# 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?

• That shape is ubiquitous and independent from the model. It’s just because the RN derivative is an exponential martingale under $P$ measure. Differences across models are in the shape of Girsanov’s kernel (yours $\lambda$). Commented Apr 8, 2021 at 10:29

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*}