Interpret a multiple linear regression when Y is log transformed
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I have the following multiple linear regression model:
Log(y) = B0 + B1X1 + B2X2 + B3x3 + e.
X1 is a dummy that can take 0 = male and 1 = female and
X2 and X3 are continuous variables.
I am not entirely sure on how to interpret the coefficients for the variables.
The coefficient for the dummy variable is 0,20. Does that mean, that changing from male to female (male is baseline) the Y will increase by an average of 20%. Is it directly translated into percentage?
And for the continuous variables, the coefficient for X2 is 0,1. Does that mean that increasing X2 with 1 unit increases Y with an average of 10%? Again is it directly translated into percentage?
Sorry for this basic question, i was not, however, able to find a clear answer if it is directly translated into percentages or not.
All the best
regression log
asked 1 hour ago
maSmaS
61
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maS is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
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I have the following multiple linear regression model:
Log(y) = B0 + B1X1 + B2X2 + B3x3 + e.
X1 is a dummy that can take 0 = male and 1 = female and
X2 and X3 are continuous variables.
I am not entirely sure on how to interpret the coefficients for the variables.
The coefficient for the dummy variable is 0,20. Does that mean, that changing from male to female (male is baseline) the Y will increase by an average of 20%. Is it directly translated into percentage?
And for the continuous variables, the coefficient for X2 is 0,1. Does that mean that increasing X2 with 1 unit increases Y with an average of 10%? Again is it directly translated into percentage?
Sorry for this basic question, i was not, however, able to find a clear answer if it is directly translated into percentages or not.
All the best
regression log
asked 1 hour ago
maSmaS
61
New contributor
maS is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Possible duplicate of Interpretation of log transformed predictor
$endgroup$
– COOLSerdash
13 mins ago
add a comment |
$begingroup$
I have the following multiple linear regression model:
Log(y) = B0 + B1X1 + B2X2 + B3x3 + e.
X1 is a dummy that can take 0 = male and 1 = female and
X2 and X3 are continuous variables.
I am not entirely sure on how to interpret the coefficients for the variables.
The coefficient for the dummy variable is 0,20. Does that mean, that changing from male to female (male is baseline) the Y will increase by an average of 20%. Is it directly translated into percentage?
And for the continuous variables, the coefficient for X2 is 0,1. Does that mean that increasing X2 with 1 unit increases Y with an average of 10%? Again is it directly translated into percentage?
Sorry for this basic question, i was not, however, able to find a clear answer if it is directly translated into percentages or not.
All the best
regression log
asked 1 hour ago
maSmaS
61
New contributor
maS is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have the following multiple linear regression model:
Log(y) = B0 + B1X1 + B2X2 + B3x3 + e.
X1 is a dummy that can take 0 = male and 1 = female and
X2 and X3 are continuous variables.
I am not entirely sure on how to interpret the coefficients for the variables.
The coefficient for the dummy variable is 0,20. Does that mean, that changing from male to female (male is baseline) the Y will increase by an average of 20%. Is it directly translated into percentage?
And for the continuous variables, the coefficient for X2 is 0,1. Does that mean that increasing X2 with 1 unit increases Y with an average of 10%? Again is it directly translated into percentage?
Sorry for this basic question, i was not, however, able to find a clear answer if it is directly translated into percentages or not.
All the best
regression log
regression log
asked 1 hour ago
maSmaS
61
New contributor
maS 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
maSmaS
61
New contributor
maS 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
maSmaS
61
New contributor
maS 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
maSmaS
61
asked 1 hour ago
maSmaS
61
61
New contributor
maS is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
maS is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
maS is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Possible duplicate of Interpretation of log transformed predictor
$endgroup$
– COOLSerdash
13 mins ago
add a comment |
$begingroup$
Possible duplicate of Interpretation of log transformed predictor
$endgroup$
– COOLSerdash
13 mins ago
$begingroup$
Possible duplicate of Interpretation of log transformed predictor
$endgroup$
– COOLSerdash
13 mins ago
$begingroup$
Possible duplicate of Interpretation of log transformed predictor
$endgroup$
– COOLSerdash
13 mins ago
add a comment |
1 Answer
1
active
oldest
votes
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Positive coefficients somehow indicate a positive effect, but they don't simply turn into percentages. There is a transformation. Let's say your model is $log y = b_0+b_1x_1$; this means $y=e^{b_0+b_1x_1}=A_0e^{b_1x_1}$. So, dummy or not, if $x_1$ increases by $1$ unit, $y$ increases by $e^{b_1}$, i.e. if $b_1=0.2$, $y$ increases by $e^{0.2}approx 1.22$, i.e. $22%$. The case is similar for your continuous variable.
answered 44 mins ago
gunesgunes
8,2611418
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$begingroup$
I have read that if the natural log is used it is approximately translated into a percentage change, if the change in x is small. Is that correct?
$endgroup$
– maS
19 mins ago
$begingroup$
Yes, but approximate is the key term here, because Taylor expansion of $e^x$ is: $e^x=1+x+x^2/2!,...approx 1+x$ when $x$ is small. Which means $x%$ increase in something. Here, to execute this idea, your coefficients should be small.
$endgroup$
– gunes
17 mins ago
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
Positive coefficients somehow indicate a positive effect, but they don't simply turn into percentages. There is a transformation. Let's say your model is $log y = b_0+b_1x_1$; this means $y=e^{b_0+b_1x_1}=A_0e^{b_1x_1}$. So, dummy or not, if $x_1$ increases by $1$ unit, $y$ increases by $e^{b_1}$, i.e. if $b_1=0.2$, $y$ increases by $e^{0.2}approx 1.22$, i.e. $22%$. The case is similar for your continuous variable.
answered 44 mins ago
gunesgunes
8,2611418
$endgroup$
$begingroup$
I have read that if the natural log is used it is approximately translated into a percentage change, if the change in x is small. Is that correct?
$endgroup$
– maS
19 mins ago
$begingroup$
Yes, but approximate is the key term here, because Taylor expansion of $e^x$ is: $e^x=1+x+x^2/2!,...approx 1+x$ when $x$ is small. Which means $x%$ increase in something. Here, to execute this idea, your coefficients should be small.
$endgroup$
– gunes
17 mins ago
add a comment |
$begingroup$
Positive coefficients somehow indicate a positive effect, but they don't simply turn into percentages. There is a transformation. Let's say your model is $log y = b_0+b_1x_1$; this means $y=e^{b_0+b_1x_1}=A_0e^{b_1x_1}$. So, dummy or not, if $x_1$ increases by $1$ unit, $y$ increases by $e^{b_1}$, i.e. if $b_1=0.2$, $y$ increases by $e^{0.2}approx 1.22$, i.e. $22%$. The case is similar for your continuous variable.
answered 44 mins ago
gunesgunes
8,2611418
$endgroup$
$begingroup$
I have read that if the natural log is used it is approximately translated into a percentage change, if the change in x is small. Is that correct?
$endgroup$
– maS
19 mins ago
$begingroup$
Yes, but approximate is the key term here, because Taylor expansion of $e^x$ is: $e^x=1+x+x^2/2!,...approx 1+x$ when $x$ is small. Which means $x%$ increase in something. Here, to execute this idea, your coefficients should be small.
$endgroup$
– gunes
17 mins ago
add a comment |
$begingroup$
Positive coefficients somehow indicate a positive effect, but they don't simply turn into percentages. There is a transformation. Let's say your model is $log y = b_0+b_1x_1$; this means $y=e^{b_0+b_1x_1}=A_0e^{b_1x_1}$. So, dummy or not, if $x_1$ increases by $1$ unit, $y$ increases by $e^{b_1}$, i.e. if $b_1=0.2$, $y$ increases by $e^{0.2}approx 1.22$, i.e. $22%$. The case is similar for your continuous variable.
answered 44 mins ago
gunesgunes
8,2611418
$endgroup$
Positive coefficients somehow indicate a positive effect, but they don't simply turn into percentages. There is a transformation. Let's say your model is $log y = b_0+b_1x_1$; this means $y=e^{b_0+b_1x_1}=A_0e^{b_1x_1}$. So, dummy or not, if $x_1$ increases by $1$ unit, $y$ increases by $e^{b_1}$, i.e. if $b_1=0.2$, $y$ increases by $e^{0.2}approx 1.22$, i.e. $22%$. The case is similar for your continuous variable.
answered 44 mins ago
gunesgunes
8,2611418
answered 44 mins ago
gunesgunes
8,2611418
answered 44 mins ago
gunesgunes
8,2611418
answered 44 mins ago
gunesgunes
8,2611418
8,2611418
$begingroup$
I have read that if the natural log is used it is approximately translated into a percentage change, if the change in x is small. Is that correct?
$endgroup$
– maS
19 mins ago
$begingroup$
Yes, but approximate is the key term here, because Taylor expansion of $e^x$ is: $e^x=1+x+x^2/2!,...approx 1+x$ when $x$ is small. Which means $x%$ increase in something. Here, to execute this idea, your coefficients should be small.
$endgroup$
– gunes
17 mins ago
add a comment |
$begingroup$
I have read that if the natural log is used it is approximately translated into a percentage change, if the change in x is small. Is that correct?
$endgroup$
– maS
19 mins ago
$begingroup$
Yes, but approximate is the key term here, because Taylor expansion of $e^x$ is: $e^x=1+x+x^2/2!,...approx 1+x$ when $x$ is small. Which means $x%$ increase in something. Here, to execute this idea, your coefficients should be small.
$endgroup$
– gunes
17 mins ago
$begingroup$
I have read that if the natural log is used it is approximately translated into a percentage change, if the change in x is small. Is that correct?
$endgroup$
– maS
19 mins ago
$begingroup$
I have read that if the natural log is used it is approximately translated into a percentage change, if the change in x is small. Is that correct?
$endgroup$
– maS
19 mins ago
$begingroup$
Yes, but approximate is the key term here, because Taylor expansion of $e^x$ is: $e^x=1+x+x^2/2!,...approx 1+x$ when $x$ is small. Which means $x%$ increase in something. Here, to execute this idea, your coefficients should be small.
$endgroup$
– gunes
17 mins ago
$begingroup$
Yes, but approximate is the key term here, because Taylor expansion of $e^x$ is: $e^x=1+x+x^2/2!,...approx 1+x$ when $x$ is small. Which means $x%$ increase in something. Here, to execute this idea, your coefficients should be small.
$endgroup$
– gunes
17 mins ago
add a comment |
maS is a new contributor. Be nice, and check out our Code of Conduct.
maS is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Possible duplicate of Interpretation of log transformed predictor
$endgroup$
– COOLSerdash
13 mins ago