Interpreting results of Linear probability models

[Tom Ford]

Hello, I am running the following regression

reg y ln_x

where y is a dummy variable taking the value of 0 or 1 and ln_x is the natural log of a continuous variable.

I am not sure about the correct way to interpret my regression coefficient.

Say for instance that ln_x has a coefficient of 0.06.

How would I interpret this? I am unsure about a number of options

1. a 1% increase in x is associated with a 6% increase in the probability of y=1
2. a 1% increase in x is associated with a 6% points increase in the probability of y=1
3. a 1% increase in x is associated with a 0.06% increase in the probability of y=1
4. a 1% increase in x is associated with a 0.06% points increase in the probability of y=1

Any guidance on this would be extremely helpful!

thanks a lot for your help

[Clyde Schechter]

Yes, it's easy to get confused by the rules of thumb. Better to work it out with algebra.

y = 0.06*ln(x) + error, with E(error) == 0.
If x increase by 1%, x* = 1.01*x.  What is y*?
y* = 0.06*ln(1.01*x) = 0.06* [ln(1.01) + ln(x)] = 0.06*.00995033 + 0.06*ln(x) = 0.00059702 + y
So, y* - y = 0.0059702 = 0.06% points rounded.

So the answer is your option 4.

The key underlying principles are:

1. An increase of p percent in x means replacing x by (1+p/100)*x.
2. The logarithm product rule: ln(a*b) = ln(a) + ln(b)
3. The approximation ln(1+epsilon) approximately equals epsilon if epsilon is small (for practical purposes, < 0.1). In the calculation above, I used an "exact" calculation of ln(1.01), but you can see that the "exact" value is very close to the result given by this approximation. In fact, when the end result is rounded off, they are the same.
4. A difference (subtraction) between percentages is denominated in .

[Tom Ford]

Wonderful this is really helpful!

[Tom Ford]

Dear Clyde,
I was going through the algebra again,
Is it possible that there is an error in the calculation
In particular it appears that
y*=0.00059702 + y but then you state y* - y = 0.0059702 = 0.06% points rounded. Which has one fewer zero than the previous line.

Is this a mistake?

apologies if the question is trivial

[Clyde Schechter]

Sorry. 0.0059702 is a typo. The one with 3 zeroes after the decimal place is correct.

[Tom Ford]

Thanks a lot.

So it appears that the rule of thumb is not working here. With a coefficient of 0.06, a 1% increase in x is associated with a 0.006% increase in the probability of y=1. If I correctly interpret this!

[Clyde Schechter]

Sorry, it's confusing because of the error I made earlier in the number of zeroes, which then propagates.

The whole calculation should be:

y = 0.06*ln(x) + error, with E(error) == 0.
If x increase by 1%, x* = 1.01*x. What is y*?
y* = 0.06*ln(1.01*x) = 0.06* [ln(1.01) + ln(x)] = 0.06*.00995033 + 0.06*ln(x) = .00059702 + y
So, y* - y = .00059702 = 0.06% points rounded.

This is, indeed, 0.06 percentage points, which is off by an order of magnitude from the rule of thumb about multiplying the coefficient by 100. That rule of thumb works, for small coefficients, when the dependent variable is log transformed and the independent variable, which increases by 1 (not 1%) is not. You have the opposite situation, and that rule does not apply.

[Tom Ford]

Dear Clyde,
thanks a lot for your comprehensive reply. And I am sorry if this is not clear to me. But also in the last iteration, I am not sure that I understand. Why is .00059702 = 0.06% points rounded? should it be 0.0006% points?

Otherwise, if a 1% increase in x is associated with a 0.06% increase in y, the results would be in line with the rule of thumb (3.) in the initial post. That you immediately suggested as correct.

I am really grateful for you taking the time of looking at this.

[Clyde Schechter]

Why is .00059702 = 0.06% points rounded? should it be 0.0006% points?

So, first, 0.0059702 rounds to 0.0006 as a number. Now let's deal with percentage points. Just as 0.15 would resacles to 15 percentage points, you move the decimal point two places to the right (i.e. multiply by 100). So 0.0006 when rescaled to percentage points would be 0.06 percentage points.

the results would be in line with the rule of thumb (3.) in the initial post. That you immediately suggested as correct.

You are incorrect, the results align with (4.) in #1 as differences in percentages are denominated in percentage points, not percents. As I was writing #7, #1 was not in view and I did not recall that you had proposed four solutions. I misremembered that you had proposed to take the coefficient, 0.06 and multiply it by 100 to get 6 percentage points. Well, in fact, you did propose that as (2), but I didn't remember that you had also proposed (1, 3, and 4).

[Tom Ford]

I am really grateful for your help Clyde, this has been extremely helpful and generous of you. I think I got it!