Why is the martingale factor a martingale in Hansen's 2012 Dynamic Valuation Decomposition?

Question

In this question, I'm continuing to explore the tools used/presented in Lars Hansen's Econometrica paper "Dynamic Valuation Decomposition within Stochastic Economies" (2012).

This might be an easy question, but I can't quite see it. In the paper linked above, the factorization is presented in which one component is a martingale. See p. 937. On this page it presents this formula and says

Given a solution to (21), I construct a martingale via $$ \widetilde M_t = \exp(-\rho t) M_t \left [ \frac{e(X_t)}{e(X_0)} \right ] $$ which is itself a multiplicative functional.

Maybe it's easy, but I just don't see right away how to show that $\widetilde M_t$ is a martingale. How can I show this?

NOTE: This question is related to the following two questions:

• Multiplicative factorization of stochastic growth time series--solving for an eigenfunction/eigenvector
• Example of the change of measure proposed in Hansen (2012)

Answer 1

The idea is that after the Perron-Frobenius problem is solved, you have $e$ such that \begin{align} E \left[\frac{M_{t+1}}{M_t} e(X_{t+1}) \middle | X_t = x \right] &= e(x) \exp(\eta) \\ E \left[\frac{M_{t+1}}{M_t} \frac{e(X_{t+1})}{e(X_t)} \exp(-\eta) \middle | X_t = x \right] &= 1 \\ E \left[\frac{\widetilde M_{t+1}}{\widetilde M_t} \middle | \mathcal F_t \right] &= 1 \\ E \left[\widetilde M_{t+1} \middle | \mathcal F_t \right] &= \widetilde M_t. \end{align}

Thus we see that $\widetilde M_t$ is a martingale.