# Interpret Regression Coefficients After various Differencing

There are few explanations I can find that describe how to interpret linear regression coefficients after differencing a time series (to eliminate a unit root). Is it just so simple that there is no need to state it formally?

(I am aware of this question, but was not sure how general it's response was).

Lets say we are interested in the model $Y_{t}=\beta_{0}+\beta_{1}X_{1t}+\beta_{2}X_{2t} + +...+\beta_{p}X_{pt}+ \delta_{t}$ where $\delta_{t}$ is possibly ARMA(p,q). It is the $\beta_{1}$, $\beta_{2}$,...$\beta_{p}$ that are of interest. Specifically the interpretation in terms of "a 1-unit change in $X_{i}$ results in an average change in $Y_{t}$ of $\beta_{i}$" for $i = 1...p.$

Now lets say we need to difference $Y_{t}$ due to suspected non-stationarity from a unit root (e.g. ADF Test). We need to then also difference in the same manner, each of the $X_{it}$.

What is the interpretation of the $\beta_{i}$ if:

1. The first difference $Y'_{t}$ is taken of $Y_{t}$ and each of the $X_{it}$?
2. The second difference (difference of the difference) ($Y''_{t}$) is taken of $Y_{t}$ and each of the $X_{it}$?
3. A seasonal difference (e.g. $(1-B^{12})$ for monthly data) is taken of $Y_{t}$ and each of the $X_{it}$?

EDIT 1

I did find one text that mentions differences and interpretation of coefficients and it sounds very similar to the linked question. This is from Alan Pankratz Forecasting with Dynamic Regression pages 119-120:

• May I assume that the time-series are monthly ? That the Y's and X's are log-transforms of economic variables ? – user83346 Sep 5 '15 at 7:00
• The question is more about general interpretation and if various forms of differencing, perhaps with ARMA errors, changes the interpretation from the undifferenced regression. So, no not logged :) – B_Miner Sep 5 '15 at 13:01
• Yes but the interpretation can be as simple as $\beta_1$ is the increase in the growth of $y$ for a unit increase in the growth of $x_1$. Where 'growth' is the month-to-month growth for your question one and the 'year-to-year' growth' for you question. Growth is the absolute growth of y but if y is the log transform of $z$ then it is the relative growth of z. Is it such a kind of interpretation you are asking for ? – user83346 Sep 5 '15 at 13:09
• This comment adds to my confusion on the topic. I find examples where the interpretation does not change at all because the betas are unchanged after differencing, but you are implying (I think) that one needs to use the word growth which implies (I think) that the interpretation changes to the differenced data (change in Y, change in X). – B_Miner Sep 5 '15 at 13:13
• Somewhat related answer here. – Richard Hardy Sep 5 '15 at 15:30

Let's take an example with one independent variable because that's easier in typing.

As you start from $y_t=\beta_0 + \beta_1 x_t$ then the same holds for $y_{t-1}=\beta_0 + \beta_1 x_{t-1}$.

So if I subtract the two then I get $\Delta y= \beta_1 \Delta x$. Therefore the interpretation of coefficient $\beta_1$ does not change, it is the same $\beta_1$ in each of these equations.

But the interpretation of the equation $y_t=\beta_0 + \beta_1 x_t$ is not the same as the interpretation of the equation $\Delta y= \beta_1 \Delta x$. That is what I mean.

So $\beta_1$ is the change in $y$ for a unit change in $x$ but is it also the change in the growth of $y$ for a unit change in the growth of $x$.

The reason for differencing is 'technical': if the series are non-stationary, then I can not estimate $y_t = \beta_0 + \beta_1 x_t$ with OLS. If the differenced series are stationary , then I can use the estimate of $\beta_1$ from the equation $\Delta y= \beta_1 \Delta x$ as as an estimate for $\beta_1$ in the equation $y_t=\beta_0 + \beta_1 x_t$, because it is the same $\beta_1$.

So differencing is a 'technical' trick for finding an estimate of $\beta_1$ in $y_t = \beta_0 + \beta_1 x_t$ when the series are non-stationary. The trick makes use of the fact that the same $\beta_1$ appears in the differenced equation.

Obviously this is not different if there are more than one independent variable.

Note: all this is a consequence of the linearity of the model, if $y=\alpha x + \beta$ then $\Delta y = \alpha \Delta x$ , so the $\alpha$ is at the same time the change in $y$ for a unit change in $x$ but also the change in the growth of y for a unit change in the growth of $x$, it is the same $\alpha$.

• So the interpretation is both ways. But the main point is that if there is differencing (any type of the three in my question or combinations thereof) the original undifferenced beta is still estimated (so the original research question of interest is still available). Correct? Does that still hold if there are Arma errors? – B_Miner Sep 5 '15 at 15:23
• Well if you estimate the $\beta_1$ from the differenced equation, then this estimated $\hat{\beta}_1$ is also an estimate for the $\beta_1$ in the undifferenced equation (because it is the same $\beta_1$). The point is that, in the equation for which you do the estimation, the series must be stationary, then all is fine (else you do not get estimators with desirable properties like unbiasedness). A drawback is of course that you can not estimate $\beta_0$ in this way, so if you want an estimate for $\beta_0$ you will have to look at co-integration. – user83346 Sep 5 '15 at 15:40
• An intercept is rarely of interest though it seems, more important is the B1 to BP that are the coefficients on continuous or dummy variables of interest. And just to clairify, nothing changes in this regard if the errors are not iid but we use ARMA errors? I would guess one needs to consider that in the interpretation with or without differences correct (as the "all else being equal" includes lagged (with AR) values of y being controlled for)? – B_Miner Sep 5 '15 at 15:56
• ARMA errors do not change anything to the interpretation. The only technical matter is that, after differencing you have to have stationary series else the estimate of $\beta_1$ is biased, so if you have ARMA errors but after differencing you get stationary series, then in my opinion all is fine. – user83346 Sep 5 '15 at 15:59
• For seasonal differencing you also get the same $\beta_1$ in the differenced equation as in the 'original' equation, so everything remains valid. In fact, whatever you do, as long as you can show that after the manipulations you have the same $\beta_1$ the reasoning remains valid. – user83346 Sep 5 '15 at 16:01

Take the final Transfer Function and re-express it as a pure right hand side equation. In this form it will be a PDL or ADL. Interpretation will then follow as usual. I implemented that option in AUTOBOX and called it the RIGHT-HAND side. If you post a data set and the model that you wish to use, I will be happy to post the results.

EDITED : TO PRESENT AN ILLUSTRATIVE EXAMPLE TO TEST HYPOTHESIS OF EQUAL COEFFICIENTS:

I took the GASX data set (X first then Y)from the Box-Jenklins text available here http://www.autobox.com/stack/GASX.ASC and estimated a Transfer Function on the undifferenced series and obtained I then introduced simple differencing on both Y and X and obtained . The hypothesis that the coefficients are the same is rejected. The coefficients are similar but definitely not the same. I then tried to introduce an MA coefficient (near 1.) to complete the algebraic exercise of multiplying through by [1-B] but that didn't reproduce the non-differenced results either.

In summary: The answer is they are different but that may be due to the omitted constant term in the undifferenced case.

I decided to simulate two white noise series (X1 and Y1 ) and to estimate an OLS model for them without a constant term and obtained. I then integrated both the X1 and the Y1 white nosie series and obtained two new series (X2 and Y2). Following is the result of an OLS model for X2 AND Y2 [ ][4 The resultant regression coefficient is nearly identical (small variation due to 1 less observation in the X2,Y2 study. Thus I can conclude that the case is proven ( or not rejected) that regression coefficients are comparable. Note that when I introduced a constant in the (X1 versus Y1) the regression coefficient was not the same. Apparently there is a requirement that no constant should be incorporated in the base case (undifferenced) . These findings agree with @f coppens .

• I dont follow - transfer function? Can you show what you mean? – B_Miner Sep 3 '15 at 0:00
• A general transfer function takes the form: Yt=μ+[(ω0−ω1B1−.....−ωsBs)/1−δ1B1−...δrBr)]Xt−b+et where et may have some arima structure – IrishStat Sep 3 '15 at 0:30
• Do I take it from your answer that the interpretation of the $\beta_{i}$ actually do change with differencing? I am not sure how to construct a transfer function from what I have in my question. – B_Miner Sep 3 '15 at 0:39
• The βi interpretation when no differencing is in effect is that the level of Y is affected while if differencing is in place the change in Y is affected. – IrishStat Sep 3 '15 at 2:26
• Look at the link in my question. It seems to say here that the interpretation for a differenced model is exactly the same as the levels. Are you suggesting this is not the case? I am confused by what seems like differences (no pun intended) in answers. – B_Miner Sep 3 '15 at 11:41