Looking at resources such as this one and this one, you see claims like
"Exponentiate the coefficient, subtract one from this number, and
multiply by 100. This gives the percent increase (or decrease) in the
response for every one-unit increase in the independent variable.
Example: the coefficient is 0.198. (exp(0.198) – 1) * 100 = 21.9. For
every one-unit increase in the independent variable, our dependent
variable increases by about 22%".
This formula for converting coefficients into percent changes seems to have come out of nowhere. I cannot see why this computes a percent change.
Consider this question, and the top answer just states the following result which appears to compute the same thing in a different way:
log(DV) = Intercept + B1 * IV + Error"One unit increase in IV is associated with a (B1 * 100) percent
increase in DV."
Furthermore, this question has an answer that says
"keep in mind that the interpretation of a "unit change in a
logarithm" as a "percent change" is a local approximation."
This just confuses me more. Why do these formulas only produce an approximation?
All of this leads to the question… Why can I interpret a log transformed dependent variable in terms of percent change in linear regression? (And why is it only an approximation?)
Best Answer
Say we have a model like this:
$$\log\hat y=\beta_0+\beta_1 x$$
Since $\exp$ is the inverse function of $\log$, we can do this:
$$\hat y = f(x)=\exp(\beta_0+\beta_1 x)$$
Now, what happens when $x$ grows by 1? $f(x)$ multiplies by $\exp(\beta_1)$:
$$\begin{align} f(x+1)&=\exp[\beta_0+\beta_1(x+1)]\\ &=\exp(\beta_0+\beta_1 x)\cdot\exp(\beta_1)\\ &=f(x)\cdot\exp(\beta_1) \end{align}$$
OK, now how much does $f(x)$ grow in percentages?
$$\left(\frac{f(x+1)}{f(x)}-1\right)\cdot100=(\exp(\beta_1)-1)\cdot100$$
This explains that formula for converting coefficients into percent changes. Until here, we used no approximations. Now, if $x$ is a small enough number, we can approximate $\exp(x)\approx1+x$. This approximation is called a first order Taylor expansion of $\exp(x)$ around $x=0$. If you replace this approximation on the $\text{coefficients}\rightarrow\text{percent change}$ we found earlier, you get:
$$\text{percent change}\approx100\cdot\beta_1$$
So, when $\beta_1$ is a small number, you can interpret it directly as a percent change - but keep in mind that this is just an approximation.