<p>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 $\eta = (F' + H'B)'(F' + H'B)$. </p>

CC BY-SA 3.0
2015-01-14T23:44:44.740Z
2015-01-20T09:42:06.003Z
2015-01-20T09:42:06.003Z
jmbejara
<p>I'm trying to understand the tools used/presented in <a href="http://www.larspeterhansen.org/documents/2012_Risk_OpMe_E_Dynamic_valuation_decomposition.pdf" rel="nofollow noreferrer">Lars Hansen's Econometrica paper "Dynamic Valuation Decomposition within Stochastic Economies."</a> 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 $\eta$.</p>

<p><strong>Progress:</strong></p>

<p>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)\} &amp;= \exp\{\eta + H' x\} \\
(D' + H'(A - I)) x + \frac 12 (F' + H'B)'(F' + H'B) &amp;= \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?</p>

<p><em>Note:</em> This problem is related to <a href="https://economics.stackexchange.com/questions/1933/decomposition-of-an-additive-functional-into-a-martingale-part-and-other">this problem about the "Decomposition of an additive functional into a Martingale part and other.</a>"</p>

CC BY-SA 3.0
2015-01-14T16:22:54.787Z
What does M(t+1)/M(t) represent?
Rusan Kax
CC BY-SA 3.0
2015-01-14T17:02:19.480Z
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.
jmbejara
<econometrics><economic-growth><risk><uncertainty><eigenvalue>
Multiplicative factorization of stochastic growth time series--solving for an eigenfunction/eigenvector
jmbejara
# Multiplicative factorization of stochastic growth time series--solving for an eigenfunction/eigenvector

(Asked by: username_0 on 2015-01-14)

I'm trying to understand the tools used/presented in [Lars Hansen's Econometrica paper "Dynamic Valuation Decomposition within Stochastic Economies."](http://www.larspeterhansen.org/documents/2012_Risk_OpMe_E_Dynamic_valuation_decomposition.pdf) 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.](https://economics.stackexchange.com/questions/1933/decomposition-of-an-additive-functional-into-a-martingale-part-and-other)"

**Tags:** econometrics, economic-growth, risk, uncertainty, eigenvalue

> **Comments:**
> **username_1**: What does M(t+1)/M(t) represent?  
> **username_0**: 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.  

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## 1 answers

### Answer by username_0 on 2015-01-14. Score: 1 | Accepted answer

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 $\eta = (F' + H'B)'(F' + H'B)$.


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