# Lag polynomial coefficients for integrated stationary processes

I'm following http://faculty.chicagobooth.edu/john.cochrane/research/papers/time_series_book.pdf to ramp up on time series (my background is Elec Eng). I'm having confusion about the meaning of some notation: When a variable that represents a lag polynomial is supplied with an integer argument. Can anyone clarify what is the prevailing interpretation?

In the above link, this occurs on page 112, where the cumulative sum of an impulse response is equated to $a(1)$. As shown in the last paragraph, $a(L)$ is a lag polynomial, so it looks like he means to take $L=1$ as the argument. I'm not sure what that means, since $L$ is not a real number.

I have a similar confusion on pages 48-49, where lag polynomials $A(L)$ and $B(L)$ are referred to with $L=0$. From equation at the top of page 51, it seems that the argument $L=0$ extracts out the coefficient for the $L^0$ term, so it presumes a specific form of the polynomial, i.e. unfactored form. If this is correct, it probably doesn't apply to the $A(L=1)$ on page 112, because the coefficient is already referred to as $a_1$, as shown at the top of page 112.

I guess a related question to the meaning of $a(1)$ on page 112 is: Assuming that it is the coeffecient for the lag-1 term (whatever the reason), why is $a(1)=0$ for a trend-stationary process? The polynomial $a(L)$ applies to the noise source (bottom of page 110), not the output signal $y$.

It might be related to the fact that the lag-1 coefficient is the long-run response, according to the last paragraph on page 52 (again, in the context of an adjacent differencing equation). But there isn't an explanation of why this is the case.

C(L) is simply short hand for evaluating the polynomial C with all the L's replaced by the argument. It doesn't necessarily means that lag L in the model becomes the argument. So that makes sense; if C(L) has no unit root, then replacing L by 1 turns C(L) into a sum of the polynomail coefficients. Similarly, replacing L by zero causes the polynomial expression to evaluate to the constant term.

Thanks to cfh at kaggle.

Consider the linear process $X_{t}=C_{0}\varepsilon_{t}+C_{1}\varepsilon_{t-1}+C_{2}\varepsilon_{t-2}+C_{3}\varepsilon_{t-3}$. We can also write this process as $X_{t}=C\left(L\right)\varepsilon_{t}$ with: $C\left(L\right)=C_{0}+C_{1}L+C_{2}L^{2}+C_{3}L^{3}$. We can then write:

$C\left(L\right)=C_{0}+C_{1}L+\left(C_{2}+C_{3}\right)L^{2}+C_{3}\left(L^{3}-L^{2}\right)$ $C\left(L\right)=C_{0}+\left(C_{1}+C_{2}+C_{3}\right)L+\left(C_{2}+C_{3}\right)\left(L^{2}-L\right)-C_{3}\left(1-L\right)L^{2}$ $C\left(L\right)=\left(C_{0}+C_{1}+C_{2}+C_{3}\right)-\left(C_{1}+C_{2}+C_{3}\right)\left(1-L\right)-\left(C_{2}+C_{3}\right)\left(1-L\right)L-C_{3}\left(1-L\right)L^{2}$ $C\left(L\right)=C\left(1\right)+\left(1-L\right)\left(-\left(C_{1}+C_{2}+C_{3}\right)-\left(C_{2}+C_{3}\right)L-C_{3}L^{2}\right)$ $C\left(L\right)=C\left(1\right)+\left(1-L\right)\left(C_{0}^{*}+C_{1}^{*}L+C_{2}^{*}L^{2}\right)$

We see that $C\left(1\right)=\sum_{i=0}^{\infty}C_{i}$, i.e. the sum of the coefficients. Note that a linear process is $I\left(0\right)$ if $C\left(1\right)=\sum_{i=0}^{\infty}C_{i}\neq0$.

The reason for $A\left(0\right)=I_{p}$ and $B\left(0\right)=I_{p}$ where $p$ is the no. of variable in the VAR is explained on p. 51: “In the original VAR, $A\left(0\right)=I_{p}$, so contemporaneous values of each variable do not appear in the othe variable's equation. Try to write down a $3$ variable VAR and you'll see the point.

If this didn't answer your question let me know so I can amend my answer.

EDIT:

The manipulation above is sometimes called the Beveridge-Nelson decomposition, $C\left(L\right)=C\left(1\right)+\left(1-L\right)\left(C_{0}^{*}+C_{1}^{*}L+C_{2}^{*}L^{2}\right)=C\left(1\right)+\left(1-L\right)C^{*}\left(L\right)$. It simply is a decomposition of the polynomial and it is useful when working with time series. The point I was making is that $C\left(1\right)=\sum_{i=0}^{3}C_{i}=C_{0}+C_{1}+C_{2}+C_{3}$, i.e. the sum of the coefficients.

On the other hand, when we write $X_{t}=C\left(L\right)\varepsilon_{t}$ then we would need to define the lag polynomial. In this case we would write: With $C\left(L\right)=C_{0}+C_{1}L+C_{2}L^{2}+C_{3}L^{3}$. Setting $L=1$ we end up with $C\left(1\right)=C_{0}+C_{1}+C_{2}+C_{3}$ as above.

• Dan, I'm having trouble following the reasoning for the manipulations. Elsewhere, someone reminded me that C(L) is simply short hand for evaluating the polynomial C with all the L's replaced by the argument. It doesn't necessarily means that lag L in the model becomes the argument. So that makes sense; if C(L) has no unit root, then replacing L by 1 turns C(L) into a sum of the polynomail coefficients. Similarly, replacing L by zero causes the polynomial expression to evaluate to the constant term. Am I missing some subtlety that motivates the manipulations above? – StatSmartWannaB May 5 '15 at 5:38
• Look at the expanded answer under EDIT:. – Plissken May 5 '15 at 10:33
• Ah, OK. Thanks. I think I'll simply replace L with the argument. I may be open to diving deeper into the math at some point, but for now, I'm just trying to get the simple intuitive view. – StatSmartWannaB May 6 '15 at 1:44
• Dan, I appreciate your response, but I was confused by the answer. My follow-up comment is the way that I most clearly see it, but it doesn't make sense to mark my own follow-up comment as an answer. I hope this doesn't dissuade you from answering future questions that I might have. – StatSmartWannaB May 12 '15 at 18:19