Interpreting regression coefficients in log models part 1

[haroldbaker]

Absolutely phenomenal. I feel like everything is slowly falling into place now. Cheers

[nahshahehsha]

At 3:20 when you are talking about interpretation… by percentage terms do you mean a percentage point or a percentage? Also, are we to assume the data to be levels or percentage data?

[coolkatiecat]

very clear and digestible explanation.

[Ytremz]

When you awaken old mathematical parts of your brain.

[ahmadghaemi]

Is the differential d(ln y) = dy/y? How do we get to that step? If d(ln y)/dx = B1/x1, then d(ln y) = B1* dx1/x1, and then what?

[ichbinirgendwer]

That´s an excellent and simple explanation!

[usernameisnowtaken]

We use a time series model at work where each variable is in a change-in-log form. I've heard disagreements regarding the interpretation of the constant in such a log-log model. Example: if an estimated coefficient for the constant in a multivariate log-log model were .0004 - is this a ceteris paribus .0004% "growth rate" or .04% "growth rate"? I think people tend to try and apply the same interpretation of the constant that is used for the independent variables in these models. Any help here?

[lastfm]

Hi Ben, thanks for your great videos. I have a question: In case of the lin-log model you state that the beta shows the increase in y that would be associated with a one percent increase in x. However, could it be that a 1% increase in x is rather associated with a beta of beta*log(1, 01) or appr. beta/100 to get the value for y for a 1% increase in x? Thanks!

[johnyf.q.]

So we have to exponentiate \beta_1 in order to the the %change in y for a 1% change in x_1?

[CaZeek]

Hey Ben- thank you again for the vids- huge help in my self study. I don't know if I'm just making a stupid mistake or what, but I'm finding something very strange that I can't explain. I fully understand the differentiating to get to the:
d/dx * dy/dy * ln(y) = d/dx * B * ln(x)
1/y * dy = B * 1/x * dx
B = (dy/y) / (dx/x)
= % chg(y) / % chg(x)

That all makes perfect sense, but for some reason, when I'm plugging in some test numbers to make sure everything's working as planned, there is a slight error that I can't get my head around. For example, set B=2:

ln(y) = 2*ln(x)

x = 200 results in y = 40, 000
x = 210 (5% increase) SHOULD result in 10% increase in y (44, 000)
However x = 210 results in y = 44, 100

I've plugged in many numbers to make sure I'm not just miscalculating. What am I missing here? Based on the differentiating, it doesn't seem we're approximating anything. B = (dy/y) / (dx/x).. if B = 2 and dx/x = 10/200, then dy/y = 2 * .05 = .1... why is this not exactly the result?

Thanks again!

[asddaasdda]

Hi Ben. You say log, but you write ln. Which one is the right way to do it, or is it log for the log log model, and ln for the semi-elasticity model?

[MrRayhan]

is it possible to only transform one variable independent into Ln? what is it called?

[edmundoinaciojr.]

But I need multiply per 100 the coeficient B1?

[martinmaldonado]

if a independent coefficient on a log-log models is .25, does that represent a 25% increase or a .25% increase?

[yogaesawibowofransiskus]

Are there any book or paper or a chapter of a book that comprehensively discuss about log model regression?

[RodrigoLopesBrazil]

Please, consider that part of your audience has a math background. (I understand that "percentage increase" is a proxy for "infinitesimal increment" but it sounds so weird). Good video overall. =)