# Multiplicative factorization of stochastic growth time series--solving for an eigenfunction/eigenvector

I'm trying to understand the tools used/presented in Lars Hansen's Econometrica paper "Dynamic Valuation Decomposition within Stochastic Economies." In a part in the paper, Hansen introduces a long-term approximation. This approximation first requires us to solve for a principle eigenfunction of a kind of conditional expectation operator. As an exercise, I'm trying to solve the following example. $\newcommand{\E}{\mathbb E}$ Suppose that $$X_{t+1} = A X_t + B W_{t+1}$$ where A has stable eigenvalues and $\{W_{t+1} : t = 0,1,... \}$ is an iid sequence of multivariate standard normally distributed random vectors. Suppose that $$\log M_{t+1} - \log M_t = D \cdot X_t + F \cdot W_{t+1}.$$ I want to show that there exists a solution to the equation $$\E \left [ \frac{M_{t+1}}{M_t} e(X_{t+1}) \mid X_t = x \right ] = \exp(\eta) e(x) \tag{1}$$ for $\log e(x) = H \cdot x.$ I also want to compute that value of $\eta$.

Progress:

The beginning of the problem is straightforward. Using the properties of log-normal random variables, we can calculate from (1) \begin{align} \exp\{ (D' + H'A) x + \frac 12 (F' + H' B)'(F' + H'B)\} &= \exp\{\eta + H' x\} \\ (D' + H'(A - I)) x + \frac 12 (F' + H'B)'(F' + H'B) &= \eta. \end{align} This is probably a simple question from here. It seems clear that a solution exists, and I have solved for $\eta.$ But from what I understand, solving this problem involves making a "guess" that $e$ was linear in $x$. I don't know what $H$ is. Thus, I need to solve for $H$ and $\eta$. How would I do this?

Note: This problem is related to this problem about the "Decomposition of an additive functional into a Martingale part and other."

• What does M(t+1)/M(t) represent? Jan 14 '15 at 16:22
• In this exercise, no interpretation is given. In the paper it is defined similarly, where it is used to "model stochastic growth, stochastic discounting, or the product of the two." The $M$ is just indicating that it is a multiplicative functional. Jan 14 '15 at 17:02

The progress given so far appears to be correct. Finishing this problem just requires us to argue that $H$ must be chosen so that the equation $$(D' + H'(A - I)) x + (F + H'B)'(F + H'B) = \eta$$ holds for all $x$. Thus, we must have $D' + H'(A - I) = 0$ and, consequently, $H' = D'(I-A)^{-1}$. This implies that $\eta = (F' + H'B)'(F' + H'B)$.