Does the direction of correlation matter for Instrumental Variable?Identifying instrumental variables for structural modelvalid instrument for oil consumption in IV modelCan you use a moving average as an instrumental variable?Continuous Instrumental Variable?Definition of validity of an instrumental variable2sls - instrumental variables mean and variance of exogenous variableTransformation of instrumental variablesSet of instrumental dummy variables for a discrete endogenous variable - monotonicity assumption, overidentification and balance testsSolving correlation between explanatory variables using instrumental variablesInstrumental variable predicts endogenous variable in an unexpected direction
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Does the direction of correlation matter for Instrumental Variable?
Identifying instrumental variables for structural modelvalid instrument for oil consumption in IV modelCan you use a moving average as an instrumental variable?Continuous Instrumental Variable?Definition of validity of an instrumental variable2sls - instrumental variables mean and variance of exogenous variableTransformation of instrumental variablesSet of instrumental dummy variables for a discrete endogenous variable - monotonicity assumption, overidentification and balance testsSolving correlation between explanatory variables using instrumental variablesInstrumental variable predicts endogenous variable in an unexpected direction
$begingroup$
For a valid instrumental variable:
(1) the instrument must be correlated with the endogenous explanatory variables, and,
(2) the instrument cannot be correlated with the error term in the explanatory equation.
My question is about point 1. Does the direction of the correlation matter? Suppose I choose Z as an instrument for the endogenous variable X. Z is highly correlated with X (and seemingly unrelated to the outcome of interest), but in the opposite direction that I expected. My hunch is that it undermines my assumptions about point 2, but I wasn't quite sure.
instrumental-variables
asked 2 days ago
Patrick SheaPatrick Shea
82
New contributor
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
For a valid instrumental variable:
(1) the instrument must be correlated with the endogenous explanatory variables, and,
(2) the instrument cannot be correlated with the error term in the explanatory equation.
My question is about point 1. Does the direction of the correlation matter? Suppose I choose Z as an instrument for the endogenous variable X. Z is highly correlated with X (and seemingly unrelated to the outcome of interest), but in the opposite direction that I expected. My hunch is that it undermines my assumptions about point 2, but I wasn't quite sure.
instrumental-variables
asked 2 days ago
Patrick SheaPatrick Shea
82
New contributor
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
By direction, do you mean sign? Correlation does not have a direction (unlike, say, causality) but has sign.
$endgroup$
– Richard Hardy
2 days ago
$begingroup$
Yes, I meant sign of the correlation. For example, I expect Z to have a positive correlation with X, but instead find a negative correlation.
$endgroup$
– Patrick Shea
2 days ago
add a comment |
$begingroup$
For a valid instrumental variable:
(1) the instrument must be correlated with the endogenous explanatory variables, and,
(2) the instrument cannot be correlated with the error term in the explanatory equation.
My question is about point 1. Does the direction of the correlation matter? Suppose I choose Z as an instrument for the endogenous variable X. Z is highly correlated with X (and seemingly unrelated to the outcome of interest), but in the opposite direction that I expected. My hunch is that it undermines my assumptions about point 2, but I wasn't quite sure.
instrumental-variables
asked 2 days ago
Patrick SheaPatrick Shea
82
New contributor
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
For a valid instrumental variable:
(1) the instrument must be correlated with the endogenous explanatory variables, and,
(2) the instrument cannot be correlated with the error term in the explanatory equation.
My question is about point 1. Does the direction of the correlation matter? Suppose I choose Z as an instrument for the endogenous variable X. Z is highly correlated with X (and seemingly unrelated to the outcome of interest), but in the opposite direction that I expected. My hunch is that it undermines my assumptions about point 2, but I wasn't quite sure.
instrumental-variables
instrumental-variables
asked 2 days ago
Patrick SheaPatrick Shea
82
New contributor
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Patrick SheaPatrick Shea
82
New contributor
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Patrick SheaPatrick Shea
82
New contributor
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Patrick SheaPatrick Shea
82
asked 2 days ago
Patrick SheaPatrick Shea
82
82
New contributor
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Patrick Shea is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
By direction, do you mean sign? Correlation does not have a direction (unlike, say, causality) but has sign.
$endgroup$
– Richard Hardy
2 days ago
$begingroup$
Yes, I meant sign of the correlation. For example, I expect Z to have a positive correlation with X, but instead find a negative correlation.
$endgroup$
– Patrick Shea
2 days ago
add a comment |
$begingroup$
By direction, do you mean sign? Correlation does not have a direction (unlike, say, causality) but has sign.
$endgroup$
– Richard Hardy
2 days ago
$begingroup$
Yes, I meant sign of the correlation. For example, I expect Z to have a positive correlation with X, but instead find a negative correlation.
$endgroup$
– Patrick Shea
2 days ago
$begingroup$
By direction, do you mean sign? Correlation does not have a direction (unlike, say, causality) but has sign.
$endgroup$
– Richard Hardy
2 days ago
$begingroup$
By direction, do you mean sign? Correlation does not have a direction (unlike, say, causality) but has sign.
$endgroup$
– Richard Hardy
2 days ago
$begingroup$
Yes, I meant sign of the correlation. For example, I expect Z to have a positive correlation with X, but instead find a negative correlation.
$endgroup$
– Patrick Shea
2 days ago
$begingroup$
Yes, I meant sign of the correlation. For example, I expect Z to have a positive correlation with X, but instead find a negative correlation.
$endgroup$
– Patrick Shea
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, direction of correlation does not matter for the purpose at hand.
In the simplest IV-estimation case, the estimator emerges from the assumption of no correlation,
$$E(Z'u) = 0 implies E[Z'(y-Xb)] = 0 implies E(Z'y) = E(Z'X)b$$
Using sample averages instead of expected values as we do with method of moments estimation, and ignoring the scaling $1/n$ we arrive at
$$hat b_IV = (Z'X)^-1Z'y = (Z'X)^-1Z'(Xb+u) = b + (Z'X)^-1Z'u$$
For this estimator to be feasible we want $E(Z'X) neq 0 $ so that we can invert the matrix. The sign of the relation does not matter.
answered 2 days ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297198
$endgroup$
add a comment |
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1 Answer
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$begingroup$
No, direction of correlation does not matter for the purpose at hand.
In the simplest IV-estimation case, the estimator emerges from the assumption of no correlation,
$$E(Z'u) = 0 implies E[Z'(y-Xb)] = 0 implies E(Z'y) = E(Z'X)b$$
Using sample averages instead of expected values as we do with method of moments estimation, and ignoring the scaling $1/n$ we arrive at
$$hat b_IV = (Z'X)^-1Z'y = (Z'X)^-1Z'(Xb+u) = b + (Z'X)^-1Z'u$$
For this estimator to be feasible we want $E(Z'X) neq 0 $ so that we can invert the matrix. The sign of the relation does not matter.
answered 2 days ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297198
$endgroup$
add a comment |
$begingroup$
No, direction of correlation does not matter for the purpose at hand.
In the simplest IV-estimation case, the estimator emerges from the assumption of no correlation,
$$E(Z'u) = 0 implies E[Z'(y-Xb)] = 0 implies E(Z'y) = E(Z'X)b$$
Using sample averages instead of expected values as we do with method of moments estimation, and ignoring the scaling $1/n$ we arrive at
$$hat b_IV = (Z'X)^-1Z'y = (Z'X)^-1Z'(Xb+u) = b + (Z'X)^-1Z'u$$
For this estimator to be feasible we want $E(Z'X) neq 0 $ so that we can invert the matrix. The sign of the relation does not matter.
answered 2 days ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297198
$endgroup$
add a comment |
$begingroup$
No, direction of correlation does not matter for the purpose at hand.
In the simplest IV-estimation case, the estimator emerges from the assumption of no correlation,
$$E(Z'u) = 0 implies E[Z'(y-Xb)] = 0 implies E(Z'y) = E(Z'X)b$$
Using sample averages instead of expected values as we do with method of moments estimation, and ignoring the scaling $1/n$ we arrive at
$$hat b_IV = (Z'X)^-1Z'y = (Z'X)^-1Z'(Xb+u) = b + (Z'X)^-1Z'u$$
For this estimator to be feasible we want $E(Z'X) neq 0 $ so that we can invert the matrix. The sign of the relation does not matter.
answered 2 days ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297198
$endgroup$
No, direction of correlation does not matter for the purpose at hand.
In the simplest IV-estimation case, the estimator emerges from the assumption of no correlation,
$$E(Z'u) = 0 implies E[Z'(y-Xb)] = 0 implies E(Z'y) = E(Z'X)b$$
Using sample averages instead of expected values as we do with method of moments estimation, and ignoring the scaling $1/n$ we arrive at
$$hat b_IV = (Z'X)^-1Z'y = (Z'X)^-1Z'(Xb+u) = b + (Z'X)^-1Z'u$$
For this estimator to be feasible we want $E(Z'X) neq 0 $ so that we can invert the matrix. The sign of the relation does not matter.
answered 2 days ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297198
answered 2 days ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297198
answered 2 days ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297198
answered 2 days ago
Alecos PapadopoulosAlecos Papadopoulos
42.8k297198
42.8k297198
add a comment |
add a comment |
Patrick Shea is a new contributor. Be nice, and check out our Code of Conduct.
Patrick Shea is a new contributor. Be nice, and check out our Code of Conduct.
Patrick Shea is a new contributor. Be nice, and check out our Code of Conduct.
Patrick Shea is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
By direction, do you mean sign? Correlation does not have a direction (unlike, say, causality) but has sign.
$endgroup$
– Richard Hardy
2 days ago
$begingroup$
Yes, I meant sign of the correlation. For example, I expect Z to have a positive correlation with X, but instead find a negative correlation.
$endgroup$
– Patrick Shea
2 days ago