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Is a regression causal if there are no omitted variables?
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A regression of $y$ on $x$ need not be causal if there are omitted variables which influence both $x$ and $y$. But if not for omitted variables and measurement error, is a regression causal? That is, if every possible variable is included in the regression?
regression bias causality
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Esha
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A regression of $y$ on $x$ need not be causal if there are omitted variables which influence both $x$ and $y$. But if not for omitted variables and measurement error, is a regression causal? That is, if every possible variable is included in the regression?
regression bias causality
asked 1 hour ago
Esha
212
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Esha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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No, even if you included every variable in the world, it could be inverse causal. For example, a planet's proximity to its nearest star could be accurately predicted by the surface temperature of the planet, but clearly the causality goes the other way
– gazza89
1 hour ago
@gazza89 - since that effectively answers the question, you might want to expand it into an answer.
– jbowman
1 hour ago
add a comment |Â
up vote
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up vote
4
down vote
favorite
A regression of $y$ on $x$ need not be causal if there are omitted variables which influence both $x$ and $y$. But if not for omitted variables and measurement error, is a regression causal? That is, if every possible variable is included in the regression?
regression bias causality
asked 1 hour ago
Esha
212
New contributor
Esha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A regression of $y$ on $x$ need not be causal if there are omitted variables which influence both $x$ and $y$. But if not for omitted variables and measurement error, is a regression causal? That is, if every possible variable is included in the regression?
regression bias causality
regression bias causality
asked 1 hour ago
Esha
212
New contributor
Esha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Esha
212
New contributor
Esha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Esha
212
New contributor
Esha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Esha
212
asked 1 hour ago
Esha
212
212
New contributor
Esha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Esha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Esha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
No, even if you included every variable in the world, it could be inverse causal. For example, a planet's proximity to its nearest star could be accurately predicted by the surface temperature of the planet, but clearly the causality goes the other way
– gazza89
1 hour ago
@gazza89 - since that effectively answers the question, you might want to expand it into an answer.
– jbowman
1 hour ago
add a comment |Â
1
No, even if you included every variable in the world, it could be inverse causal. For example, a planet's proximity to its nearest star could be accurately predicted by the surface temperature of the planet, but clearly the causality goes the other way
– gazza89
1 hour ago
@gazza89 - since that effectively answers the question, you might want to expand it into an answer.
– jbowman
1 hour ago
1
1
No, even if you included every variable in the world, it could be inverse causal. For example, a planet's proximity to its nearest star could be accurately predicted by the surface temperature of the planet, but clearly the causality goes the other way
– gazza89
1 hour ago
No, even if you included every variable in the world, it could be inverse causal. For example, a planet's proximity to its nearest star could be accurately predicted by the surface temperature of the planet, but clearly the causality goes the other way
– gazza89
1 hour ago
@gazza89 - since that effectively answers the question, you might want to expand it into an answer.
– jbowman
1 hour ago
@gazza89 - since that effectively answers the question, you might want to expand it into an answer.
– jbowman
1 hour ago
add a comment |Â
1 Answer
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No, it's not, I will show you some counterexamples.
The first is reverse causation. Consider the causal model is $Y rightarrow X$, where $X$ and $Y$ are standard gaussian random variables. Then $E[Y|do(x)] = 0$, since $X$ does not cause $Y$, but $E[Y|x]$ will depend on $X$.
The second example is controlling for colliders. Consider the causal model $X rightarrow Z leftarrow Y$, that is $X$ does not cause $Y$ and $Z$ is a common cause. But note that, if you run a regression including $Z$, the regression coefficient of $X$ will not be zero, because conditioning on the common cause will induce association between $Y$ and $X$ (you may want to see here as well Path Analysis in the Presence of a Conditioned-Upon Collider).
More generally, the regression of $Y$ on $X$ will be causal if the variables included in the regression satisfy the backdoor criterion.
answered 1 hour ago
Carlos Cinelli
4,20331844
1
Highly recommend the Book of Why, by Judea Pearl. Explains thoroughly what Carlos refers to.
– Markos Kashiouris
1 hour ago
add a comment |Â
1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
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up vote
4
down vote
No, it's not, I will show you some counterexamples.
The first is reverse causation. Consider the causal model is $Y rightarrow X$, where $X$ and $Y$ are standard gaussian random variables. Then $E[Y|do(x)] = 0$, since $X$ does not cause $Y$, but $E[Y|x]$ will depend on $X$.
The second example is controlling for colliders. Consider the causal model $X rightarrow Z leftarrow Y$, that is $X$ does not cause $Y$ and $Z$ is a common cause. But note that, if you run a regression including $Z$, the regression coefficient of $X$ will not be zero, because conditioning on the common cause will induce association between $Y$ and $X$ (you may want to see here as well Path Analysis in the Presence of a Conditioned-Upon Collider).
More generally, the regression of $Y$ on $X$ will be causal if the variables included in the regression satisfy the backdoor criterion.
answered 1 hour ago
Carlos Cinelli
4,20331844
1
Highly recommend the Book of Why, by Judea Pearl. Explains thoroughly what Carlos refers to.
– Markos Kashiouris
1 hour ago
add a comment |Â
up vote
4
down vote
No, it's not, I will show you some counterexamples.
The first is reverse causation. Consider the causal model is $Y rightarrow X$, where $X$ and $Y$ are standard gaussian random variables. Then $E[Y|do(x)] = 0$, since $X$ does not cause $Y$, but $E[Y|x]$ will depend on $X$.
The second example is controlling for colliders. Consider the causal model $X rightarrow Z leftarrow Y$, that is $X$ does not cause $Y$ and $Z$ is a common cause. But note that, if you run a regression including $Z$, the regression coefficient of $X$ will not be zero, because conditioning on the common cause will induce association between $Y$ and $X$ (you may want to see here as well Path Analysis in the Presence of a Conditioned-Upon Collider).
More generally, the regression of $Y$ on $X$ will be causal if the variables included in the regression satisfy the backdoor criterion.
answered 1 hour ago
Carlos Cinelli
4,20331844
1
Highly recommend the Book of Why, by Judea Pearl. Explains thoroughly what Carlos refers to.
– Markos Kashiouris
1 hour ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
No, it's not, I will show you some counterexamples.
The first is reverse causation. Consider the causal model is $Y rightarrow X$, where $X$ and $Y$ are standard gaussian random variables. Then $E[Y|do(x)] = 0$, since $X$ does not cause $Y$, but $E[Y|x]$ will depend on $X$.
The second example is controlling for colliders. Consider the causal model $X rightarrow Z leftarrow Y$, that is $X$ does not cause $Y$ and $Z$ is a common cause. But note that, if you run a regression including $Z$, the regression coefficient of $X$ will not be zero, because conditioning on the common cause will induce association between $Y$ and $X$ (you may want to see here as well Path Analysis in the Presence of a Conditioned-Upon Collider).
More generally, the regression of $Y$ on $X$ will be causal if the variables included in the regression satisfy the backdoor criterion.
answered 1 hour ago
Carlos Cinelli
4,20331844
No, it's not, I will show you some counterexamples.
The first is reverse causation. Consider the causal model is $Y rightarrow X$, where $X$ and $Y$ are standard gaussian random variables. Then $E[Y|do(x)] = 0$, since $X$ does not cause $Y$, but $E[Y|x]$ will depend on $X$.
The second example is controlling for colliders. Consider the causal model $X rightarrow Z leftarrow Y$, that is $X$ does not cause $Y$ and $Z$ is a common cause. But note that, if you run a regression including $Z$, the regression coefficient of $X$ will not be zero, because conditioning on the common cause will induce association between $Y$ and $X$ (you may want to see here as well Path Analysis in the Presence of a Conditioned-Upon Collider).
More generally, the regression of $Y$ on $X$ will be causal if the variables included in the regression satisfy the backdoor criterion.
answered 1 hour ago
Carlos Cinelli
4,20331844
edited 57 mins ago
answered 1 hour ago
Carlos Cinelli
4,20331844
answered 1 hour ago
Carlos Cinelli
4,20331844
answered 1 hour ago
Carlos Cinelli
4,20331844
4,20331844
1
Highly recommend the Book of Why, by Judea Pearl. Explains thoroughly what Carlos refers to.
– Markos Kashiouris
1 hour ago
add a comment |Â
1
Highly recommend the Book of Why, by Judea Pearl. Explains thoroughly what Carlos refers to.
– Markos Kashiouris
1 hour ago
1
1
Highly recommend the Book of Why, by Judea Pearl. Explains thoroughly what Carlos refers to.
– Markos Kashiouris
1 hour ago
Highly recommend the Book of Why, by Judea Pearl. Explains thoroughly what Carlos refers to.
– Markos Kashiouris
1 hour ago
add a comment |Â
Esha is a new contributor. Be nice, and check out our Code of Conduct.
Esha is a new contributor. Be nice, and check out our Code of Conduct.
Esha is a new contributor. Be nice, and check out our Code of Conduct.
Esha is a new contributor. Be nice, and check out our Code of Conduct.
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1
No, even if you included every variable in the world, it could be inverse causal. For example, a planet's proximity to its nearest star could be accurately predicted by the surface temperature of the planet, but clearly the causality goes the other way
– gazza89
1 hour ago
@gazza89 - since that effectively answers the question, you might want to expand it into an answer.
– jbowman
1 hour ago