In DSGE models, if you shut all shocks to zero, then the variables have zero variations. Thus, they just equal their steady-state values for all periods. So the series for all variables are just flat lines.

Is that the same concept for VAR models? Thus, zero shocks mean zero variations? And is that why (decomposition of the forecast error in Y) = (decomposition of variations in Y)?


2 Answers 2


My short answer is that if shocks are zero over the whole horizon (history, present, and future) then the variables in a VAR will not move at all, just as in a DSGE. You cannot assume a starting point off-equilibrium without implicitly assuming that at least one historical shock was different from zero. If only all future shocks are zero, then the impact of historical shocks will still play out over the future and have an impact on dynamics, due to the auto-regressive structure. My long answer is below.

Assume the variables of interest are stationary (suggested by the comparison with DSGEs, i.e. looking at deviations from steady state or balanced growth path) and are contained in the vector $y$. A VAR for $y$ can be written as $$ y_t = c + A y_{t-1} +\varepsilon_t $$ if it is stable, the VAR has an infinite moving-average represenation $$ y_t = \mu + \varepsilon_t + \phi_1 \varepsilon_{t-1} +\phi_2 \varepsilon_{t-2} +\cdots = \mu +\sum_{i=0}^{\infty}\phi_i \varepsilon_{t-i} $$ which is obtained via recursion of the expression at the top. Here, $\mu$ is the cumulative effect of the constant, that is $\mu:=c\sum_{i=0}^{\infty}\phi_i$, and the $\phi_i$ can be calculated recursively from $A$.

The MA representation shows that if all shocks are zero over the whole horizon $(-\infty,\infty)$, $y_t$ will not move for any $t$. We will have $y_t=\mu$ for all $t$. So you have the same result as in your DSGE model if you are setting shocks to zero over the whole horizon.

Because your comments also suggest that you are interested in another case, let's look at the dynamics of the system over $[t+1,t+h]$, with all shocks set to zero over this forecast horizon. If you assume a starting point $y_t$ that is off steady state (or equilibrium), in our case $y_t \neq \mu$, this implies that not all shocks prior to $t$ can have been zero. Otherwise, you couldn't have gotten to an off-equilibrium point in the first place. In that case, the history of shocks prior to $t+1$ still has an impact, even at the end of the forecast horizon $t+h$. This is due to the auto-regressive structure of the system. To see this, rewrite the MA representation for $t+h$: $$ y_{t+h} = \mu + \sum_{i=0}^{\infty}\phi_i \varepsilon_{t+h-i} = \mu + \sum_{i=h}^{\infty}\phi_i \varepsilon_{t-i} $$ where the expression on the very right explicitly incorporates your assumptions of zero shocks over the forecast horizon, that is $\varepsilon_s=0$ for $s \in [t+1, t+h]$. Note that $i$ now starts at $h$. So, the history of shocks prior to $t+1$ is still at play at $t+h$ (and beyond).

Finally, regarding the relationship between DSGE model and VARs: Generally, the solutions to log-linearised DSGE models can be, and often are, written in VAR form, and the same dynamic considerations apply.

  • $\begingroup$ Thanks!!!!!!!!!!!!!!!!!!! Nicely explained. $\endgroup$ Jul 3, 2021 at 13:25

Zero shocks do not imply zero variations in a VAR model.

In a stationary VAR model, if the shocks starting from a time period $t=\tau$ are all permanently zero, the variables will converge to their unconditional means, but this will not happen immediately. Due to the autoregressive structure, the convergence will be gradual and will never cease to happen, as the variation will get ever smaller.

If you start the process from the stationary point (steady state) and all of the shocks are permanently equal to zero, the process will not move. In that sense all variations come solely from shocks. But if you start elsewhere, then there will be variations as described above, and they will not be due to shocks.

In a nonstationary VAR model, there may be divergence instead of convergence, but again there will be variation. The variables will not just equal their steady state values for all periods.

  • $\begingroup$ excellent explanation. $\endgroup$
    – mark leeds
    Jun 29, 2021 at 23:58
  • $\begingroup$ @Richard and Mark. Thanks!! But I don't understand yet. If you start the process from stationary point (steady state) and all of the shocks are permanently equal to zero like: X_1 = 2X_0 + 3Y_0 (no shocks), where initial values are X_0 = 1, Y_0 = 1. Why won't it move? Here X_1 = 5, and we can also calculate Y_1. I see the system moving with no shocks, or no? Also you say if all of the shocks are permanently equal to zero, then all variations come solely from shocks? I did not get that. Lemme ask though, by steady state, you mean X_0 = 0, Y_0 = 0? And X_0 = 1, Y_0 = 1 is elsewhere? Thanks! $\endgroup$ Jul 1, 2021 at 2:03
  • $\begingroup$ @EmmanuelAmeyaw, the equation you suggest as an example does not look like one coming form a stationary system, and such systems do not have stationary points / steady states. Also, why do you think $(x_0=1,y_0=1)$ is the steady state? $\endgroup$ Jul 1, 2021 at 10:55
  • $\begingroup$ @Richard. Yes you are right. It isn't stationary. My point, however, was that for a non-zero initial values, the system will move even if shock are permanently zero. Or maybe every VAR model has zero stationary values so that the system doesnt move if shocks are zero? $\endgroup$ Jul 2, 2021 at 6:50
  • $\begingroup$ @EmmanuelAmeyaw, if the system moves, the initial values are not the stationary point (by definition). I suppose if all initial values are zeros but the model has nonzero intercepts, the system does move, so this is not a stationary point either. I would try the unconditional means of the variables as a candidate stationary point. $\endgroup$ Jul 2, 2021 at 7:20

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