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  • 2SLS interpretation when instrument is log variable

    Hi there,

    I hope you can help me with the interpretation of the coefficients in a 2SLS model.

    DV: ln GDP per capita lgdp
    EV: linear index of protection against expropriation risk (0 to 10) avexpr
    IV: ln mortality (not a rate, continuous variable) lsetmort

    My professor explained to me that you can interpret the coefficients of a IV 2SLS model the same was as you do with an OLS model.
    So when you have ln(y) = a + b*x -> 1 additional unit of x increases y by (100 * (exp(b) - 1)%, ceteris paribus.
    I was wondering if this is still true in a 2SLS model when the instrumental variable is a log variable as well. I was wondering this because the effect would be extremely big in my case (the output is correct, we're replicating a previous study).

    Is it correct to say: 1 additional point on the index increases the GDP per capita by 157.1%, ceteris paribus.


    Thank you in advance (:


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  • #2
    Hi Sandra
    so, in my opinion
    1. it doesnt mother how your IV is measured, the interpretation of your endogenous variables is still the same.
    2. While the "technical" measure of the effect is large, i wonder if you have to rely more on the practical effect.
    In other words, what is the range of -avexpr-. Is a 1 unit change on this variable something that makes sense, if this variable ranges, say, between 5 to 6 ?
    HTH
    Fernando

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    • #3
      FernandoRios

      Thank you for the input! (:

      Yes, that's what I'm trying to figure out as well. The index ranges from 0 to 10, but in the sample this ranges from 3.5 to 10. It's not a very skewed distribution. However, it's not explained to us what this index really means. We don't know whether an increase of 1 point means a world of a difference. I will do some more research as to what this measure exactly means. Also, most countries in the sample are LMIC, and the GDP is measured in 1995 USD PPP. Anyways, thank you for the help and I'm sure I will figure the index variable out (:

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      • #4
        Sandra: What is the OLS estimate? Can you show summary statistics?

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        • #5
          Jeff Wooldridge

          Yes, of course (:

          These are the summary statistics

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          This is the estimated OLS regression

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          • #6
            Also (in case it's interesting), these are the detailed summary statistics

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            • #7
              I also think that the nature of the instrumental variable is irrelevant.

              From the summary statistics I see that one unit change in the endogenous regressor is not unreasonable to consider: the standard deviation of the endogenous regressor is 1.5. One can also consider what happens when the endogenous regressor changes by 1.5, i.e., by one standard deviation.

              I also think that this interpretation you are using is for dummy variables.

              I would have interpreted the coefficient from d(ln y)/dx = (dy/y)/dx, hence one unit increase in the endogenous regressor leads to 94.4% increase in GDP.

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              • #8
                Joro Kolev

                Okay, that makes a lot of sense (:

                About the interpretation of the coefficient: In previous courses I indeed learned the interpretation you gave. For some reason with this course they say to interpret the coefficient like I described in #1... I guess I will keep using my interpretation for this course, but after that continue using your interpretation.

                Thank you so much for the help!

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                • #9
                  Sandra's calculation makes sense even if x is not binary. A one unit increase in x is changing log(GDP) by a lot. Plus, it is about 2/3 of an SD, as noted above. For such changes the calculus approximation is not very good.

                  If we compare the mean at x + 1 and x using the exponential mean for GPD then the proportionate change (holding other variables fixed, and they cancel out) is

                  (exp(b*(x + 1)) - exp(b*x))/exp(b*x) = exp(b*(x + 1))/exp(b*x) - 1 = exp(b*(x + 1) - b*x)) - 1 = exp(b) - 1,

                  and then multiply by 100.

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                  • #10
                    Jeff Wooldridge

                    Okay, thank you so much for the explanation. I looked at the video lecture again and she said that the 100% calculation is an approximation indeed, and that the formula you wrote out is the correct way. It's always nice to know the reason behind these things, so thank you so much (:

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