<p>Usually <em>additive functionals</em> are defined for (strong) Markov processes with continuous sample paths (diffusions) but I suppose you do have a Markov---AR(1)---time series and $\{ Y_t \}$ is indeed additive. So, in your case,</p>

<p>$$
X_t = a_0 + a_1 X_{t-1} + W_t,
$$</p>

<p>and you would like</p>

<p>\begin{align*}
Y_t - Y_{t-1} &amp;= r_1 + (M_t - M_{t-1}) + r_2 (X_t - X_{t-1}) \\
              &amp;= a_0 + W_t + a_1 X_{t-1}.
\end{align*}</p>

<p>If such a decomposition exists, then conditioning on $\sigma(X_1,...X_{t-1})$ gives necessary conditions</p>

<p>\begin{align*}
r_1 + r_2 (a_0 + (a_1 -1) X_{t-1}) &amp;= a_0 + a_1 X_{t-1}  \\
\end{align*}</p>

<p>with solutions
\begin{align*}
r_1 = (1-\frac{a_1}{a_1 -1})a_0, \;r_2 = \frac{a_1}{a_1 -1}.
\end{align*}</p>

<p>One can then substitute $(r_1, r_2)$ into </p>

<p>$$
a_0 + W_t + a_1 X_{t-1} - r_1 -  r_2 (X_t - X_{t-1})
$$</p>

<p>and check that it gives a martingale difference sequence $\frac{-1}{a_1 -1} W_t$.</p>

CC BY-SA 3.0
2017-04-14T14:23:42.890Z
2017-04-14T14:23:42.890Z
Michael
<p>Using the given formulas (inserting the expression for the $X$'s in the sum), we arrive at</p>

<p>$$Y_t = (1-\beta_0)\alpha_0\cdot t + \beta_0\sum_{j=0}^{t-1}X_j +\sum_{j=1}^{t}W_j $$</p>

<p>We do not need a theorem to obtain that, given the description of the problem.</p>

<p>Manipulating, </p>

<p>$$Y_t= (1-\beta_0)\alpha_0\cdot t  - \beta_0(X_t-X_0) + \beta_0Y_{t}  +\sum_{j=1}^{t}W_j$$</p>

<p>$$\implies Y_t = \alpha_0t +\frac{1}{1-\beta_0}\sum_{j=1}^{t}W_j-\frac{\beta_0}{1-\beta_0}(X_t-X_0)$$</p>

<p>and we want to match this with </p>

<p>$$Y_t = r_1 t + M_t - r_2 (X_t - X_0)$$</p>

<p>We immediately get that we must have  $r_1 =  \alpha_0$, $r_2 = \beta_0/(1-\beta_0)$, and 
$$M_t = \frac{1}{1-\beta_0}\sum_{j=1}^{t}W_j$$</p>

<p>Is $M_t$ as obtained a martingale? We have</p>

<p>$$E[M_t\mid \sigma(t-1,t-2,..)] = \\=\frac{1}{1-\beta_0}E[W_t\mid \sigma(t-1,t-2,..)] + \frac{1}{1-\beta_0}\sum_{j=1}^{t-1}W_j$$</p>

<p>$$= \frac{1}{1-\beta_0}E[W_t\mid \sigma(t-1,t-2,..)] + M_{t-1}$$</p>

<p>So <em>if</em> $E[W_t\mid \sigma(t-1,t-2,..)] = 0$, namely, if the original disturbance is mean-independent from the past (in a more fancy term, if $\{W_t\}$ is a "Martingale difference"), then $M_t$ is a martingale and we have our mapping.</p>

CC BY-SA 3.0
2017-04-14T17:26:05.950Z
2017-04-24T11:13:46.950Z
2017-04-24T11:13:46.950Z
Alecos Papadopoulos
<p>This question relates to a theorem about the decomposition of additive functionals---a technique especially useful in macroeconomics and finance. This question has two objective. First, I don't have a reference to the theorem that shows when this is possible, and I would like to find it. Second, I would like to know how to actually find this decomposition.</p>

<p>Consider the following autoregressive model:
$$
X_{t+1} = \alpha_0 + \beta_0 (X_t - \alpha_0) + W_{t+1},
$$
where $-1 &lt; \beta_0 &lt; 1$ and $W_{t+1}$ is distributed as a normal with mean zero and variance one. Now, consider the additive functional
$$
Y_t = \sum_{j=1}^t X_j.
$$
How would I produce the decomposition of the form
$$
Y_t = r_1 t + M_t - r_2 (X_t - X_0)
$$
where $\{M_t \}$ is a Martingale? (How can obtain the values $r_1$, $r_2$, and $M_t - M_{t-1}$?)</p>

CC BY-SA 3.0
2015-01-14T15:22:20.957Z
I think a useful reference for this might be Lars Hansens Econometrica paper "Dynamic Valuation Decomposition within Stochastic Economies."  http://www.larspeterhansen.org/documents/2012_Risk_OpMe_E_Dynamic_valuation_decomposition.pdf
jmbejara
<econometrics><martingales><stochastic-processes>
Decomposition of an additive functional into a Martingale part and other
jmbejara
# Decomposition of an additive functional into a Martingale part and other

(Asked by: username_0 on 2015-01-06)

This question relates to a theorem about the decomposition of additive functionals---a technique especially useful in macroeconomics and finance. This question has two objective. First, I don't have a reference to the theorem that shows when this is possible, and I would like to find it. Second, I would like to know how to actually find this decomposition.

Consider the following autoregressive model:
$$
X\_{t+1} = \alpha\_0 + \beta\_0 (X\_t - \alpha\_0) + W\_{t+1},
$$
where $-1 < \beta\_0 < 1$ and $W\_{t+1}$ is distributed as a normal with mean zero and variance one. Now, consider the additive functional
$$
Y\_t = \sum\_{j=1}^t X\_j.
$$
How would I produce the decomposition of the form
$$
Y\_t = r\_1 t + M\_t - r\_2 (X\_t - X\_0)
$$
where $\{M\_t \}$ is a Martingale? (How can obtain the values $r\_1$, $r\_2$, and $M\_t - M\_{t-1}$?)

**Tags:** econometrics, martingales, stochastic-processes
> **Comments:**
> **username_0**: I think a useful reference for this might be Lars Hansens Econometrica paper "Dynamic Valuation Decomposition within Stochastic Economies."  http://www.larspeterhansen.org/documents/2012_Risk_OpMe_E_Dynamic_valuation_decomposition.pdf  

---

## 2 answers

### Answer by username_2 on 2017-04-14. Score: 2 | Accepted answer

Using the given formulas (inserting the expression for the $X$'s in the sum), we arrive at

$$Y\_t = (1-\beta\_0)\alpha\_0\cdot t + \beta\_0\sum\_{j=0}^{t-1}X\_j +\sum\_{j=1}^{t}W\_j $$

We do not need a theorem to obtain that, given the description of the problem.

Manipulating,

$$Y\_t= (1-\beta\_0)\alpha\_0\cdot t - \beta\_0(X\_t-X\_0) + \beta\_0Y\_{t} +\sum\_{j=1}^{t}W\_j$$

$$\implies Y\_t = \alpha\_0t +\frac{1}{1-\beta\_0}\sum\_{j=1}^{t}W\_j-\frac{\beta\_0}{1-\beta\_0}(X\_t-X\_0)$$

and we want to match this with

$$Y\_t = r\_1 t + M\_t - r\_2 (X\_t - X\_0)$$

We immediately get that we must have $r\_1 = \alpha\_0$, $r\_2 = \beta\_0/(1-\beta\_0)$, and
$$M\_t = \frac{1}{1-\beta\_0}\sum\_{j=1}^{t}W\_j$$

Is $M\_t$ as obtained a martingale? We have

$$E[M\_t\mid \sigma(t-1,t-2,..)] = \\=\frac{1}{1-\beta\_0}E[W\_t\mid \sigma(t-1,t-2,..)] + \frac{1}{1-\beta\_0}\sum\_{j=1}^{t-1}W\_j$$

$$= \frac{1}{1-\beta\_0}E[W\_t\mid \sigma(t-1,t-2,..)] + M\_{t-1}$$

So *if* $E[W\_t\mid \sigma(t-1,t-2,..)] = 0$, namely, if the original disturbance is mean-independent from the past (in a more fancy term, if $\{W\_t\}$ is a "Martingale difference"), then $M\_t$ is a martingale and we have our mapping.


---

### Answer by username_1 on 2017-04-14. Score: 1

Usually *additive functionals* are defined for (strong) Markov processes with continuous sample paths (diffusions) but I suppose you do have a Markov---AR(1)---time series and $\{ Y\_t \}$ is indeed additive. So, in your case,

$$
X\_t = a\_0 + a\_1 X\_{t-1} + W\_t,
$$

and you would like

\begin{align\*}
Y\_t - Y\_{t-1} &= r\_1 + (M\_t - M\_{t-1}) + r\_2 (X\_t - X\_{t-1}) \\
&= a\_0 + W\_t + a\_1 X\_{t-1}.
\end{align\*}

If such a decomposition exists, then conditioning on $\sigma(X\_1,...X\_{t-1})$ gives necessary conditions

\begin{align\*}
r\_1 + r\_2 (a\_0 + (a\_1 -1) X\_{t-1}) &= a\_0 + a\_1 X\_{t-1} \\
\end{align\*}

with solutions
\begin{align\*}
r\_1 = (1-\frac{a\_1}{a\_1 -1})a\_0, \;r\_2 = \frac{a\_1}{a\_1 -1}.
\end{align\*}

One can then substitute $(r\_1, r\_2)$ into

$$
a\_0 + W\_t + a\_1 X\_{t-1} - r\_1 - r\_2 (X\_t - X\_{t-1})
$$

and check that it gives a martingale difference sequence $\frac{-1}{a\_1 -1} W\_t$.


---

