<p>What would be the (expected) mean at each period <span class="math-container">$t$</span> of a (in)finite aggregation of <span class="math-container">$AR(1)$</span> processes generated by the same data generating process? 
How would the resulting plot look?
For <span class="math-container">$i.i.d.$</span> data generating process, expected mean would always be zero, in finite or infinite cases. 
What about <span class="math-container">$AR(1)$</span>?</p>

CC BY-SA 4.0
2020-05-29T16:27:46.190Z
If $y_t= \alpha y_{t-1}+\epsilon_t$ where $\epsilon_t$~iid with mean zero and variance $\sigma^2$ then $\mathbb{E}(y_t)=0$. Similarly  $\mathbb{E}(\sum y_t)=0$. All of this follows from the fact that $\epsilon_t$ has mean zero and the aggregation is irrelevant (for the mean). The results change if you assume that the iid term has a non-zero mean. Is that what you want?
Andrew M
CC BY-SA 4.0
2020-06-01T14:32:36.710Z
No, you answered my question,thx!

I just don't get why the following statement is not also true for an AR(1) process, as it is for an idd process:

"
Assenza et al. (2012) elicit forecasts of both future inflation and the future output gap in accordance with the NK model. Moreover, Pfajfar and Zakelj (2011) assume AR(1) processes for the exogenous shocks gt and ut, while Assenza et al. (2012) use IID shocks with the consequence that the rational expectation fundamental solution is an IID process and any observed fluctuations in aggregate variables are endogenously driven by ind. exp.
"
Beck Batucada
CC BY-SA 4.0
2020-06-01T14:34:19.700Z
From this paper: Assenza, T., Bao, T., Hommes, C., & Massaro, D. (2014). Experiments on expectations in macroeconomics and finance. Research in Experimental Economics, 17(320278), 11–70. https://doi.org/10.1108/S0193-230620140000017002


it seems if mean is still zero, statement should hold for both idd and AR(1) processes (with alpha <1)
Beck Batucada
CC BY-SA 4.0
2020-06-02T15:10:33.187Z
Im not sure I understand your issue. $y_t$ as I defined it is an AR(1) process and does have a zero mean if the underlying shocks to that process also have a zero mean and there is no constant in the process. Adding persistence to the process does not change this However, if there is a non-zero short-run mean, then persistence will affect the long-run mean (hence non-stationarity if $\alpha\geq1$). It seems you may be confusing the IID vs. AR(1) shocks for the affect that they have on the model. Those are two different things and depend on the model itself.
Andrew M
CC BY-SA 4.0
2020-06-02T15:16:35.363Z
For example, assume that $\epsilon_t = \rho \epsilon_t + v_t$ where $v_t$~iid mean zero with variance $\sigma^2$. This is an AR(1) process of the shocks to $y_t$. Now since $\mathbb{E}$($\epsilon_t$)=0 from the fact that $v_t$ has mean zero then we also know that $\mathbb{E}$($y_t$)=0 and $\mathbb{E}$($ \sum y_t$)=0 for the same reason. All of this changes if you allow for a non-zero mean in any of these processes but the dynamics in this special case are irrelevant for the mean.
Andrew M
CC BY-SA 4.0
2020-06-02T16:29:45.327Z
Yes I'm trying to understand their effect on the model, as the authors imply idd vs AR(1) shocks matter a lot. However I don't undertand why it would matter if the exected mean is zero in both cases.
Beck Batucada
CC BY-SA 4.0
2020-06-02T21:41:54.597Z
The shocks may have an expected mean of zero but if the model has a non-zero expected mean then the shocks will propagate through the model and possibly alter its mean. However, this really depends on the structure and dynamics of the model itself.
Andrew M
CC BY-SA 4.0
2020-06-03T13:09:27.617Z
Get the concept/idea, but not the mechanics. Anyway, thanks!
Beck Batucada
CC BY-SA 4.0
2020-06-03T17:19:01.900Z
Try this. Assume the model takes following form $y_t=\mu+\epsilon_t$ where we first assume the shocks are $\epsilon_t$~iid mean zero. Then the expected value of the model is $\mathbb{E}(y_t)=\mu$. Now assume $\epsilon_t = \rho \epsilon_t + v_t$ where $v_t$~iid mean zero but the $\epsilon_t$ is an AR(1) process. Then the original model can be rewritten as this model: $y_t = (1-\rho)\mu +\rho y_{t-1} + v_t$. So in this case $\mathbb{E}(y_t)=\frac{(1-\rho)\mu}{(1-\rho)}=\mu$ if $\rho \neq 1$.
Andrew M
CC BY-SA 4.0
2020-06-03T17:25:23.133Z
However, they are no longer equal if the model includes its own dynamics. For example, assume  $y_t=\mu + \alpha y_{t-1}+\epsilon_t$. Then if the shocks are iid we get $\mathbb{E}(y_t)=\frac{\mu}{(1-\alpha)}$ on the other hand if the shocks follow an AR(1) process we get that $\mathbb{E}(y_t)=\frac{(1-\rho) \mu }{ (1-\alpha-\rho+\rho\alpha) }$ so the difference really depends on the dynamics of both the model and of the shocks themselves.
Andrew M
CC BY-SA 4.0
2020-06-04T13:33:05.647Z
Aha, now I get it. Thanks a lot for writing it out!
Beck Batucada
<statistics><time-series>
What would be the mean of a finite aggregation of AR(1) processes?
Beck Batucada
# What would be the mean of a finite aggregation of AR(1) processes?

(Asked by: username_0 on 2020-05-29)

What would be the (expected) mean at each period $t$ of a (in)finite aggregation of $AR(1)$ processes generated by the same data generating process?
How would the resulting plot look?
For $i.i.d.$ data generating process, expected mean would always be zero, in finite or infinite cases.
What about $AR(1)$?

**Tags:** statistics, time-series
> **Comments:**
> **username_1**: If $y_t= \alpha y_{t-1}+\epsilon_t$ where $\epsilon_t$~iid with mean zero and variance $\sigma^2$ then $\mathbb{E}(y_t)=0$. Similarly  $\mathbb{E}(\sum y_t)=0$. All of this follows from the fact that $\epsilon_t$ has mean zero and the aggregation is irrelevant (for the mean). The results change if you assume that the iid term has a non-zero mean. Is that what you want?  
> **username_0**: No, you answered my question,thx!  I just don't get why the following statement is not also true for an AR(1) process, as it is for an idd process:  " Assenza et al. (2012) elicit forecasts of both future inflation and the future output gap in accordance with the NK model. Moreover, Pfajfar and Zakelj (2011) assume AR(1) processes for the exogenous shocks gt and ut, while Assenza et al. (2012) use IID shocks with the consequence that the rational expectation fundamental solution is an IID process and any observed fluctuations in aggregate variables are endogenously driven by ind. exp. "  
> **username_0**: From this paper: Assenza, T., Bao, T., Hommes, C., & Massaro, D. (2014). Experiments on expectations in macroeconomics and finance. Research in Experimental Economics, 17(320278), 11–70. https://doi.org/10.1108/S0193-230620140000017002   it seems if mean is still zero, statement should hold for both idd and AR(1) processes (with alpha <1)  
> **username_1**: Im not sure I understand your issue. $y_t$ as I defined it is an AR(1) process and does have a zero mean if the underlying shocks to that process also have a zero mean and there is no constant in the process. Adding persistence to the process does not change this However, if there is a non-zero short-run mean, then persistence will affect the long-run mean (hence non-stationarity if $\alpha\geq1$). It seems you may be confusing the IID vs. AR(1) shocks for the affect that they have on the model. Those are two different things and depend on the model itself.  
> **username_1**: For example, assume that $\epsilon_t = \rho \epsilon_t + v_t$ where $v_t$~iid mean zero with variance $\sigma^2$. This is an AR(1) process of the shocks to $y_t$. Now since $\mathbb{E}$($\epsilon_t$)=0 from the fact that $v_t$ has mean zero then we also know that $\mathbb{E}$($y_t$)=0 and $\mathbb{E}$($ \sum y_t$)=0 for the same reason. All of this changes if you allow for a non-zero mean in any of these processes but the dynamics in this special case are irrelevant for the mean.  
> **username_0**: Yes I'm trying to understand their effect on the model, as the authors imply idd vs AR(1) shocks matter a lot. However I don't undertand why it would matter if the exected mean is zero in both cases.  
> **username_1**: The shocks may have an expected mean of zero but if the model has a non-zero expected mean then the shocks will propagate through the model and possibly alter its mean. However, this really depends on the structure and dynamics of the model itself.  
> **username_0**: Get the concept/idea, but not the mechanics. Anyway, thanks!  
> **username_1**: Try this. Assume the model takes following form $y_t=\mu+\epsilon_t$ where we first assume the shocks are $\epsilon_t$~iid mean zero. Then the expected value of the model is $\mathbb{E}(y_t)=\mu$. Now assume $\epsilon_t = \rho \epsilon_t + v_t$ where $v_t$~iid mean zero but the $\epsilon_t$ is an AR(1) process. Then the original model can be rewritten as this model: $y_t = (1-\rho)\mu +\rho y_{t-1} + v_t$. So in this case $\mathbb{E}(y_t)=\frac{(1-\rho)\mu}{(1-\rho)}=\mu$ if $\rho \neq 1$.  
> **username_1**: However, they are no longer equal if the model includes its own dynamics. For example, assume  $y_t=\mu + \alpha y_{t-1}+\epsilon_t$. Then if the shocks are iid we get $\mathbb{E}(y_t)=\frac{\mu}{(1-\alpha)}$ on the other hand if the shocks follow an AR(1) process we get that $\mathbb{E}(y_t)=\frac{(1-\rho) \mu }{ (1-\alpha-\rho+\rho\alpha) }$ so the difference really depends on the dynamics of both the model and of the shocks themselves.  
> **username_0**: Aha, now I get it. Thanks a lot for writing it out!  

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## 0 answers

