# Decomposition of an additive functional into a Martingale part and other

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 $M_t - M_{t-1}$?)

• I think a useful reference for this might be Lars Hansens Econometrica paper "Dynamic Valuation Decomposition within Stochastic Economies." larspeterhansen.org/documents/… – jmbejara Jan 14 '15 at 15:22

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.

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 $\frac{-1}{a_1 -1} W_t$.