# Making sense of the first difference regression model

There must be a fundamental error in my approach. Let's start by stating we have a simple regression with two variables $X_t$ and $Y_t$:

$Y_t = BX_t + e_t$

Where $B$ is the coefficient and $e_t$ is the error term. Next, take the first difference of the said equation by removing $Y_{t-1}$ from both sides:

$Y_t-Y_{t-1} = BX_t+ e_t - Y_{t-1}$

Substitute $Y_{t-1}$ from the first equation:

$Y_t-Y_{t-1} = BX_t+ e_t -BX_{t-1}-e_{t-1}$

=> $ΔY_t = BΔX_t + Δe_t$

The first difference regression is often presented this way, but then when it is actually ran, it is ran by replacing $X_t$ and $Y_t$ by their differences, and not by subtracting $Y_{t-1}$ from both sides:

$ΔY_t = B_1ΔX_t + v_t$

Where $v_t$ is the new error term of the equation. Now, these procedures are not equivalent, so why are they described as such? Further why is the error term of the first difference model often described as $\Delta e_t$, when likewise this isn't true as the error term is not related to the original error term, since the estimated equation is simply different. Finally, why isn't the first difference regression performed by subtracting the $Y_{t-1}$ from both sides, giving equivalent results to the first equation (in this case without cross sectional panel data)?

Actually, the two procedures are the same. The difference between $$\Delta Y_t = B\Delta X_t + \Delta \epsilon_t$$ and $$\Delta Y_t = B\Delta X_t + v_t$$ is that you can estimate the second but not the first because you don't observe $\epsilon_t$. So the first equation is rather a theoretical model whilst the second is the estimating equation that you would use in practice. If you wanted to directly subtract $Y_{t-1}$ from both sides manually then this can only be done if you observe the true errors. You will notice that $v_t$ is an estimate of $\epsilon_t$. Re-arrange the theoretical model and the regression equation, if $\Delta Y_t - B\Delta X_t = \Delta \epsilon_t$ and $\Delta Y_t - B\Delta X_t = v_t$, then it must be true that $\Delta \epsilon_t = v_t$. Consider a simple example with two time periods and $B=0.3$ being constant over time.
$$\begin{array}{c|lc|r} time & Y_t & X_t & Y_t - BX_t =v_t \\ \hline 1 & 10 & 17 & \\ 2 & 13 & 21 & \\ \hline \Delta & 3 & 4 & 3 - 0.3\cdot 4 = 1.8 \end{array}$$
Suppose that $v_t$ was a consistent estimate of $\epsilon_t$ in all periods (which is true here because we have deterministically specified the data generating process by fixing $B$), then $\widehat{v}_t = \Delta \epsilon_t = 1.8$ is the residual from our second regression as an estimate of the error of the first equation.
• You assumed $B$ to be constant over time as well because it has no time subscript. And in general you cannot just subtract $Y_{t-1}$ from both sides because you need to observe $e_t$ for that. – Andy May 24 '15 at 12:22
• And what does $B_1$ mean? If $B$ isn't constant you cannot difference the time periods in the way you did because $B_2 X_t - B_1 X_{t-1} = (B_2 - B_1)\Delta X_t$. – Andy May 24 '15 at 12:33