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I happened to question the rationale of employing VECM, since some empirical studies like Basu (2017) employed a VAR model to obtain impulse-response analysis. As far as I know, one should consider using the VECM if the multivariate time-series of interest consists of cointegrated I(1) processes. However, the study did not use the VECM, but estimated a VAR model.

One more question: consider two cointegrated I(1) processes. One could estimate a VAR with differenced processes of these processes, since they become I(0) after differencing, without including the error correction term. Would this be misleading? What would be the consequences of not including the error correction term, though differencing the original series makes it a vector of I(0) processes?

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To formalize and generalize dlnB's +1 answer a little:

Cointegration implies that the deviations from the equilibrium are $I(0)$. Hence, some mechanism must bring back deviations back to the long-run relationship. This idea is formalized using error correction models (ECM).

Assume the following $VAR(p)$ \begin{equation}\tag{1}\label{1} y_t = \alpha + \Phi_1{y_{t-1}} + \ldots + \Phi_p{y_{t - p}} + \epsilon_t \end{equation} Using the lag operators we can write this as $$(I - \Phi_1{L^1} - \ldots - \Phi_pL^p) \cdot y_t = \Phi(L) y_t = \alpha + \epsilon_t$$ Define \begin{equation}\tag{2}\label{2} \rho \equiv \Phi_1 + \Phi_2 + \ldots + \Phi_p \end{equation} and \begin{equation}\tag{3}\label{3} \zeta_s \equiv - [\Phi_{s + 1} + \Phi_{s + 2} + \ldots + \Phi_p] \end{equation} Rewrite $I - \Phi_1{L^1} - \ldots - \Phi_pL^p$ by adding and immediately subtracting the coefficient matrices of order $j+1$ to $p$ on the lag operator of order $j$. We get $$ \begin{gathered} I - [(\Phi_1 + \Phi_2 + \ldots + \Phi_p) - (\Phi_2 + \Phi_3 + \ldots + \Phi_p)]L \hfill \\ - [(\Phi_2 + \ldots + \Phi_p) - (\Phi_3 + \ldots + \Phi_p)]L^2 \hfill \\ - [{\Phi_{p-1}} + \Phi_p - \Phi_p]L^{p-1} - \Phi_pL^p \hfill \\ \end{gathered} $$

Using \eqref{2} and \eqref{3} yields $$I - (\rho + \zeta_1)L - (\zeta_2 - \zeta_1)L^2 - \ldots - (\zeta_{p-1} - {\zeta_{p-2}})L^{p-1} - ( - \zeta_{p-1})L^p_ \cdot $$ Solving the terms in brackets gives \begin{equation}\tag{4}\label{4} I - \rho L - \zeta_1L - \zeta_2L^2+\zeta_1L^2 - \ldots - \zeta_{p-1}L^{p-1} + {\zeta_{p-2}}L^{p-1} - ( - \zeta_{p-1})L^p \end{equation} The $\zeta_i$-matrices appear both before the $i$th lag operator and, with reverse sign, before the $i+1$th lag operator. We can hence rewrite \eqref{4} as $$I - \rho L - (\zeta_1L + \zeta_2L^2 + \ldots + \zeta_{p-1}L^{p-1})(1-L)$$ Hence, we have rewritten \eqref{1} as $$\left[ I - \rho L - \left( \zeta_1L + \zeta_2L^2 + \ldots + \zeta_{p-1}L^{p-1} \right)(1-L) \right]y_t = \alpha + \epsilon_t$$ Multiplying out the square brackets, using $\Delta=1-L$, applying the lag operators and rearranging yields $$y_t = \alpha + \rho y_{t-1} + \zeta_1\Delta y_{t-1} + \zeta_2\Delta y_{t-2} + \ldots + \zeta_{p-1}\Delta y_{t-p+1} + \epsilon_t$$ Subtract $y_{t-1}$ from either side to get \begin{equation}\tag{5}\label{5} \Delta y_t = \alpha - (I - \rho )y_{t-1} + \zeta_1\Delta y_{t-1} + \zeta_2\Delta y_{t-2} + \ldots + \zeta_{p-1}\Delta y_{t-p+1} + \epsilon_t \end{equation} Note $I-\rho=\Phi(1)$. This matrix is of reduced rank, say $h$. To see this, note the $I(1)$ assumption on $y_t$ implies that $\Phi(L)$ has a unit root, i.e. $$ |I - \Phi_1{1^1} - \ldots - \Phi_p1^p|=|I - \Phi_1 - \ldots - \Phi_p|=0 $$ (note the determinant is 0 as matrix does not have full rank). That is, $\Phi(1)$ can be decomposed into two $(n\times h)$ matrices $B$ and $A$ such that $$\Phi(1) = BA'$$ The $h$ rows of $A'$ are the cointegrating relationships. Linear combinations of cointegrating vectors and variables $A'y_{t-1}=e_{t-1}$ are stationary. We can thus rewrite \eqref{5} as \begin{equation}\tag{6}\label{6} \Delta y_t = \alpha + \zeta_1\Delta y_{t-1} + \zeta_2\Delta y_{t-2} + \ldots + \zeta_{p-1}\Delta y_{t-p+1}-Be_{t-1}+\epsilon_t \end{equation} Estimating the VAR in first differences implies omitting $Be_{t-1}$, which is relevant under cointegration.

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  • $\begingroup$ I never saw such a parsimonious derivation of VECM form, thanks! $\endgroup$ – Young Mar 25 '19 at 17:32
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Yes, omitting the error-correction term and estimating a VAR in first-differences when the series are cointegrated is problematic. If the series are cointegrated and you omit the error correction term (once-lagged),the estimates will be biased and inconsistent. The logic is that if the two series are cointegrated, the two series will correct toward the long-run equilibrium. For example, suppose $y_t$ and $x_t$ are two $I(1)$ series that are cointegrated such that $$y_t-a-bx_t=\epsilon_t$$ is a stationary process. This means, if $y_t>a+bx_t$, the resulting forces push the two series closer to the equilibrium level $y-a-bx=0.$ Likewise, if $y_t<a+bx_t$, the resulting forces push the two series farther apart toward to the equilibrium level $y-a-bx=0.$ Hence, it must be the case that $\Delta y_t$ and $\Delta x_t$ depend not only of lagged values of the two series, but also on the distance from the equilibrium $y-a-bx=0$ in the previous period $t-1$. Omitting the lagged error correction term leads to omitted variable bias.

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  • $\begingroup$ Thanks a lot. Would this mean that the expectation of the error term vector in the VAR in first differences is non-zero, though covariance-stationary? I understand that the estimates will be biased in that case, but how can we show that the estimates are inconsistent as well? $\endgroup$ – Young Mar 15 '19 at 7:29
  • $\begingroup$ Ah never mind. Omitting such error correction term would make the lagged differences be correlated with the error term. I got it, thanks. $\endgroup$ – Young Mar 15 '19 at 7:33

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