My instructor stated that when the dependent variable is 1st differenced, the constant term represents the deterministic change or trend in the dependent variable.

When I search for information about deterministic trends, I am getting conflicting/confusing results. The site below describes the differences between trend-stationarity and difference-stationarity.

What I am taking from this website is that differencing the data does not result in a constant term that can be represented as a deterministic trend. And if the series contains a deterministic trend, differencing is usually not performed.

Can someone help clarify this for me?



1 Answer 1


Let's say you have a model $$ y_t = \alpha + \beta x_t + \epsilon_t $$ where $y_t$ and $x_t$ are your variables and $\alpha$ and $\beta$ are your intercept (constant) and slope, respectively, then obviously $$ y_{t-1} = \alpha + \beta x_{t-1} + \epsilon_{t-1} $$ From first-differencing you get $$ y_t - y_{t-1} = (\alpha + \beta x_t + \epsilon_t) - ( \alpha + \beta x_{t-1} + \epsilon_{t-1} ) $$ which simplifies to $$ \Delta y_t = \beta \Delta x_t + \eta_t $$ where $\Delta$ is the difference operator and $\eta_t$ is the new error term. Obviously the constant $\alpha$ disappeared, because it is time-invariant and cancels out during the differencing. But now assume your model was $$ y_t = \alpha + \beta x_t + \delta t + \epsilon_t $$ which means it included a trend term $\delta t$. Then first-differencing will give you $$ y_t - y_{t-1} = (\alpha + \beta x_t + \delta t + \epsilon_t) - ( \alpha + \beta x_{t-1} + \delta [t-1] + \epsilon_{t-1} ) $$ which again simplifies to $$ \Delta y_t = \delta + \beta \Delta x_t + \eta_t $$ Notice that the only difference between the first "1st-dif" model and this one is $\delta$, which now "looks" like a constant. But remember from the original model that it was the trend term. So your instructor is right.

Think about it like this: how much does $y_t$ change from one period to the next one (which is what $\Delta y_t$ means)? It changes by $\beta$ times the change in $x_t$ plus $\delta$, viz your trend.

  • $\begingroup$ Thank you very much. So if there is no trend, there will be no constant term. If there is a trend, regardless if it is stochastic or deterministic, it will be represented by the constant. Is this correct? $\endgroup$
    – Amaziah
    Mar 22, 2015 at 5:41
  • $\begingroup$ @Amaziah yes, as long as you are careful about what you mean by "constant term" (in this case it should mean $\delta$) $\endgroup$ Mar 22, 2015 at 13:05
  • $\begingroup$ Thank you very much for the clarification and clear explanations. $\endgroup$
    – Amaziah
    Mar 22, 2015 at 15:11

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