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?
$begingroup$
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?
time-series var cointegration vecm
New contributor
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add a comment |
$begingroup$
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?
time-series var cointegration vecm
New contributor
$endgroup$
add a comment |
$begingroup$
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?
time-series var cointegration vecm
New contributor
$endgroup$
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
time-series var cointegration vecm
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Christoph Hanck
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YoungYoung
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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.
New contributor
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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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2 Answers
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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.
New contributor
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
New contributor
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
New contributor
$endgroup$
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.
New contributor
New contributor
answered yesterday
dlnBdlnB
75310
75310
New contributor
New contributor
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited yesterday
answered yesterday
Christoph HanckChristoph Hanck
17.6k34174
17.6k34174
add a comment |
add a comment |
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Young is a new contributor. Be nice, and check out our Code of Conduct.
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