Why do we need a VECM specification if the I(1) processes are cointegrated?Constructing a VECM with a mix of I(0) and I(1) variablesWhy use vector error correction model?Why Are Impulse Responses in VECM Permanent?How to forecast from VECM (in R)?Panel VECM interpretation and specificationImpulse response for cointegrated variablesTesting VECM coefficientsIRF with I(1) variables? VECM needed?Which model to use between VAR and VECM for the following problems (conditions)?Cointegration: comparing IRF for the univariate ECT, versus for the multivariate VECM?

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Why do we need a VECM specification if the I(1) processes are cointegrated?


Constructing a VECM with a mix of I(0) and I(1) variablesWhy use vector error correction model?Why Are Impulse Responses in VECM Permanent?How to forecast from VECM (in R)?Panel VECM interpretation and specificationImpulse response for cointegrated variablesTesting VECM coefficientsIRF with I(1) variables? VECM needed?Which model to use between VAR and VECM for the following problems (conditions)?Cointegration: comparing IRF for the univariate ECT, versus for the multivariate VECM?













2












$begingroup$


I happened to question the rationale of employing VECM, since some empirical studies as in 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 there are 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 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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    2












    $begingroup$


    I happened to question the rationale of employing VECM, since some empirical studies as in 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 there are 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 the consequences of not including the error correction term, though differencing the original series makes it a vector of I(0) processes?










    share|cite|improve this question









    New contributor




    Young is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      2












      2








      2





      $begingroup$


      I happened to question the rationale of employing VECM, since some empirical studies as in 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 there are 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 the consequences of not including the error correction term, though differencing the original series makes it a vector of I(0) processes?










      share|cite|improve this question









      New contributor




      Young is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I happened to question the rationale of employing VECM, since some empirical studies as in 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 there are 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 the consequences of not including the error correction term, though differencing the original series makes it a vector of I(0) processes?







      time-series var cointegration vecm






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      edited yesterday









      Christoph Hanck

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          3












          $begingroup$

          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.






          share|cite|improve this answer








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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
            yesterday











          • $begingroup$
            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
            yesterday


















          1












          $begingroup$

          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 $VAR(p)$
          beginequationtag1label1
          y_t = alpha + Phi_1y_t-1 + ldots + Phi_py_t - p + epsilon_t
          endequation

          Using the lag operators we can write this as
          $$(I - Phi_1L^1 - ldots - Phi_pL^p) cdot y_t = Phi(L) y_t = alpha + epsilon_t$$
          Define
          beginequationtag2label2
          rho equiv Phi_1 + Phi_2 + ldots + Phi_p
          endequation

          and
          beginequationtag3label3
          zeta_s equiv - [Phi_s + 1 + Phi_s + 2 + ldots + Phi_p]
          endequation

          Rewrite $I - Phi_1L^1 - ldots - Phi_pL^p$ by adding and immediately subtracting the coefficient matrices of order $j+1$ to $p$ on the lag operator of order $j$. We get
          $$
          begingathered
          I - [(Phi_1 + Phi_2 + ldots + Phi_p) - (Phi_2 + Phi_3 + ldots + Phi_p)]L hfill \
          - [(Phi_2 + ldots + Phi_p) - (Phi_3 + ldots + Phi_p)]L^2 hfill \
          - [Phi_p-1 + Phi_p - Phi_p]L^p-1 - Phi_pL^p hfill \
          endgathered
          $$



          Using eqref2 and eqref3 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
          beginequationtag4label4
          I - rho L - zeta_1L - zeta_2L^2+zeta_1L^2 - ldots - zeta_p-1L^p-1 + zeta_p-2L^p-1 - ( - zeta_p-1)L^p
          endequation

          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 eqref4 as
          $$I - rho L - (zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1)(1-L)$$
          Hence, we have rewritten eqref1 as
          $$left[ I - rho L - left( zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1 right)(1-L) right]y_t = alpha + epsilon_t$$
          Multiplying out the square brackets, using $Delta=1-L$, applying the lag operators and rearranging yields
          $$y_t = alpha + rho y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t$$
          Subtract $y_t-1$ from either side to get
          beginequationtag5label5
          Delta y_t = alpha - (I - rho )y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t
          endequation

          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.
          $$
          |I - Phi_11^1 - ldots - Phi_p1^p|=|I - Phi_1 - ldots - Phi_p|=0
          $$

          (note the determinant is 0 as matrix does not have full rank). That is, $Phi(1)$ can be decomposed into two $(ntimes h)$ 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 eqref5 as
          beginequationtag6label6
          Delta y_t = alpha + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1-Be_t-1+epsilon_t
          endequation

          Estimating the VAR in first differences implies omitting $Be_t-1$, which is relevant under cointegration.






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            $begingroup$

            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.






            share|cite|improve this answer








            New contributor




            dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$












            • $begingroup$
              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
              yesterday











            • $begingroup$
              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
              yesterday















            3












            $begingroup$

            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.






            share|cite|improve this answer








            New contributor




            dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$












            • $begingroup$
              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
              yesterday











            • $begingroup$
              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
              yesterday













            3












            3








            3





            $begingroup$

            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.






            share|cite|improve this answer








            New contributor




            dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            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.







            share|cite|improve this answer








            New contributor




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            answered yesterday









            dlnBdlnB

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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
              yesterday











            • $begingroup$
              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
              yesterday
















            • $begingroup$
              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
              yesterday











            • $begingroup$
              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
              yesterday















            $begingroup$
            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
            yesterday





            $begingroup$
            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
            yesterday













            $begingroup$
            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
            yesterday




            $begingroup$
            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
            yesterday













            1












            $begingroup$

            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 $VAR(p)$
            beginequationtag1label1
            y_t = alpha + Phi_1y_t-1 + ldots + Phi_py_t - p + epsilon_t
            endequation

            Using the lag operators we can write this as
            $$(I - Phi_1L^1 - ldots - Phi_pL^p) cdot y_t = Phi(L) y_t = alpha + epsilon_t$$
            Define
            beginequationtag2label2
            rho equiv Phi_1 + Phi_2 + ldots + Phi_p
            endequation

            and
            beginequationtag3label3
            zeta_s equiv - [Phi_s + 1 + Phi_s + 2 + ldots + Phi_p]
            endequation

            Rewrite $I - Phi_1L^1 - ldots - Phi_pL^p$ by adding and immediately subtracting the coefficient matrices of order $j+1$ to $p$ on the lag operator of order $j$. We get
            $$
            begingathered
            I - [(Phi_1 + Phi_2 + ldots + Phi_p) - (Phi_2 + Phi_3 + ldots + Phi_p)]L hfill \
            - [(Phi_2 + ldots + Phi_p) - (Phi_3 + ldots + Phi_p)]L^2 hfill \
            - [Phi_p-1 + Phi_p - Phi_p]L^p-1 - Phi_pL^p hfill \
            endgathered
            $$



            Using eqref2 and eqref3 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
            beginequationtag4label4
            I - rho L - zeta_1L - zeta_2L^2+zeta_1L^2 - ldots - zeta_p-1L^p-1 + zeta_p-2L^p-1 - ( - zeta_p-1)L^p
            endequation

            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 eqref4 as
            $$I - rho L - (zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1)(1-L)$$
            Hence, we have rewritten eqref1 as
            $$left[ I - rho L - left( zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1 right)(1-L) right]y_t = alpha + epsilon_t$$
            Multiplying out the square brackets, using $Delta=1-L$, applying the lag operators and rearranging yields
            $$y_t = alpha + rho y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t$$
            Subtract $y_t-1$ from either side to get
            beginequationtag5label5
            Delta y_t = alpha - (I - rho )y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t
            endequation

            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.
            $$
            |I - Phi_11^1 - ldots - Phi_p1^p|=|I - Phi_1 - ldots - Phi_p|=0
            $$

            (note the determinant is 0 as matrix does not have full rank). That is, $Phi(1)$ can be decomposed into two $(ntimes h)$ 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 eqref5 as
            beginequationtag6label6
            Delta y_t = alpha + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1-Be_t-1+epsilon_t
            endequation

            Estimating the VAR in first differences implies omitting $Be_t-1$, which is relevant under cointegration.






            share|cite|improve this answer











            $endgroup$

















              1












              $begingroup$

              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 $VAR(p)$
              beginequationtag1label1
              y_t = alpha + Phi_1y_t-1 + ldots + Phi_py_t - p + epsilon_t
              endequation

              Using the lag operators we can write this as
              $$(I - Phi_1L^1 - ldots - Phi_pL^p) cdot y_t = Phi(L) y_t = alpha + epsilon_t$$
              Define
              beginequationtag2label2
              rho equiv Phi_1 + Phi_2 + ldots + Phi_p
              endequation

              and
              beginequationtag3label3
              zeta_s equiv - [Phi_s + 1 + Phi_s + 2 + ldots + Phi_p]
              endequation

              Rewrite $I - Phi_1L^1 - ldots - Phi_pL^p$ by adding and immediately subtracting the coefficient matrices of order $j+1$ to $p$ on the lag operator of order $j$. We get
              $$
              begingathered
              I - [(Phi_1 + Phi_2 + ldots + Phi_p) - (Phi_2 + Phi_3 + ldots + Phi_p)]L hfill \
              - [(Phi_2 + ldots + Phi_p) - (Phi_3 + ldots + Phi_p)]L^2 hfill \
              - [Phi_p-1 + Phi_p - Phi_p]L^p-1 - Phi_pL^p hfill \
              endgathered
              $$



              Using eqref2 and eqref3 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
              beginequationtag4label4
              I - rho L - zeta_1L - zeta_2L^2+zeta_1L^2 - ldots - zeta_p-1L^p-1 + zeta_p-2L^p-1 - ( - zeta_p-1)L^p
              endequation

              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 eqref4 as
              $$I - rho L - (zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1)(1-L)$$
              Hence, we have rewritten eqref1 as
              $$left[ I - rho L - left( zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1 right)(1-L) right]y_t = alpha + epsilon_t$$
              Multiplying out the square brackets, using $Delta=1-L$, applying the lag operators and rearranging yields
              $$y_t = alpha + rho y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t$$
              Subtract $y_t-1$ from either side to get
              beginequationtag5label5
              Delta y_t = alpha - (I - rho )y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t
              endequation

              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.
              $$
              |I - Phi_11^1 - ldots - Phi_p1^p|=|I - Phi_1 - ldots - Phi_p|=0
              $$

              (note the determinant is 0 as matrix does not have full rank). That is, $Phi(1)$ can be decomposed into two $(ntimes h)$ 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 eqref5 as
              beginequationtag6label6
              Delta y_t = alpha + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1-Be_t-1+epsilon_t
              endequation

              Estimating the VAR in first differences implies omitting $Be_t-1$, which is relevant under cointegration.






              share|cite|improve this answer











              $endgroup$















                1












                1








                1





                $begingroup$

                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 $VAR(p)$
                beginequationtag1label1
                y_t = alpha + Phi_1y_t-1 + ldots + Phi_py_t - p + epsilon_t
                endequation

                Using the lag operators we can write this as
                $$(I - Phi_1L^1 - ldots - Phi_pL^p) cdot y_t = Phi(L) y_t = alpha + epsilon_t$$
                Define
                beginequationtag2label2
                rho equiv Phi_1 + Phi_2 + ldots + Phi_p
                endequation

                and
                beginequationtag3label3
                zeta_s equiv - [Phi_s + 1 + Phi_s + 2 + ldots + Phi_p]
                endequation

                Rewrite $I - Phi_1L^1 - ldots - Phi_pL^p$ by adding and immediately subtracting the coefficient matrices of order $j+1$ to $p$ on the lag operator of order $j$. We get
                $$
                begingathered
                I - [(Phi_1 + Phi_2 + ldots + Phi_p) - (Phi_2 + Phi_3 + ldots + Phi_p)]L hfill \
                - [(Phi_2 + ldots + Phi_p) - (Phi_3 + ldots + Phi_p)]L^2 hfill \
                - [Phi_p-1 + Phi_p - Phi_p]L^p-1 - Phi_pL^p hfill \
                endgathered
                $$



                Using eqref2 and eqref3 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
                beginequationtag4label4
                I - rho L - zeta_1L - zeta_2L^2+zeta_1L^2 - ldots - zeta_p-1L^p-1 + zeta_p-2L^p-1 - ( - zeta_p-1)L^p
                endequation

                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 eqref4 as
                $$I - rho L - (zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1)(1-L)$$
                Hence, we have rewritten eqref1 as
                $$left[ I - rho L - left( zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1 right)(1-L) right]y_t = alpha + epsilon_t$$
                Multiplying out the square brackets, using $Delta=1-L$, applying the lag operators and rearranging yields
                $$y_t = alpha + rho y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t$$
                Subtract $y_t-1$ from either side to get
                beginequationtag5label5
                Delta y_t = alpha - (I - rho )y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t
                endequation

                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.
                $$
                |I - Phi_11^1 - ldots - Phi_p1^p|=|I - Phi_1 - ldots - Phi_p|=0
                $$

                (note the determinant is 0 as matrix does not have full rank). That is, $Phi(1)$ can be decomposed into two $(ntimes h)$ 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 eqref5 as
                beginequationtag6label6
                Delta y_t = alpha + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1-Be_t-1+epsilon_t
                endequation

                Estimating the VAR in first differences implies omitting $Be_t-1$, which is relevant under cointegration.






                share|cite|improve this answer











                $endgroup$



                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 $VAR(p)$
                beginequationtag1label1
                y_t = alpha + Phi_1y_t-1 + ldots + Phi_py_t - p + epsilon_t
                endequation

                Using the lag operators we can write this as
                $$(I - Phi_1L^1 - ldots - Phi_pL^p) cdot y_t = Phi(L) y_t = alpha + epsilon_t$$
                Define
                beginequationtag2label2
                rho equiv Phi_1 + Phi_2 + ldots + Phi_p
                endequation

                and
                beginequationtag3label3
                zeta_s equiv - [Phi_s + 1 + Phi_s + 2 + ldots + Phi_p]
                endequation

                Rewrite $I - Phi_1L^1 - ldots - Phi_pL^p$ by adding and immediately subtracting the coefficient matrices of order $j+1$ to $p$ on the lag operator of order $j$. We get
                $$
                begingathered
                I - [(Phi_1 + Phi_2 + ldots + Phi_p) - (Phi_2 + Phi_3 + ldots + Phi_p)]L hfill \
                - [(Phi_2 + ldots + Phi_p) - (Phi_3 + ldots + Phi_p)]L^2 hfill \
                - [Phi_p-1 + Phi_p - Phi_p]L^p-1 - Phi_pL^p hfill \
                endgathered
                $$



                Using eqref2 and eqref3 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
                beginequationtag4label4
                I - rho L - zeta_1L - zeta_2L^2+zeta_1L^2 - ldots - zeta_p-1L^p-1 + zeta_p-2L^p-1 - ( - zeta_p-1)L^p
                endequation

                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 eqref4 as
                $$I - rho L - (zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1)(1-L)$$
                Hence, we have rewritten eqref1 as
                $$left[ I - rho L - left( zeta_1L + zeta_2L^2 + ldots + zeta_p-1L^p-1 right)(1-L) right]y_t = alpha + epsilon_t$$
                Multiplying out the square brackets, using $Delta=1-L$, applying the lag operators and rearranging yields
                $$y_t = alpha + rho y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t$$
                Subtract $y_t-1$ from either side to get
                beginequationtag5label5
                Delta y_t = alpha - (I - rho )y_t-1 + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1 + epsilon_t
                endequation

                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.
                $$
                |I - Phi_11^1 - ldots - Phi_p1^p|=|I - Phi_1 - ldots - Phi_p|=0
                $$

                (note the determinant is 0 as matrix does not have full rank). That is, $Phi(1)$ can be decomposed into two $(ntimes h)$ 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 eqref5 as
                beginequationtag6label6
                Delta y_t = alpha + zeta_1Delta y_t-1 + zeta_2Delta y_t-2 + ldots + zeta_p-1Delta y_t-p+1-Be_t-1+epsilon_t
                endequation

                Estimating the VAR in first differences implies omitting $Be_t-1$, which is relevant under cointegration.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday

























                answered yesterday









                Christoph HanckChristoph Hanck

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