Solved – What’s the difference in growth of Y in a linear regression model when using a log-lin model or a lin-log model

econometricslog-linearlogarithm

Description

I'm currently studying a chapter on linear regression analysis. I have come to a section where we study the interpretation of the coefficients with logarithmically transformed variables. I would like to know what happens to Y when an absolute change in the value of X, or a relative change in the X occurs.

Linear regression

The formula presented here is the basic linear representation of my dataset.

$ Y = \alpha + \beta X + \epsilon$

When you take the first derivative of the formula you get:

$dY = \beta * dX $

$\beta $ is the slope of the formula, thus when X's value increases with an increment of 1 $ Y $ increases with a value of $\beta $. On the other hand when X increases with 1%, Y increases with $\beta\%$.

Log-Lin model

What I don't understand is how much Y grows when we transform the original formula logarithmically.

$\log(Y) = \alpha + \beta X + \epsilon$

When I take the first derivative of this formula I become:

$\ dY/Y = \beta*dX $

How do I interpret this formula?
Does this mean that when $ dX = 1$ that $Y $ grows with $ \beta \% $ ?
What happens when to Y when X grows with 1%?

Lin-Log model

Converserly, when I apply the same reasoning to the following transformation, is my conclusion still valid?

The first derivative of:

$ Y = \alpha + \beta*log(X) + \epsilon$

is:

$ dY = \beta * (dX/X) $

So that, when there is an in increase in X of 1%, Y increases with $\beta$ What if X increases its value with 1, how much does Y increase?

This was my first question. If the format or the content of the question can be improved please let me know.

Many thanks in advance!

Best Answer

So in the top model, $Y=\alpha+\beta X+u$ a 1 unit change in X relates to a 1$\beta$ unit change in Y. So whatever units you are using, its unit change in both.

With $\ln(Y)=\alpha+\beta X+u$ then we have that a 1 unit change in X relates to a $\beta*100%$ percent changes in Y. That is because the LHS in the derivative is the growth rate in Y.

With $Y=\alpha+\beta \ln X+u$ , we now have the opposite, so a 1 percent change in X relates to a $\beta/100$ unit change in Y.

The reason why we multiply by 100 in the log-lin case is because as X changes by 1% the $\beta$ needs to be converted from percent to units. The opposite in the lin-log case.

Edit: I should add that in the log-lin model, that interpretation is only an approximation that works for small $\beta$. The exact percentage difference is:

$100*[exp(\beta*\Delta X)-1]$ This is easily seen when you do a percentage change between the model for 2 different values of X.

Related Solutions

Solved – Are log difference time series models better than growth rates

One major advantage of log-differences is symmetry: if you have a log difference of $0.1$ today and one of $-0.1$ tomorrow, you are back from where you started. In contrast, 10% growth today and 10% decline tomorrow will not bring you back to the initial value.

Solved – Difference in Differences estimation in a log linear model

The treatment effect in your case is in percent and not in percentage points. To see this, consider the following simple example with two groups and two time periods for before and after the intervention. Let the group means be

  y   time    group         lny
110      0        0   4.7004805
111      1        0   4.7095304
110      0        1   4.7004805
115      1        1   4.7449322

where $y$ is the outcome, time indexes the periods (0 before, 1 after treatment), and group indexes the control (0) and treatment (1) group, and $\ln y$ is the log transformed outcome.

Suppose that the outcome trends for the two groups was the same before the treatment, then the difference in differences estimate would be $$\beta_{y,DiD} = [115 - 110] - [111 - 110] = 4 $$

so the treatment group increased their outcome by 4 units more than the control (which is the same as saying that the treatment group increased their outcome by 4 more units compared to the counterfactual when the treatment had not happened).

When you compare the growth rates in each group, you get a percent increase of $\frac{111-110}{110}\cdot 100 = 0.91$% in the control, and $\frac{115-110}{110}\cdot 100 = 4.54$% in the treatment group. This is approximately the same as taking the differences in the logs between period 1 and 0 in each group (this approximation works well because the differences are not too big). So the difference-in-differences in the growth rates are 4.54% - 0.91% = 3.6%. The DiD estimate using the log transformed outcome is $$\beta_{\ln y,DiD} = [4.7449322 - 4.7004805] - [4.7095304 - 4.7004805] = 0.0354 $$

and multiplying this by hundred gives you approximately again the difference that we had before. The exact result is given by (exp(.0354018)-1)*100 = 3.6

In the absence of the treatment, the treated group would have had 111 in the second period instead of 115. The percentage change from 111 to 115 is $\frac{115 - 111}{111} \cdot 100 = 3.6$

So the papers that you were reading are right. If you want to estimate a DiD coefficient that estimates a percentage point increase, this works in case of a $y \in [0,100]$, i.e. an outcome on a percent scale. Regressing such an outcome on the time and group indicator, and their interaction will yield a DiD coefficient that shows a percentage point increase for the treatment group. Your other point is correct when you say that regressing $$y = \alpha + \beta T + \epsilon$$ in each group separately gives the time change for the outcome in each group. When you then compute $\beta_1 - \beta_0$ this will again give you the DiD estimate.