I am reading this paper https://mpra.ub.uni-muenchen.de/4969/1/MPRA_paper_4969.pdf pp.6-7 on discrete-time bond pricing. The model adopted is a a common affine model,

the short rate follows \begin{equation} r_t = a + b' X_t \end{equation}

the state variables $X_t$, under $Q$, follows \begin{equation} X_t = \bar{\mu} + \bar{\rho} X_{t-1} + \Sigma \eta_t \end{equation} so that the price at time $t$ of a bond paying a unitary amount at time $t+n$ equals \begin{equation} p_t^n = E_t^Q(\exp{(-r_t) p_{t+1}^{n-1})} \end{equation} According to the authors "The link between the risk-neutral distribution Q and the historical distribution P is given by the prices of risk, denoted by $\lambda_0 = \Sigma^{-1} (\mu-\bar{\mu})$ and $\lambda_1 = \Sigma^{-1} (\rho-\bar{\rho})$:

\begin{equation} \frac{dQ}{dP}\vert_t = \frac{\xi_{t+1}}{E(\xi_{t+1})} \end{equation} with \begin{equation} \xi_{t+1} = \prod_{j=1}^{\infty} \exp{((-\lambda_0 + \lambda_1 X_{t+j-1}) \epsilon_{t+j})}. \end{equation}

However, I do not understand the representation of the pricing kernel: where this last product comes from?

Usually, for instance in https://web.stanford.edu/~piazzesi/AP.pdf, the pricing kernel is \begin{equation} M_{t,t+1}= \exp{(-r_t -\frac{1}{2} \lambda_t' \lambda_t - \lambda_t' \epsilon_t)} \end{equation} so that

\begin{equation} E_t[M_{t,t+1}]= \exp{(-r_t)} \end{equation}

And defining

\begin{equation} \xi_{t,t+1}=\exp{(r_t M_{t,t+1}})=\exp{(-\frac{1}{2} \lambda_t' \lambda_t - \lambda_t'\epsilon_{t+1})} \end{equation}

we can define a new measure $Q$ equivalent to $P$ since $E[\xi_{t,t+1}]=1$ and $\xi_{t,t+1}>0$, and clearly,$\frac{dQ}{dP}\vert_t = \xi_{t,t+1} $ is the Radon-Nikodym derivative

Are these two representation equivalent?

thank you very much

  • 2
    $\begingroup$ I think the notion is confusing though I have seen both forms before. Apparently the Radon-Nikodym derivative is defined over a random variable with arbitrary time horizons versus the latter definition concerns time $t+1$ variables. This is similar to arbitrary horizon vs period by period stochastic discount factors. $\endgroup$ – fesman Apr 18 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.