0
$\begingroup$

Hi All: In time series econometrics, if one has a relationship say

$y_{t} = \frac{x_{t}}{1- \rho L} + \epsilon_{t} $ where L is the backshift operator, then the expression can be re-written as the following infinite series:

$y_{t} = \sum_{i=0}^{\infty} \rho^{i} x_{t-i} + \epsilon_{t}$.

But, what is the expression ( and can it be derived easily ) when the original expression is

$y_{t} = \frac{x_{t}}{1- \rho (L)^{k}}$ where $k is some positive integer greater than 1. Just to be clear, since this is coming out so small when I look at it, the rho is not raised to k but the backshift operator is.

Thanks for the wisdom. Also, if it's too complicated or too long to explain, then any reference that explains it is appreciated.

$\endgroup$
  • $\begingroup$ How about $\sum_{i=0}^{\infty} \rho^i x_{t-i-k}$? $\endgroup$ – T. G. Jul 14 '18 at 7:48
  • $\begingroup$ @T.G. I was thinking that possibly also but is there a way to derive it. I'm not even clear on how to derive the more basic result where k = 1 !!! thanks a lot. $\endgroup$ – mark leeds Jul 14 '18 at 7:55
  • $\begingroup$ @ T. G. : I argued to myself heuristically by multiplying both sides of the first equation by $(1-\rho L)$ on both sides. (assume no epsilon). Then rewrite that as difference of $y$ at time t and $y$ at time t-1 = $x_{t}$. Then initialize to $y_{0} = 0$ and just keep predicting 1 step ahead out say n times. Then sub everything and get func of the latest so $y_{t+n}$. You end up with your expression by taking limit as $n$ goes $\infty$. The real way is to use basic dfeqs which I barely recall anyway. your answer was obviously quite helpful in this approach. any other k is similar. $\endgroup$ – mark leeds Jul 14 '18 at 9:40
0
$\begingroup$

From your last comment it seems that you might have already figured out how to proceed, although I'm not following some of your reasoning. I just want to point out that there is no need to initialize $y_0$.

The absolute value of ρ is typically assumed to be lower than 1 (otherwise, the series would be "explosive"). Because of that, the terms $ρ^ny_{t-n*k}$ and $ρ^nϵ_{t-n*k}$ will be zero when taking the limit of $n->∞$ in the expansion of $y_t$.

The expression for $y_t$ is $$ y_t= ∑_{i=0}^{∞} ρ^ix_{t−i*k}+ϵ_t.$$

This means that the effect of $ρL^k$ is to model $y_t$ from $x_t$, $x_{t-k}$, $x_{t-2k}$, $x_{t-3k}$, and so forth. $k=1$ is just a special case of this expression.

I feel too lazy to plug my scanner and scan my derivation right now, but I'll do so if you really want to see it.

Edited to add scan and explanation of the derivation derivation Here's a walk-through of the derivation. Hope my handwriting is legible enough; if not, it's the scanner's fault.

  • (1) shows the equivalent way of writing $y_t$ by applying the backshift operator $L^k$ (of course, after multiplying everything by $1-ρL^k$ ).
  • (2) and (3) apply the backshift operator to $y_{t-k}$ and $y_{t-2*k}$, since each one appears in each one's preceding equation.
  • (4) substitutes in (1) the expressions obtained from (2) and (3).
  • Multiplications by ρ are developed in (5). Notice the resulting pairs of (underlined) terms offsetting each other.
  • (6) is (5) after removing the underlined/offsetting terms.
  • The pattern is discernible from (6) and expressed in (7) for $n$.
  • (8) is (7) but with summation notation for the terms involving $x$.
  • (9) is the limit of (8) when $n$ tends to infinity. There, the assumption of |ρ|<1 is essential for convergence of the summation (as justified in the 2nd paragraph I wrote this morning).

Since $|ρ|<1$, then $|ρ^n|<<1$ for sufficient large $n$. Therefore, the term $ρ^ny_{t-nk}$ in (8) will tend to zero regardless of the value of $y_0$. In other words, the influence of $y_0$ on $y_t$ decreases as $t$ departs from 0.

$\endgroup$
  • $\begingroup$ @~Inaki Viggers. Thanks for your comments and I was similar in that I didn't want to type all in latex so it wasn't clear. What I meant was: ( leaving out error term for clarity ) :A) $ y_1 = \rho y_{0} + x_{1}$ and B) $y_2 = \rho y_{1} + x_{2}$, and c) $y_{i} = \rho y_{i-1} + x_{i} $. Then (assuming $y_{0} = 0$) substitute $y_1$ in A ) into $y_{1}$ in B) and do this over and over and you get the infinite sum that you wrote. The error term depends on the model for $x_{t}$ which is usually some kind of expectations model. IIf what you did is pretty different, then scan is appreciated. $\endgroup$ – mark leeds Jul 14 '18 at 17:21
  • $\begingroup$ Also, if you used diiff eqs, the scan is appreciated because I don't remember how to do it that way. thanks. $\endgroup$ – mark leeds Jul 14 '18 at 17:32
  • $\begingroup$ @markleeds See the scan. Hope that helps. From the explanation you'll see why it isn't necessary to set $y_0$ to zero. Other than that, the derivation is similar to what you outline. $\endgroup$ – Iñaki Viggers Jul 14 '18 at 20:22
  • $\begingroup$ thanks. I'll print it out and see if I can make it out and more importantly, understand it. I'm pretty sure there's a very short way of doing it using diff eqs.. If I happen to come across that solution, I'll put it up somehow. I find that typing latex in comments is much harder than typing latex in an answer. and I don't provide many complete answers !!!!!!!!! thanks so much. $\endgroup$ – mark leeds Jul 14 '18 at 21:20
  • 1
    $\begingroup$ any time. I got a lot out of that derivation. If you like these types of models, you would love Harvey's text. Nerlove's "analysis of economic time series" is also good. Harvey's is better for PA versus AE but Nerlove's coverage is broader and more organized and descriptive. I've read all of Harvey's books and I find his style to be too terse in general. He's big on equation after equation after equation after equation without much discussion. But for what we have been talking about, his discussion is excellent squared. $\endgroup$ – mark leeds Jul 15 '18 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.