<p>Zero shocks do not imply zero variations in a VAR model.</p>
<p>In a stationary VAR model, if the shocks starting from a time period <span class="math-container">$t=\tau$</span> 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.</p>
<p>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.</p>
<p>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.</p>

CC BY-SA 4.0
2021-07-01T02:03:50.713Z
@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!
Emmanuel Ameyaw
CC BY-SA 4.0
2021-07-01T10:55:22.420Z
@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?
Richard Hardy
CC BY-SA 4.0
2021-07-02T06:50:26.220Z
@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?
Emmanuel Ameyaw
CC BY-SA 4.0
2021-07-02T07:20:50.843Z
@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.
Richard Hardy
CC BY-SA 4.0
2021-07-03T18:12:00.643Z
Emmanuel: If there are no shocks, then the variables can't move because the shocks are what become the variables of the next period. ( atleast for some variable. who knows which one but they have to become some variable ). Each shock is from a variable.
mark leeds
CC BY-SA 4.0
2021-07-03T18:15:43.623Z
Emmanauel: Let me give a simple example of what I'm saying since above may not be clear. Suppose one has an AR(1) so that $y_t = \phi y_{t-1}   + \epsilon_t$. Then, at time $t$,  this means that $\epsilon_t$ goes into $y_t$. But, in the next period, the $y_t$ becomes the $y_{t-1}$ for period t+1. So, the shocks in some sense are the variables. Of course, in a VAR, it becomes more complicated but the idea is the same. I hope this helps.
mark leeds
CC BY-SA 4.0
2021-07-03T19:35:57.943Z
@markleeds, if $\epsilon_t=0$, then $y_{t+1}=\phi y_t\neq y_t$ unless $\phi=1$ or $y_t=0$. So even when the shocks are zero, the system moves unless it is in a stationary point.
Richard Hardy
CC BY-SA 4.0
2021-07-04T02:41:49.017Z
Hi Richard: Right. I meant zero shocks and stationary point. I was specifically addressing Emmanuel's first comment where he mentioned my name. My apologies for confusion. Actually I've always wondered the following: When someone comments, is there a way to comment right under their comment even if there are other comments there already ?  Thanks.
mark leeds
CC BY-SA 4.0
2021-07-04T09:17:12.797Z
@markleeds, I am not aware of such functionality. I think the comments just come one after another over time.
Richard Hardy
CC BY-SA 4.0
2021-06-29T23:58:27.273Z
excellent explanation.
mark leeds
CC BY-SA 4.0
2021-06-29T12:05:40.323Z
2021-06-29T12:13:09.117Z
2021-06-29T12:13:09.117Z
Richard Hardy
<p>My <strong>short answer</strong> 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 <strong>long answer</strong> is below.</p>
<p>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 <span class="math-container">$y$</span>. A VAR for <span class="math-container">$y$</span> can be written as
<span class="math-container">$$
y_t = c  + A y_{t-1} +\varepsilon_t
$$</span>
if it is stable, the VAR has an infinite moving-average represenation
<span class="math-container">$$
y_t = \mu + \varepsilon_t + \phi_1 \varepsilon_{t-1} +\phi_2 \varepsilon_{t-2} +\cdots = \mu +\sum_{i=0}^{\infty}\phi_i \varepsilon_{t-i}
$$</span>
which is obtained via recursion of the expression at the top. Here, <span class="math-container">$\mu$</span> is the cumulative effect of the constant, that is <span class="math-container">$\mu:=c\sum_{i=0}^{\infty}\phi_i$</span>, and the <span class="math-container">$\phi_i$</span> can be calculated recursively from <span class="math-container">$A$</span>.</p>
<p>The MA representation shows that if all shocks are zero over the whole horizon <span class="math-container">$(-\infty,\infty)$</span>, <span class="math-container">$y_t$</span> will <strong>not move</strong> for any <span class="math-container">$t$</span>. We will have <span class="math-container">$y_t=\mu$</span> for all <span class="math-container">$t$</span>. So you have the same result as in your DSGE model if you are setting shocks to zero over the whole horizon.</p>
<p>Because your comments also suggest that you are interested in another case,  let's look at the dynamics of the system over <span class="math-container">$[t+1,t+h]$</span>, with all shocks set to zero over this forecast horizon. If you assume a starting point <span class="math-container">$y_t$</span> that is off steady state (or equilibrium), in our case <span class="math-container">$y_t \neq \mu$</span>, this implies that not all shocks prior to <span class="math-container">$t$</span> 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 <span class="math-container">$t+1$</span> still has an impact, even at the end of the forecast horizon <span class="math-container">$t+h$</span>. This is due to the auto-regressive structure of the system. To see this, rewrite the MA representation for <span class="math-container">$t+h$</span>:
<span class="math-container">$$
y_{t+h} = \mu + \sum_{i=0}^{\infty}\phi_i \varepsilon_{t+h-i} = \mu + \sum_{i=h}^{\infty}\phi_i \varepsilon_{t+h-i} = \mu + \sum_{j=0}^{\infty}\phi_{h+j} \varepsilon_{t-j}
$$</span>
where the expression on the very right explicitly incorporates your assumptions of zero shocks over the forecast horizon, that is <span class="math-container">$\varepsilon_s=0$</span> for <span class="math-container">$s \in [t+1, t+h]$</span>. Note that <span class="math-container">$i$</span> now starts at <span class="math-container">$h$</span>. So, the history of shocks prior to <span class="math-container">$t+1$</span> is still at play at <span class="math-container">$t+h$</span> (and beyond).</p>
<p>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.</p>

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2021-07-03T13:25:17.397Z
Thanks!!!!!!!!!!!!!!!!!!! Nicely explained.
Emmanuel Ameyaw
CC BY-SA 4.0
2024-03-13T07:17:23.580Z
I don't understand the last equation where you illustrate the effect of zero shocks over the forecast horizon h. Why does time in the expression on the very right start at i=h? If e.g. the forecast horizon h=5, why do zero shocks over the five periods from t+1 to t+5 force the expression on the very right to start at t-5 instead of t? Shocks should be zero starting at t+1, so i should go from zero to infinity.
Pavel Filip
CC BY-SA 4.0
2024-03-14T17:57:00.220Z
Thanks @PavelFilip. There was an h index omitted, which I have now corrected. I also added a term for clarity.
BrsG
CC BY-SA 4.0
2021-07-02T09:33:54.640Z
2024-03-14T17:55:04.450Z
2024-03-14T17:55:04.450Z
BrsG
<p>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.</p>
<p>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)?</p>

<macroeconomics><applied-econometrics>
In VAR models, do variations in the variables come solely from shocks?
Emmanuel Ameyaw
# In VAR models, do variations in the variables come solely from shocks?

(Asked by: username_0 on 2021-06-25)

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)?

**Tags:** macroeconomics, applied-econometrics


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## 2 answers

### Answer by username_3 on 2021-07-02. Score: 5 | Accepted answer

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 $t+h$:
$$
y\_{t+h} = \mu + \sum\_{i=0}^{\infty}\phi\_i \varepsilon\_{t+h-i} = \mu + \sum\_{i=h}^{\infty}\phi\_i \varepsilon\_{t+h-i} = \mu + \sum\_{j=0}^{\infty}\phi\_{h+j} \varepsilon\_{t-j}
$$
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.

> **Comments:**
> **username_0**: Thanks!!!!!!!!!!!!!!!!!!! Nicely explained.  
> **username_4**: I don't understand the last equation where you illustrate the effect of zero shocks over the forecast horizon h. Why does time in the expression on the very right start at i=h? If e.g. the forecast horizon h=5, why do zero shocks over the five periods from t+1 to t+5 force the expression on the very right to start at t-5 instead of t? Shocks should be zero starting at t+1, so i should go from zero to infinity.  
> **username_3**: Thanks @PavelFilip. There was an h index omitted, which I have now corrected. I also added a term for clarity.  

---

### Answer by username_1 on 2021-06-29. Score: 4

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.

> **Comments:**
> **username_2**: excellent explanation.  
> **username_0**: @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!  
> **username_1**: @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?  
> **username_0**: @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?  
> **username_1**: @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.  
> **username_2**: Emmanuel: If there are no shocks, then the variables can't move because the shocks are what become the variables of the next period. ( atleast for some variable. who knows which one but they have to become some variable ). Each shock is from a variable.  
> **username_2**: Emmanauel: Let me give a simple example of what I'm saying since above may not be clear. Suppose one has an AR(1) so that $y_t = \phi y_{t-1}   + \epsilon_t$. Then, at time $t$,  this means that $\epsilon_t$ goes into $y_t$. But, in the next period, the $y_t$ becomes the $y_{t-1}$ for period t+1. So, the shocks in some sense are the variables. Of course, in a VAR, it becomes more complicated but the idea is the same. I hope this helps.  
> **username_1**: @markleeds, if $\epsilon_t=0$, then $y_{t+1}=\phi y_t\neq y_t$ unless $\phi=1$ or $y_t=0$. So even when the shocks are zero, the system moves unless it is in a stationary point.  
> **username_2**: Hi Richard: Right. I meant zero shocks and stationary point. I was specifically addressing Emmanuel's first comment where he mentioned my name. My apologies for confusion. Actually I've always wondered the following: When someone comments, is there a way to comment right under their comment even if there are other comments there already ?  Thanks.  
> **username_1**: @markleeds, I am not aware of such functionality. I think the comments just come one after another over time.  

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