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Suppose I know the true DGP is a VAR(1) process. Instead of fitting a VAR model, I can still fit univariate ARMA models to each of its components.

Does anyone know whether it will result in biased forecasts or unbiased forecasts?

update:

require(MTS)
require(forecast)

# VAR(1) coefficients
phi <- matrix(c(0.5, -0.5, -0.66, -0.3), 2, 2) 
sigma <- diag(2)
n <- 500
data <- VARMAsim(nobs = n, arlags = 1, phi = phi,
                 skip = 200, sigma = sigma)$series

# one univariate series is ARMA(2, 1) X[t] = 0.2X[t-1] + 0.48X[t-2] + v[t] + 0.204v[t-1]

fit <- arima(data[, 1], order = c(2, 0 , 1), fixed = c(0.2, 0.48, 0.204), 
             include.mean = FALSE)

forecast(fit, h = 1)$mean

phi[1, ] %*% t(tail(data, 1))
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Consider the so-called final equation representation of a VAR in terms of an $ARMA(p,q)$. This builds on Zellner and Palm, J Econometrics 1974.

Multiply the $n$-dimensional $VAR(p)$ $\Phi(L)y_t=\epsilon_t$ with $\Phi(L)^{adj}$, the adjoint matrix (or the adjugate) associated with $\Phi(L)$, in order to obtain the "final equations" $$ \det[\Phi(L)]y_t=\Phi(L)^{adj}\epsilon_t $$ Then, each row of the resulting system gives an $ARMA(p^*,q^*)$ representation for each $y_{it}$: $$ \det[\Phi(L)]y_{it}=\Phi_{i\cdot}(L)^{adj}\epsilon_t, $$ where $\Phi_{i\cdot}(L)^{adj}$ is the $i$th row of $\Phi(L)^{adj}$.

Then, all $y_{it}$ have the same $AR$ polynomial. This follows directly from the fact that $\det[\Phi(L)]$ is a scalar polynomial.

We further have that $p^*\leqslant np$ and $q^*\leqslant (n-1)p$. As for $p^*$, the determinant polynomial has (at most) powers of $n\cdot p$ in the lag operator. As for $q^*$, the adjoint matrix is a collection of cofactors, which are determinants of $(n-1)\times(n-1)$ matrices. Each of the elements of the determinants can contain elements up to $L^p$, hence the result follows because the determinant then has (at most) powers of $(n-1)\cdot p$ in the lag operator.

Hence, if you approximate each equation of a VAR(1) with a marginal AR(1), you will (likely) underspecify the dynamics of that process, as this derivation shows that it will likely have ARMA dynamics. Hence, your forecasts will at least also not be optimal anymore.

As an example, consider the $VAR$ $$y_{t}=\begin{pmatrix} -0.1 & -0.21 \\ 0.7 & 0.76 \\ \end{pmatrix}y_{t-1}+\begin{pmatrix} 0.3 & 0.8 \\ 0.3 & -0.8 \\ \end{pmatrix}y_{t-2}+\epsilon_t$$ We may then show that $$ (1-0.66L+0.571L^2-0.189L^3-0.48L^4)y_{1t}=(1 -0.76L + 0.8L^2)\epsilon_{1t}+(-0.21L + 0.8L^2)\epsilon_{2t} $$ Also, we can verify that $y_{1t}$ is, whatever the correlation of $\epsilon_{1t}$ and $\epsilon_{2t}$, an $ARMA(4,2)$.

Write \begin{eqnarray*} \Phi(L)y_{t}&=&\left[\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}- \begin{pmatrix} -0.1 & -0.21 \\ 0.7 & 0.76 \\ \end{pmatrix}L-\begin{pmatrix} 0.3 & 0.8 \\ 0.3 & -0.8 \\ \end{pmatrix}L^2\right]y_{t}=\epsilon_t\\ &=&\begin{pmatrix} 1+0.1L-0.3L^2&0.21L-0.8L^2\\ -0.7L-0.3L^2&1-0.76L+0.8L^2 \end{pmatrix}y_{t}=\epsilon_t, \end{eqnarray*} so that $$ \det[\Phi(L)]=1-0.66L+0.571L^2-0.189L^3-0.48L^4 $$ and from the relationship $A^{-1}=\text{adj}(A)/\det(A)$ we can easily deduce for the $2\times2$ case that $$ \Phi(L)^{adj}=\begin{pmatrix} 1-0.76L+0.8L^2&-0.21L+0.8L^2\\ 0.7L+0.3L^2&1+0.1L-0.3L^2 \end{pmatrix} $$ One can directly calculate that all autocovariances of the $MA$ part cut off after the first two lags, so that the $MA$ part is indeed of order two.

If you had access to the errors of the VAR, the optimal forecasts of the two representations---the VAR(2) and the ARMA(4,2)---would also be identical. Consider this numerical illustration:

library(tsDyn)
n <- 5
eps <- matrix(rnorm(n*2), ncol=2)
B <- matrix(c(-0.1, -0.21, 0.3, 0.8, 0.7, 0.76, 0.3, -0.8), nrow=2, ncol=4, byrow = T)
y <- VAR.sim(B, n=5, lag=2, include = "none", innov = eps)

(VAR.forecast.y1 <- B[1,1:2]%*%y[n,] + B[1,3:4]%*%y[n-1,])

        [,1]
[1,] 1.029909  
(ARMA.forecast.y1 <- 0.66*y[n,1] - 0.571*y[n-1,1] + 0.189*y[n-2,1] + 0.48*y[n-3,1] 
                          -0.76*eps[n,1] + 0.8*eps[n-1,1] - 0.21*eps[n,2] + 0.8*eps[n-1,2])
[1] 1.029909
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  • $\begingroup$ Yes, I agree with your derivation. I am not fitting univariate AR(1) processes for the components in the VAR(1) process. I am fitting the implied ARMA process. For example, you have derived the univariate process for $y_{1t}$ as ARMA(4, 2). What I want to know is whether fitting ARMA(4, 2) causes any bias? To me, the optimal predictor of VAR(1) for $y_{1t}$ and the optimal predictor under ARMA(4, 2) looks differernt. So I am thinking whether this means, optimal univariate forecasts are biased? $\endgroup$ – shani Feb 16 at 10:27
  • $\begingroup$ OK, it would be helpful to clarify what you are precisely doing in the question. I made an edit showing that, if you had access to the errors of both equations (which you typically do not in practice) the implied forecasts of the two models would be identical. In practice you would however estimate an ARMA(4,2) (assuming you are lucky to know the right model order) and the estimated coefficients for the MA part would somehow estimate a mixture of the coefficients for $\epsilon_{1t}$ and $\epsilon_{2t}$, with likely consequences for the optimal forecast. I am not sure what these exactly are. $\endgroup$ – Christoph Hanck Feb 16 at 15:57
  • $\begingroup$ I derived the coefficients of one of the implied VAR(1) processes and fitted an ARMA model by fixing the coefficients. When I compute its forecast, it is very different from the VAR(1) forecast. I am not sure whether to call this discrepancy as bias or something else. $\endgroup$ – shani Feb 17 at 8:29
  • $\begingroup$ An unbiased forecast usually is understood to be one whose forecast errors have zero mean. I am not sure where exactly what you describe differs from what I illustrated in my edit (where the forecasts are identical). $\endgroup$ – Christoph Hanck Feb 17 at 9:27
  • $\begingroup$ Please refer to my updated post. $\endgroup$ – shani Feb 17 at 9:33

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