<p>Let <span class="math-container">$\mathbf{X}$</span> be conditionally iid on <span class="math-container">$\{0,1\}$</span> with distribution <span class="math-container">$(P,1-P)$</span>, with <span class="math-container">$P$</span> being a nontrivial random variable. Let <span class="math-container">$Y_t=\limsup_{T\to\infty} T^{-1}\sum_{i=1}^TX_{t-i}$</span>.
Then <span class="math-container">$\textbf{X}\overset{N}{\to}\textbf{Y}$</span> but not <span class="math-container">$\textbf{X}\overset{\text{Sims}}{\to}\textbf{Y}$</span>.</p>
<p>Here is an example that shows that the other direction need not hold either:</p>
<p>Let <span class="math-container">$\mathbf{X}$</span> be iid uniform on <span class="math-container">$\{0,1\}$</span> and let <span class="math-container">$Y_{t+1}=X_t+X_{t-1}\mod 2$</span>. Then <span class="math-container">$\textbf{X}\overset{\text{Sims}}{\to}\textbf{Y}$</span> but not <span class="math-container">$\textbf{X}\overset{N}{\to}\textbf{Y}$</span>.</p>

CC BY-SA 4.0
2023-09-14T05:05:18.743Z
The $\mathbf{X}$ variable is normally known as Bernoulli.
Richard Hardy
CC BY-SA 4.0
2023-09-14T10:49:28.450Z
Yeah, this counterexample seems valid, thank you for it. But it only holds because $Y_{t}$ is independent with $X_s$ for any $s,t$ which seems as an extremely untraditional situation that cant hold for example in continuous processes. However, I mainly asked this question because of the other implication, which is more interesting (that I feel it should hold). Can $\textbf{X}\overset{N}{\to} \textbf{Y}$ but $\textbf{X}\overset{Sims}{\not\to} \textbf{Y}$?
Albert Paradek
CC BY-SA 4.0
2023-09-14T11:44:06.370Z
That direction need not hold either; see my updated answer.
Michael Greinecker
CC BY-SA 4.0
2023-09-14T15:24:12.593Z
Thank you for the update. However, I am not sure if I understand. Won't it be just $Y_t = constant$ due to the law of large numbers?
Albert Paradek
CC BY-SA 4.0
2023-09-14T15:59:28.323Z
@AlbertParadek It will be constant conditional on the realization of $P$. But since $P$ is random, the value of this constant is not known. $X_t$ is slightly informative, and the complete history is perfectly informative.
Michael Greinecker
CC BY-SA 4.0
2023-09-14T16:33:22.187Z
But since we condition on $Y_{past(t)}$, all this information is already known from the past. Therefore, X does not add anything new, right?
Albert Paradek
CC BY-SA 4.0
2023-09-14T19:11:14.783Z
Of course; you are right. I'll think some more about it.
Michael Greinecker
CC BY-SA 4.0
2023-09-12T14:49:14.153Z
2023-09-14T11:43:37.617Z
2023-09-14T11:43:37.617Z
Michael Greinecker
<p>For a bivariate process  <span class="math-container">$(\textbf{X},\textbf{Y})=( (X_t, Y_t)^\top, t\in\mathbb{Z})$</span>, we say that the process <span class="math-container">$\textbf{X}$</span> Sims-causes the process <span class="math-container">$\textbf{Y}$</span> (notation <span class="math-container">$\textbf{X}\overset{Sims}{\to}\textbf{Y}$</span>), if
<span class="math-container">$$
  (Y_{t+s}, s\geq 1)\not\perp X_t \mid    X_{past(t-1)}, Y_{past(t)}, Z_{past(t)},  \text{       for all }t\in\mathbb{Z},
$$</span>
where <span class="math-container">$past(t) = (t, t-1, \dots)$</span>, <span class="math-container">$\perp$</span> means independence and <span class="math-container">$Z$</span> are all revelant (causallty sufficient) variables.</p>
<p>Consider my new definition: we say that the process <span class="math-container">$\textbf{X}$</span> N-causes the process <span class="math-container">$\textbf{Y}$</span> (notation <span class="math-container">$\textbf{X}\overset{N}{\to}\textbf{Y}$</span>), if
<span class="math-container">$$
 (Y_{t+s}, s\geq 1)\not\perp X_t \mid    Y_{past(t)}, Z_{past(t)}, \text{       for all }t\in\mathbb{Z}. 
 $$</span></p>
<p>The difference is that we omit <span class="math-container">$\textbf{X}$</span> in the conditioning set.</p>
<p>Can you give me an example when <span class="math-container">$\textbf{X}\overset{N}{\to}\textbf{Y}$</span> but <span class="math-container">$\textbf{X}\overset{Sims}{\not\to}\textbf{Y}$</span>? Or vice-versa? I feel that in the vast majority of random processes, these definitions should be equivalent. Can you find a condition under which the definitions are equivalent?</p>

CC BY-SA 4.0
2023-09-12T07:23:11.623Z
Where does the process $\mathbf{Z}$ come in?
Michael Greinecker
CC BY-SA 4.0
2023-09-12T11:17:28.533Z
$Z$ are possible confounders. In the most general form, $Z$ should contain all variables in the universe. But for the sake of argument, say that there is at most one common confounder. For example, if $(\textbf{X},\textbf{Y},\textbf{Z})$ follow some statistical model such as VAR model, we assume that neither component of $Z$ is hidden. That is, $Y_{t+1}$ is a function of $(X_{past(t)}, Y_{past(t)}, Z_{past(t)})$. 

How does $Z$ come in in the definition? We condition on $Z_{past(t)}$, because if we wont, we can not distinguish between dependence due to $X\to Y$ or $Y\leftarrow Z\to X$.
Albert Paradek
CC BY-SA 4.0
2023-09-12T11:44:53.817Z
In that case, the definitions should be applied to $(\textbf{X},\textbf{Y},\textbf{Z})$ instead.
Michael Greinecker
CC BY-SA 4.0
2023-09-14T10:55:30.277Z
Whatt do you mean by 'applied'? The definiton relies on $Z$, but it is not specified explicitely
Albert Paradek
CC BY-SA 4.0
2023-09-14T11:21:12.303Z
The definition as stated, is about trivariate processes, not bivariate processes.
Michael Greinecker
<time-series><causality><stochastic-processes><causal-inference>
Granger-Sims causality and subtle differences
Albert Paradek
# Granger-Sims causality and subtle differences

(Asked by: username_0 on 2023-09-11)

For a bivariate process $(\textbf{X},\textbf{Y})=( (X\_t, Y\_t)^\top, t\in\mathbb{Z})$, we say that the process $\textbf{X}$ Sims-causes the process $\textbf{Y}$ (notation $\textbf{X}\overset{Sims}{\to}\textbf{Y}$), if
$$
(Y\_{t+s}, s\geq 1)\not\perp X\_t \mid X\_{past(t-1)}, Y\_{past(t)}, Z\_{past(t)}, \text{ for all }t\in\mathbb{Z},
$$
where $past(t) = (t, t-1, \dots)$, $\perp$ means independence and $Z$ are all revelant (causallty sufficient) variables.

Consider my new definition: we say that the process $\textbf{X}$ N-causes the process $\textbf{Y}$ (notation $\textbf{X}\overset{N}{\to}\textbf{Y}$), if
$$
(Y\_{t+s}, s\geq 1)\not\perp X\_t \mid Y\_{past(t)}, Z\_{past(t)}, \text{ for all }t\in\mathbb{Z}.
$$

The difference is that we omit $\textbf{X}$ in the conditioning set.

Can you give me an example when $\textbf{X}\overset{N}{\to}\textbf{Y}$ but $\textbf{X}\overset{Sims}{\not\to}\textbf{Y}$? Or vice-versa? I feel that in the vast majority of random processes, these definitions should be equivalent. Can you find a condition under which the definitions are equivalent?

**Tags:** time-series, causality, stochastic-processes, causal-inference

> **Comments:**
> **username_1**: Where does the process $\mathbf{Z}$ come in?  
> **username_0**: $Z$ are possible confounders. In the most general form, $Z$ should contain all variables in the universe. But for the sake of argument, say that there is at most one common confounder. For example, if $(\textbf{X},\textbf{Y},\textbf{Z})$ follow some statistical model such as VAR model, we assume that neither component of $Z$ is hidden. That is, $Y_{t+1}$ is a function of $(X_{past(t)}, Y_{past(t)}, Z_{past(t)})$.   How does $Z$ come in in the definition? We condition on $Z_{past(t)}$, because if we wont, we can not distinguish between dependence due to $X\to Y$ or $Y\leftarrow Z\to X$.  
> **username_1**: In that case, the definitions should be applied to $(\textbf{X},\textbf{Y},\textbf{Z})$ instead.  
> **username_0**: Whatt do you mean by 'applied'? The definiton relies on $Z$, but it is not specified explicitely  
> **username_1**: The definition as stated, is about trivariate processes, not bivariate processes.  

---

## 1 answers

### Answer by username_1 on 2023-09-12. Score: 1

Let $\mathbf{X}$ be conditionally iid on $\{0,1\}$ with distribution $(P,1-P)$, with $P$ being a nontrivial random variable. Let $Y\_t=\limsup\_{T\to\infty} T^{-1}\sum\_{i=1}^TX\_{t-i}$.
Then $\textbf{X}\overset{N}{\to}\textbf{Y}$ but not $\textbf{X}\overset{\text{Sims}}{\to}\textbf{Y}$.

Here is an example that shows that the other direction need not hold either:

Let $\mathbf{X}$ be iid uniform on $\{0,1\}$ and let $Y\_{t+1}=X\_t+X\_{t-1}\mod 2$. Then $\textbf{X}\overset{\text{Sims}}{\to}\textbf{Y}$ but not $\textbf{X}\overset{N}{\to}\textbf{Y}$.

> **Comments:**
> **username_2**: The $\mathbf{X}$ variable is normally known as Bernoulli.  
> **username_0**: Yeah, this counterexample seems valid, thank you for it. But it only holds because $Y_{t}$ is independent with $X_s$ for any $s,t$ which seems as an extremely untraditional situation that cant hold for example in continuous processes. However, I mainly asked this question because of the other implication, which is more interesting (that I feel it should hold). Can $\textbf{X}\overset{N}{\to} \textbf{Y}$ but $\textbf{X}\overset{Sims}{\not\to} \textbf{Y}$?  
> **username_1**: That direction need not hold either; see my updated answer.  
> **username_0**: Thank you for the update. However, I am not sure if I understand. Won't it be just $Y_t = constant$ due to the law of large numbers?  
> **username_1**: @AlbertParadek It will be constant conditional on the realization of $P$. But since $P$ is random, the value of this constant is not known. $X_t$ is slightly informative, and the complete history is perfectly informative.  
> **username_0**: But since we condition on $Y_{past(t)}$, all this information is already known from the past. Therefore, X does not add anything new, right?  
> **username_1**: Of course; you are right. I'll think some more about it.  

---

