logit interpretation / Marginal effects

[Ihab Man]

Dear all,
Please, I have two questions related to the logit and marginal effects interpretations.

1. ) one of the pervious literatures (the author used logit model) and the coefficient on X is = -1.08. So in this case, he must say that a 1% increase in x is associated with a decrease of 1.08 in the odds of y. right? But he said a 10% increase in X is associated with a decrease of 11% in the odds of y. Well, my question is that there is something or any idea behind his interpretation instead of the normal interpretation? I know that at the end the same amount of probability, but I would like to know from you if there is any idea behind that. Thanks
2. I would like to get the Average Marginal Effects for a Continuous variable after run a binary logit regression. Please, do you agree/advice with that or it’s better to get the Marginal effects at means? Thank you so much.

[Clyde Schechter]

Well, assuming you are reporting everything correctly in 1, both you and the author you are reporting about have it wrong.

In a logistic regression, the coefficient of X is the expected difference in log odds y associated with a change of 1 (not 1%) in X. This absolute difference in log odds also corresponds to a proportional difference in the odds itself. So with a coefficient of -1.08, a unit change in X would be associated with decrease in the log odds of y by 1.08. If the log decreases by 1.08, then the odds of y themselves are multiplied by a factor of exp(-1.08) = 0.34 (to 2 decimal places). This could also be stated as a 66% decrease in the odds of y being associated with a unit increase in X.

The percent decrease in odds of Y associated with a 1% (or any %) change in X is not calculable just from the coefficient--a starting odds must also be specified in order to do that.

However, I think I know what your author meant to say. If we look at a change in X of 0.1 (not 10%--but an easy mistake to make), then the same reasoning tells us that the odds of y get multiplied by a factor of exp(-1.08*0.1) = 0.898 (to three decimal places), which corresponds to a 10.2% decrease in the odds of y. Now, if one were to round that 0.898 incorrectly as 0.89 (another easy mistake to make), you would then conclude you have an 11% decrease in the odds of y.

As for your second question, there is no general answer. The average marginal effect and marginal effect at means are two different things, with different meanings and different uses. Which is appropriate for a particular purpose depends on the purpose.

[Ihab Man]

Dear Clyde Schechter
Thank you so much for your corrections. From you we learn. Thanks
Please, I have three following questions.
1- ) Can you please explain me how 0.898 (to three decimal places) corresponding to a 10.2%, I mean the steps.
2- ) According to your sentence (The percent decrease in odds of Y associated with a 1% (or any %) change in X is not calculable just from the coefficient--a starting odds must also be specified in order to do that). Do you mean if I said that a unit change in X would be associated with decrease in the log odds of y by 1.08 is incorrect? I mean why/ and when should I multiplied by a factor ?
3- ) please, the third one is related to the second question. When the author said a 10% increase in X is associated with a decrease of 11% in the odds of y. is he enforced to said the instead of (a unit change in X would be associated with decrease in the log odds of y by 1.08) , there are any reasons behind that or this topic consider based on preferences ? Thank you so much.

[Clyde Schechter]

1) 0.898 = 89.8% = 100% - 10.2%

2)

Do you mean if I said that a unit change in X would be associated with decrease in the log odds of y by 1.08 is incorrect?

No, I don't mean that. If you said that, that would be correct. Well, almost. A unit change in X could be in either direction, so it's a unit increase in X which is associated with a decrease in log odds of y by 1.08.

3) There are various ways of describing these findings. The one associating a unit increase in X with a 1.08 decrease in log odds of y stays closest to the original finding. It ha the disadvantage that most people will find it difficult to understand: few people have intuition about the meaning of a decrease of 1.08 in the log odds of something. As I showed in #2, the finding is equivalent to a unit increase in X being associated with a 66% decrease in the odds of y. Most people would find this version more intuitive.

The problem with the statement that "a 10% increase in X is associated with a decrease of 11% in the odds of y" is that it's wrong. It simply isn't true. It is not equivalent to the original finding in the coefficient. It's a false calculation.

[Ihab Man]

Dear Clyde Schechter

I appreciated that too much. Now everything is clear . Many Thanks.
well, please, only two things . 1-) its wrong because the author round 0.898 as 0.89 and he must make it (to three decimal places) correct? not because he multiplied by 0.1 ? well, if we assume he made it with a decreases by 10.2% as you have mentioned will be OK and equivalent to the original finding in the coefficient (a unit change in X would be associated with decrease in the log odds of y by 1.08) correct ?.
2-) please, why you did not say a 34% decrease in the odds of y being associated with a unit increase in X. Why you said 66% decrease in the odds of y etc.. if the exp(-1.08) = 0.34. because when I checked for you posts in this site i have seen this answer for you (To two decimal places, exp(-1.0954) == 0.33. So one way to interpret the results is that a unit increase in whatever variable (let's call it x) this is the coefficient of is associated with decreased odds of whatever the positive outcome in your model is (let's call it y) by a factor of 0.33 (or roughly 1/3) Many Thanks in advance.

[Clyde Schechter]

1-) its wrong because the author round 0.898 as 0.89 and he must make it (to three decimal places) correct? not because he multiplied by 0.1 ? well, if we assume he made it with a decreases by 10.2% as you have mentioned will be OK and equivalent to the original finding in the coefficient (a unit change in X would be associated with decrease in the log odds of y by 1.08) correct ?.

Well, the rounding was done incorrectly, but even before getting to that point, the author referred to this as arising with a 10% change in X. But that is not what happens with a 10% change in X. In fact, from the coefficient you cannot say at all what happens with a 10% change in X. You cannot even get to the 0.898 that way. You get to the 0.898 by talking about a 0.1 change in X. That is not the same thing as a 10% change in X. But I think this is what your author (mistakenly) meant.

Why you said 66% decrease in the odds of y etc.. if the exp(-1.08) = 0.34.

So a unit change in X results in multiplying the odds of y by 0.34. If you multiply something by 0.34, then the decrease in that thing is 100*(1-0.34) percent, which is 66%.

[Ihab Man]

Dear Clyde Schechter

You told me a very important information and help. I understand now . Thank you so much. I appreciated that .

[Ihab Man]

Dear Clyde Schechter

I am sorry for the inconvenience, but please, I have one question related to #1 second question.

Well, my logit command looks like this ( logit Y L.X1_w L.X2_w L.X3 L.X4_w L.X5_w i.Year i.Country , vce(cluster CompanyID)).

So, I want to use marginal command as well after running the previous command . All the independent varietals are Continuous (like financial ratio) . In this case, please, I have two question . First, when I used this one ( margins, dydx( L.X1_w) atmeans) the other variables will be at the mean values already ? Second , are the lagg operator , fixed effects( i.Year i.Country), and vce(cluster CompanyID) make any problems for the marginal results ? I mean when I will use this command ( margins, dydx( L.X1_w) atmeans) shoude I remove the lagg or something like that ? Thank you so much.

[Clyde Schechter]

Yes, the other variables will be set to their means when calculating the marginal effect of L.X1_w. There are no special problems created by the lag operator, fixed effects, or variance clustering. And you not only shouldn't remove the lag operator in your -margins- command, you can't -- you will get no results except for an error message if you try.

[Ihab Man]

Dear Clyde Schechter


Thank you so much , Yes I tried now and everything is clear for first and second question. Many Thanks.