<p>As I explained in the edited question, the comment-question made by @caverac made me write the models out in their respective long form versions. Doing this made it obvious that an MA(1) structure can be imposed and the model is shown below.  As far as I can tell, the long form of this model cannot be transformed to a lagged dependent variable model (what I refer to as the short form ) nicely like can be in the  AR(1) and pure noise cases.</p>

<p>Long Form of Koyck Distributed Lag Model With MA(1) error term:</p>

<p><span class="math-container">$y_{t} = \beta \times \sum_{i=0}^{\infty} \rho^{i} x_{t-i} + \epsilon_t + \rho \times \epsilon_{t-1}$</span></p>

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2018-12-03T03:42:22.227Z
2018-12-03T03:42:22.227Z
mark leeds
<p>In econometrics, ( this material is in a lot of the literature but, IMHO, can most clearly be found in Harvey's text, "econometric analysis of time series"), there are generally two versions of a specific lagged dependent variable which is usually  termed the koyck distributed lag. ( It's a very special case of a distributed lag ). </p>

<p>I write down version 1 and version 2 below.</p>

<p>Version 1 is shown below.</p>

<p>1):   <span class="math-container">$y_t = \rho \times y_{t-1} + \beta \times x_t + \epsilon_t$</span></p>

<p>where <span class="math-container">$\epsilon_t = (v_t - \rho v_{t-1}) \sim N(0, \sigma^2) $</span>.</p>

<p>There is also version 2 which is</p>

<p>2):     <span class="math-container">$y_t = \rho \times y_{t-1} + \beta \times x_t + \epsilon_t$</span></p>

<p>where <span class="math-container">$\epsilon_t \sim N(0, \sigma^2) $</span>.</p>

<p>So, in version 1, the error term is such that <span class="math-container">$v_t$</span> is AR(1) and in version 2), the error term is pure noise.</p>

<p>My question is the following: Is there a way to write to a version where the error term is MA(1) ? I initially figured there must be some kind of symmetry because  we have the pure noise case and the AR(1) case so I figured there must be a version with an MA(1) error term. But now I'm not so sure and actually don't think do.  Thanks for any insights.</p>

<p>EDIT BASED ON GOOD QUESTION IN COMMENT:</p>

<p>Hi: <span class="math-container">$\frac{\epsilon_t}{(1 - \rho L)} = v_t$</span> where <span class="math-container">$L$</span> is the lag operator.</p>

<p>Therefore, <span class="math-container">$v_{t}$</span> can be thought of as an exponentially smoothed average of the past white noise error terms <span class="math-container">$\epsilon_{i}$</span>.</p>

<p>Now, version 1) of the model can be re-written in the following way:</p>

<p><span class="math-container">$y_t = \rho \times y_{t-1} + \beta \times x_t + (1-\rho L) v_{t} $</span></p>

<p>which can be re-written as</p>

<p><span class="math-container">$y_{t}(1 - \rho L) = \beta \times x_t + (1 - \rho L) v_{t} $</span></p>

<p>Then, dividing the whole the equation by <span class="math-container">$(1 - \rho L)$</span> results in what I call the long form of version 1 of the model:</p>

<p><span class="math-container">$y_{t} =  \beta \times \sum_{i=0}^{\infty} \rho^{i} x_{t-i} +  v_{t}$</span></p>

<p>So, the model implies that response is <span class="math-container">$\beta$</span> times an exponentially smoothed average of the past <span class="math-container">$x_{i}$</span>  plus v_{t}. But, we can re-write the last term which results in</p>

<p><span class="math-container">$y_{t} =  \beta \times \sum_{i=0}^{\infty} \rho^{i} \times x_{t-i} + \sum_{i=0}^\infty \rho^{i} \epsilon_{t-i}$</span></p>

<p>So, in version 1), the response can be thought of as an exponentially smoothed version of the past <span class="math-container">$x_{t}$</span> plus the exponentially smoothed version of the past error terms, namely the <span class="math-container">$\epsilon_{i}$</span>.</p>

<p>I won't write it out, but, in  the long form of version 2, the error term is not an exponentially smoothed average of the past <span class="math-container">$\epsilon_{i}$</span>. The past error terms  are not involved and the error term is just <span class="math-container">$\epsilon_{t}$</span>.</p>

<p>Thanks for good question.</p>

<p>SECOND EDIT BASED ON GOOD QUESTION IN COMMENT:</p>

<p>When your question made me write it out in the long form, it hit me that
an MA(1) structure could just be:</p>

<p><span class="math-container">$y_{t} =  \beta \times \sum_{i=0}^{\infty} \rho^{i} \times x_{t-i} + \epsilon_t + \rho \epsilon_{t-1} $</span></p>

<p>This would be an MA(1) error term for the koyck distributed lag. I don't think it simplifies in any way by writing it in the short form but it's still a nice way of thinking of a third possibility. In one case, the error term lasts on period. in another case, the error term is  is an exp smoothed average of past errors and in the last case, the error is linear combination of the last two errors. Clearly your question totally led to this and I think what you're saying in your comment is the same as here, so thanks much for insight.</p>

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2018-12-02T10:12:14.877Z
Where does $v_t$ come from?
caverac
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2018-12-02T19:20:02.100Z
@caverac: I added some explanation at at the bottom. Thanks for good question.
mark leeds
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2018-12-02T22:35:22.977Z
Thanks for the update. So what you are looking for is something of the form $\epsilon_t = \eta_{t} + \theta \eta_{t-1}$, where $\eta_t \sim N(0, s^2)$, is that correct?
caverac
CC BY-SA 4.0
2018-12-02T23:45:39.500Z
@caverac: I added another update but I'm pretty certain that we're on the same page as far as what you said in your latest comment. funny how that worked out in terms of your question giving me the answer. thanks a lot..
mark leeds
CC BY-SA 4.0
2018-12-02T23:47:53.300Z
I don't think I did much ... but am happy you solved your question
caverac
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2018-12-02T23:48:40.853Z
you answered the question because I would not have written it out in that better way if you didn't ask where the $v_t$ came from.
mark leeds
CC BY-SA 4.0
2018-12-03T00:06:00.297Z
does anyone know how to say that the question was answered in the question so that something can be checked. If someone can tell me what to do so that this thread doesn't keep existing in unanswered, I can try to do it.
mark leeds
CC BY-SA 4.0
2018-12-03T00:07:38.330Z
You can write an answer to your question
caverac
CC BY-SA 4.0
2018-12-03T03:43:10.953Z
just did that now but can't check it immediately. thanks for help.
mark leeds
<time-series>
2 versions of the koyck distributed lag
mark leeds
# 2 versions of the koyck distributed lag

(Asked by: username_0 on 2018-12-02)

In econometrics, ( this material is in a lot of the literature but, IMHO, can most clearly be found in Harvey's text, "econometric analysis of time series"), there are generally two versions of a specific lagged dependent variable which is usually termed the koyck distributed lag. ( It's a very special case of a distributed lag ).

I write down version 1 and version 2 below.

Version 1 is shown below.

1): $y\_t = \rho \times y\_{t-1} + \beta \times x\_t + \epsilon\_t$

where $\epsilon\_t = (v\_t - \rho v\_{t-1}) \sim N(0, \sigma^2) $.

There is also version 2 which is

2): $y\_t = \rho \times y\_{t-1} + \beta \times x\_t + \epsilon\_t$

where $\epsilon\_t \sim N(0, \sigma^2) $.

So, in version 1, the error term is such that $v\_t$ is AR(1) and in version 2), the error term is pure noise.

My question is the following: Is there a way to write to a version where the error term is MA(1) ? I initially figured there must be some kind of symmetry because we have the pure noise case and the AR(1) case so I figured there must be a version with an MA(1) error term. But now I'm not so sure and actually don't think do. Thanks for any insights.

EDIT BASED ON GOOD QUESTION IN COMMENT:

Hi: $\frac{\epsilon\_t}{(1 - \rho L)} = v\_t$ where $L$ is the lag operator.

Therefore, $v\_{t}$ can be thought of as an exponentially smoothed average of the past white noise error terms $\epsilon\_{i}$.

Now, version 1) of the model can be re-written in the following way:

$y\_t = \rho \times y\_{t-1} + \beta \times x\_t + (1-\rho L) v\_{t} $

which can be re-written as

$y\_{t}(1 - \rho L) = \beta \times x\_t + (1 - \rho L) v\_{t} $

Then, dividing the whole the equation by $(1 - \rho L)$ results in what I call the long form of version 1 of the model:

$y\_{t} = \beta \times \sum\_{i=0}^{\infty} \rho^{i} x\_{t-i} + v\_{t}$

So, the model implies that response is $\beta$ times an exponentially smoothed average of the past $x\_{i}$ plus v\_{t}. But, we can re-write the last term which results in

$y\_{t} = \beta \times \sum\_{i=0}^{\infty} \rho^{i} \times x\_{t-i} + \sum\_{i=0}^\infty \rho^{i} \epsilon\_{t-i}$

So, in version 1), the response can be thought of as an exponentially smoothed version of the past $x\_{t}$ plus the exponentially smoothed version of the past error terms, namely the $\epsilon\_{i}$.

I won't write it out, but, in the long form of version 2, the error term is not an exponentially smoothed average of the past $\epsilon\_{i}$. The past error terms are not involved and the error term is just $\epsilon\_{t}$.

Thanks for good question.

SECOND EDIT BASED ON GOOD QUESTION IN COMMENT:

When your question made me write it out in the long form, it hit me that
an MA(1) structure could just be:

$y\_{t} = \beta \times \sum\_{i=0}^{\infty} \rho^{i} \times x\_{t-i} + \epsilon\_t + \rho \epsilon\_{t-1} $

This would be an MA(1) error term for the koyck distributed lag. I don't think it simplifies in any way by writing it in the short form but it's still a nice way of thinking of a third possibility. In one case, the error term lasts on period. in another case, the error term is is an exp smoothed average of past errors and in the last case, the error is linear combination of the last two errors. Clearly your question totally led to this and I think what you're saying in your comment is the same as here, so thanks much for insight.

**Tags:** time-series

> **Comments:**
> **username_1**: Where does $v_t$ come from?  
> **username_0**: @username_1: I added some explanation at at the bottom. Thanks for good question.  
> **username_1**: Thanks for the update. So what you are looking for is something of the form $\epsilon_t = \eta_{t} + \theta \eta_{t-1}$, where $\eta_t \sim N(0, s^2)$, is that correct?  
> **username_0**: @username_1: I added another update but I'm pretty certain that we're on the same page as far as what you said in your latest comment. funny how that worked out in terms of your question giving me the answer. thanks a lot..  
> **username_1**: I don't think I did much ... but am happy you solved your question  
> **username_0**: you answered the question because I would not have written it out in that better way if you didn't ask where the $v_t$ came from.  
> **username_0**: does anyone know how to say that the question was answered in the question so that something can be checked. If someone can tell me what to do so that this thread doesn't keep existing in unanswered, I can try to do it.  
> **username_1**: You can write an answer to your question  
> **username_0**: just did that now but can't check it immediately. thanks for help.  

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## 1 answers

### Answer by username_0 on 2018-12-03. Score: 1 | Accepted answer

As I explained in the edited question, the comment-question made by @username_1 made me write the models out in their respective long form versions. Doing this made it obvious that an MA(1) structure can be imposed and the model is shown below. As far as I can tell, the long form of this model cannot be transformed to a lagged dependent variable model (what I refer to as the short form ) nicely like can be in the AR(1) and pure noise cases.

Long Form of Koyck Distributed Lag Model With MA(1) error term:

$y\_{t} = \beta \times \sum\_{i=0}^{\infty} \rho^{i} x\_{t-i} + \epsilon\_t + \rho \times \epsilon\_{t-1}$


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