# 2 versions of the koyck distributed lag

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.

• Where does $v_t$ come from? – caverac Dec 2 '18 at 10:12
• @caverac: I added some explanation at at the bottom. Thanks for good question. – mark leeds Dec 2 '18 at 19:20
• 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 Dec 2 '18 at 22:35
• @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 Dec 2 '18 at 23:45
• I don't think I did much ... but am happy you solved your question – caverac Dec 2 '18 at 23:47

$$y_{t} = \beta \times \sum_{i=0}^{\infty} \rho^{i} x_{t-i} + \epsilon_t + \rho \times \epsilon_{t-1}$$