I do not understand why mean levels of the state variables under the risk-neutral measure, $\theta^{\mathbb{Q}}$, in Arbitrage-free Nelson-Siegel is set to zero. It should follow from the following relationship:

The relationship between the factor dynamics under the real-world probability measure $\mathbb{P}$ and the risk-neutral measure $\mathbb{Q}$ is given by \begin{equation} \label{eq11} dW_{t}^{\mathbb{Q}}=dW_{t}^{\mathbb{P}}+\Gamma_{t}dt \end{equation} where $\Gamma_{t}$ represents the risk premium.

Christensen et al. [$2011$] article here want to preserve affine dynamics under the both, risk-neutral measure and the real-world probability measure, thus the risk premium parameter must be an affine function of factors: \begin{equation} \label{eq12} \Gamma_{t}=\begin{pmatrix} \gamma_{1}^{0}\\ \gamma_{2}^{0}\\ \gamma_{3}^{0} \end{pmatrix}+\begin{pmatrix} \gamma_{1,1}^{1} & \gamma_{1,2}^{1} & \gamma_{1,3}^{1} \\ \gamma_{2,1}^{1} & \gamma_{2,2}^{1} & \gamma_{2,3}^{1}\\ \gamma_{3,1}^{1} & \gamma_{3,2}^{1} & \gamma_{3,3}^{1} \end{pmatrix}\begin{pmatrix} X_{t}^{1}\\ X_{t}^{2}\\ X_{t}^{3} \end{pmatrix} \end{equation}

Any help?


1 Answer 1


The reason for setting $\theta^{\mathbb{Q}}=0$ is that, along with other restrictions, it identifies the parameters of the model uniquely, which means that the model is then well-defined, and there is then a one-to-one relationship between the model parameters and the probability distribution of the data. If the model is uniquely identified, there is only one set of parameter values that can generate the observed data, given the specified model and vice versa.

This is described in Section 3 of the paper you linked to:

Because the latent state variables may rotate without changing the probability distribution of bond yields, not all parameters in the above model can be identified. Singleton (2006) imposes identifying restrictions under the $\mathbb{Q}$-measure.


  • the mean $\theta^{\mathbb{Q}}=0$,
  • the volatility matrix $\Sigma$ equal to the identity matrix and
  • the mean-reversion matrix $K^{\mathbb{Q}}$ equal to the triangular matrix

makes it possible to identify all the other model parameters uniquely from the data.

  • $\begingroup$ Thank you so much $\endgroup$
    – Martin N.
    Commented Apr 5, 2023 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.