When running log-log model in differences to correct for unit root:
D(logY) = c + b1 D(logX1)
what would be the correct interpretation of b1? :
- a 1% increase in X1 is associated with a b1 percent increase in Y,
OR
- a 1% increase in X1 is associated with a b1 percent increase in the rate of change of Y? Thanks.
D(logY) = c + b1 D(logX1), How to interpret b1?
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Re: D(logY) = c + b1 D(logX1), How to interpret b1?
I think:
a 1% increase in X1 is associated with a b1 percent increase in Y,
because that difference one is not relating to interprating differently, you got my point?
you take difference to correct your model and fit it into better model.
Hope that helps...
a 1% increase in X1 is associated with a b1 percent increase in Y,
because that difference one is not relating to interprating differently, you got my point?
you take difference to correct your model and fit it into better model.
Hope that helps...
-
frankmachianno
- Posts: 2
- Joined: Thu Dec 08, 2011 2:48 am
Re: D(logY) = c + b1 D(logX1), How to interpret b1?
I am wondering about the same issue as johnyboy... and as nannigia didnt sound very convinced of his answer, I wonder whether anyone else here can give some explanation?
This explanation is the same as the elasticity. However, the elasticity is also given when regressing to logs on each other (not log-differences). So are these two ways to estimate the elasticity or do we have to use the following explanation...what would be the correct interpretation of b1? :
- a 1% increase in X1 is associated with a b1 percent increase in Y,
I appreciate every reply...- a 1% increase in X1 is associated with a b1 percent increase in the rate of change of Y? Thanks.
Re: D(logY) = c + b1 D(logX1), How to interpret b1?
Guys,
D(logY) is not equal to the "rate of change of logY"! The rate of change (=growth rate) of logY is equal to: log(logYt)-log(logYt-1). In EViews language: log(logY)-log(logY(-1)). If you're not convinced, just graph in the same graph the growth rate of Y and its first difference. You will see that you do not have the same results.
D(logY) is the first difference of logY. If you want to interpret the elasticity coefficient b1, you have to do it this way:
A 1% change in D(logX1) leads to a b1% change in D(logY). What does it mean? It means that you have only a short run relationship between the two variables (due to the differenciation). It's the shortcome of differenciating... You got only short run relationships... But if your series are cointegrated, you could apply an ECM (Error Correction Model) in order to measure both the short run relationship and the long run relationship.
I do not know what is X1, but I guess Y is the GDP. So, you have a short run relationship between X1 and Y (if the series are not cointegrated).
Hope I was convincing.
D(logY) is not equal to the "rate of change of logY"! The rate of change (=growth rate) of logY is equal to: log(logYt)-log(logYt-1). In EViews language: log(logY)-log(logY(-1)). If you're not convinced, just graph in the same graph the growth rate of Y and its first difference. You will see that you do not have the same results.
D(logY) is the first difference of logY. If you want to interpret the elasticity coefficient b1, you have to do it this way:
A 1% change in D(logX1) leads to a b1% change in D(logY). What does it mean? It means that you have only a short run relationship between the two variables (due to the differenciation). It's the shortcome of differenciating... You got only short run relationships... But if your series are cointegrated, you could apply an ECM (Error Correction Model) in order to measure both the short run relationship and the long run relationship.
I do not know what is X1, but I guess Y is the GDP. So, you have a short run relationship between X1 and Y (if the series are not cointegrated).
Hope I was convincing.
-
frankmachianno
- Posts: 2
- Joined: Thu Dec 08, 2011 2:48 am
Re: D(logY) = c + b1 D(logX1), How to interpret b1?
Yep, that was convincing. Thank you Dim.
Further, I read that in such a log-log model in differences (series not cointegrated, thus no ECM), repeated substitution might be a way to isolate a long-run effect. However, I cannot see how to apply repeated substitution with such a dlog as the dependent variable. Could you also give me a hint on this?
Cheers
Further, I read that in such a log-log model in differences (series not cointegrated, thus no ECM), repeated substitution might be a way to isolate a long-run effect. However, I cannot see how to apply repeated substitution with such a dlog as the dependent variable. Could you also give me a hint on this?
Cheers
Re: D(logY) = c + b1 D(logX1), How to interpret b1?
By isolate, the author means remove the long-run relationship. There is no way to detect or measure a long-run relationship among time series if they are not cointegrated. The maximum you can do is measure the short-run relationship between them by interpreting elasticities of differenciated variables. I know you differenciated some series because they were not stationnary at level (=I(1)). But if you want to interprete elasticities with series which are not stationnary (then using them without differenciating), than you will get spurious regressions...
I'm not an expert in econometrics, but what I know is written here... I have a master degree in Economics and I'm writting my final dissertation on Econometrics...
I also have a question: see the topic "Johansen approach for testing cointegration" posted yesterday... If someone could help me, I will appreciate.
Thanks
I'm not an expert in econometrics, but what I know is written here... I have a master degree in Economics and I'm writting my final dissertation on Econometrics...
I also have a question: see the topic "Johansen approach for testing cointegration" posted yesterday... If someone could help me, I will appreciate.
Thanks
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