It doesn't make sense to log transform my x-variable (for a more intuitive elasticity interpretation), since it is already in a % format, but with a log transformed outcome:

ln(y) = B0 + B1X1 where X1 in its raw form is a %, say B1 is 0.2

Is the correct interpretation for a 1% increase in X1, Y increases by 0.2 percentage points or by 20 percentage points?

  • $\begingroup$ (1) How many percentage points is 1%? (2) what is 0.2 times that? $\endgroup$ – whuber Jul 8 '16 at 17:58

Taking the derivative of $\ln(y) = b_0 + b_1 \cdot x$ with respect to $x$, you get

$$\frac{1}{y} \cdot \frac{dy}{dx}=b_1.$$ You can think of $$\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x},$$ the change in $y$ for a small change in $x$. This means that

$$b_1=\frac{\Delta y}{y} \cdot \frac{1}{\Delta x}=\frac{\frac{\Delta y}{y}}{\Delta x}.$$

Typically we would multiply $\frac{\Delta y}{y}$ by 100 to get a percentage change, which we can also do on the left hand side as well.

This means that $100 \cdot b_1$ is the approximate percentage change in $y$ for a small change in $x$. If you set $\Delta x=1$, that becomes the percentage change for a one unit change in $x$.

As long as your $x$ is not measured on a $[0,1]$ scale, the second interpretation is almost correct. Percentage points are used for arithmetic differences of two percentages. You have a mere percent change.

If your $x$ is measured on a $[0,1]$ scale, then a one unit change is a $100\%$ increase, not $1\%$, so then the effect is $0.2\%$

  • 1
    $\begingroup$ Just to be explicitly clear for the questioner, an increase in $x$ from $0.02$ to $0.03$ would correspond to roughly a $0.2\%$ increase in $y$. $\endgroup$ – Matthew Gunn Jul 8 '16 at 18:48

Another derivation of the same thing as Dimitriy . Assuming you have a regression equation:

$$ \ln y_i = b_0 + b_1 x_i + \epsilon_i $$

Take the expectation conditional on $X$ and then the difference between two observations:

$$\mathbf{E}\left[ \ln(y_2) - \ln(y_1) \mid X \right]= b_1(x_2 - x_1)$$

The difference in logarithms is approximately the percent change for small differences. Hence:

$$\mathbf{E}\left[ \left. \frac{y_2 - y_1}{y_1} \, \right| \, X \right] \approx b_1(x_2 - x_1)$$ Why is difference in natural logarithm approximately the percent change? Observe that the first order taylor expansion of the natural logarithm around 1 is: $$ \log(x) \approx x - 1$$ Hence if $\frac{y_2}{y_1}$ near 1 then $\log\left(\frac{y_2}{y_1} \right) \approx \frac{y_2}{y_1} - 1 = \frac{y_2 - y_1}{y_1}$ This argument that difference in logs is approximately percent change comes from linearizing the logarithm, and so it won't be accurate for larger differences where the curvature of the logarithm really matters.


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