# The constant term after 1st differencing

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?

http://www.mathworks.com/help/econ/trend-stationary-vs-difference-stationary.html

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.
• @Amaziah yes, as long as you are careful about what you mean by "constant term" (in this case it should mean $\delta$) – shadowtalker Mar 22 '15 at 13:05