<p>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 $y_0$.</p>

<p>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 $n-&gt;∞$ in the expansion of $y_t$.</p>

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

<p>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.</p>

<p>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.</p>

<p><strong>Edited to add scan and explanation of the derivation</strong>
<a href="https://i.sstatic.net/9zF0S.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9zF0S.jpg" alt="derivation"></a>
Here's a walk-through of the derivation. Hope my handwriting is legible enough; if not, it's the scanner's fault. </p>

<ul>
<li>(1) shows the equivalent way of writing $y_t$ by applying the
backshift operator $L^k$ (of course, after multiplying everything by $1-ρL^k$ ).</li>
<li>(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.</li>
<li>(4) substitutes in (1) the expressions obtained from (2) and (3).</li>
<li>Multiplications by ρ are developed in (5). Notice the resulting pairs of
(underlined) terms offsetting each other.</li>
<li>(6) is (5) after removing the underlined/offsetting terms.</li>
<li>The pattern is discernible from (6) and expressed in (7) for $n$.</li>
<li>(8) is (7) but with summation notation for the terms involving $x$.</li>
<li>(9) is the limit of (8) when $n$ tends to infinity. There, the
assumption of |ρ|&lt;1 is essential for convergence of the summation
(as justified in the 2nd paragraph I wrote this morning).</li>
</ul>

<p>Since $|ρ|&lt;1$, then $|ρ^n|&lt;&lt;1$ 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.</p>

CC BY-SA 4.0
2018-07-14T17:21:07.853Z
@~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.
mark leeds
CC BY-SA 4.0
2018-07-14T17:32:01.857Z
Also, if you used diiff eqs, the scan is appreciated because I don't remember how to do it that way. thanks.
mark leeds
CC BY-SA 4.0
2018-07-14T20:22:38.520Z
@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.
Iñaki Viggers
CC BY-SA 4.0
2018-07-14T21:20:55.537Z
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.
mark leeds
CC BY-SA 4.0
2018-07-14T21:26:16.533Z
just one other quick thing in case you're interested. you "inserted" epsilon_{t} at the end and the model you derived is equivalent to an "adaptive expectatons"  model. if you had inserted it at the beginning and included it in the derivation as another observable variable ( which in certain cases, is possible ), you get the "partial adjustment" model. Andrew Harvey's text "econometric analysis of time series" covers these two "competing" models  quite nicely. all the best.
mark leeds
CC BY-SA 4.0
2018-07-14T21:40:32.417Z
your derivation is actually REALLY NEAT because it shows explicitly how, the process for the error term being AR(1) causes the final summation expression to only include the  pure error term. The other competing model ( partial adjustment. ) ends up with an infinite sum of the epsilon_{t} with the same weights as x_{t}. Great stuff and thanks.
mark leeds
CC BY-SA 4.0
2018-07-14T22:59:44.023Z
@markleeds You are very welcome. I'm glad it helped. I actually included $ϵ_t$ from the start (see leftmost equation at (1)) to make it a "thorough" generalization of the case k=1 you formulated in the inquiry. Thanks for the reference of the book by Andrew Harvey.
Iñaki Viggers
CC BY-SA 4.0
2018-07-15T00:23:24.017Z
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.
mark leeds
CC BY-SA 4.0
2018-07-15T00:28:35.513Z
and yes, I made a mistake when saying you inserted $\epsilon_{t}$ at the end. I was dead wrong there. actually quite the opposite. my bad.
mark leeds
CC BY-SA 4.0
2018-07-14T11:29:30.377Z
2018-07-14T20:20:01.963Z
2018-07-14T20:20:01.963Z
Iñaki Viggers
<p>Hi All: In time series econometrics, if one has a relationship say</p>

<p>$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:</p>

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

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

<p>$y_{t} = \frac{x_{t}}{1- \rho (L)^{k}}$ 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.</p>

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

CC BY-SA 4.0
2018-07-14T07:48:47.877Z
How about $\sum_{i=0}^{\infty} \rho^i x_{t-i-k}$?
T. G.
CC BY-SA 4.0
2018-07-14T07:55:08.283Z
@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.
mark leeds
CC BY-SA 4.0
2018-07-14T09:40:43.870Z
@ 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.
mark leeds
<time-series>
backshift operator in time series confusion
mark leeds
# backshift operator in time series confusion

(Asked by: username_0 on 2018-07-14)

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.

**Tags:** time-series
> **Comments:**
> **username_1**: How about $\sum_{i=0}^{\infty} \rho^i x_{t-i-k}$?  
> **username_0**: @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.  
> **username_0**: @ 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.  

---

## 1 answers

### Answer by username_2 on 2018-07-14. Score: 0 | Accepted answer

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](https://i.sstatic.net/9zF0S.jpg)](https://i.sstatic.net/9zF0S.jpg)
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.

> **Comments:**
> **username_0**: @~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.  
> **username_0**: Also, if you used diiff eqs, the scan is appreciated because I don't remember how to do it that way. thanks.  
> **username_2**: @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.  
> **username_0**: 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.  
> **username_0**: just one other quick thing in case you're interested. you "inserted" epsilon_{t} at the end and the model you derived is equivalent to an "adaptive expectatons"  model. if you had inserted it at the beginning and included it in the derivation as another observable variable ( which in certain cases, is possible ), you get the "partial adjustment" model. Andrew Harvey's text "econometric analysis of time series" covers these two "competing" models  quite nicely. all the best.  
> **username_0**: your derivation is actually REALLY NEAT because it shows explicitly how, the process for the error term being AR(1) causes the final summation expression to only include the  pure error term. The other competing model ( partial adjustment. ) ends up with an infinite sum of the epsilon_{t} with the same weights as x_{t}. Great stuff and thanks.  
> **username_2**: @markleeds You are very welcome. I'm glad it helped. I actually included $ϵ_t$ from the start (see leftmost equation at (1)) to make it a "thorough" generalization of the case k=1 you formulated in the inquiry. Thanks for the reference of the book by Andrew Harvey.  
> **username_0**: 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.  
> **username_0**: and yes, I made a mistake when saying you inserted $\epsilon_{t}$ at the end. I was dead wrong there. actually quite the opposite. my bad.  

---

