{\sigma}$ and \begin{align*} A_t=\frac{\mathrm{d}\mathbb Q}{\mathrm{d}\mathbb P}= \exp\bigg(- \frac{\mu-r}{\sigma}B_{t}-\frac{1}{2}\left(\frac{\mu-r}{\sigma}\right)^2 t\bigg). \end{align*}

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.