<p>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]
 &amp;= 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]
 &amp;= 1 \\
E \left[\frac{\widetilde M_{t+1}}{\widetilde M_t} \middle | \mathcal F_t \right]
 &amp;= 1 \\
E \left[\widetilde M_{t+1} \middle | \mathcal F_t \right]
 &amp;= \widetilde M_t.
\end{align}</p>

<p>Thus we see that $\widetilde M_t$ is a martingale.</p>

CC BY-SA 3.0
2015-01-27T05:20:03.067Z
2015-01-27T05:20:03.067Z
jmbejara
<p>In this question, I'm continuing to explore 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> (2012).</p>

<p>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 </p>

<blockquote>
  <p>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.</p>
</blockquote>

<p>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?</p>

<p><strong>NOTE:</strong> This question is related to the following two questions:</p>

<ul>
<li><a href="https://economics.stackexchange.com/questions/3000/multiplicative-factorization-of-stochastic-growth-time-series-solving-for-an-ei">Multiplicative factorization of stochastic growth time series--solving for an eigenfunction/eigenvector</a></li>
<li><a href="https://economics.stackexchange.com/questions/3009/example-of-the-change-of-measure-proposed-in-hansen-2012?lq=1">Example of the change of measure proposed in Hansen (2012)</a></li>
</ul>

<econometrics><finance><time-series><martingales>
Why is the martingale factor a martingale in Hansen's 2012 Dynamic Valuation Decomposition?
jmbejara
# Why is the martingale factor a martingale in Hansen's 2012 Dynamic Valuation Decomposition?

(Asked by: username_0 on 2015-01-20)

In this question, I'm continuing to explore 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) (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](https://economics.stackexchange.com/questions/3000/multiplicative-factorization-of-stochastic-growth-time-series-solving-for-an-ei)
* [Example of the change of measure proposed in Hansen (2012)](https://economics.stackexchange.com/questions/3009/example-of-the-change-of-measure-proposed-in-hansen-2012?lq=1)

**Tags:** econometrics, finance, time-series, martingales


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## 1 answers

### Answer by username_0 on 2015-01-27. Score: 0 | Accepted answer

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.


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