# Interpretation of the regression coefficient of a proportion II

In my model I have a log dependet variable. As indenpendet variables I have one proportion $X_1$ that lies within [0.001, 0.30] and a second proportion $X_2$ that lies within [0.12, 0.99].

$E(\log(y))$ = $\alpha + \beta_1X_1$ + $\beta_2X_2$

How do I interpret the coefficients? Let´s say, the estimated $\beta_1$ is 1.5 , the estimated $\beta_1$ is 0.8.

Or do I have to adjust the range before running the regression?

• The question is rather unclear. In case $\hat\beta_1 = 1.5$: If $X_1$ increases by one percentage point (by 0.01), average $\log(y)$ increases by 1.5/100, i.e. the geometric mean of $y$ increases by about 1.5%. – Michael M Nov 6 '13 at 13:29
• The maximum by which $X_1$ can increase is 0.30-0.001. Following your explanation, if I want to compare the max. value with the lowest value, does this mean that the average $log(y)$ increases by 1.5/100 * 0.3-0.001, i.e. around 45 \% ? – Jules Nov 6 '13 at 14:51
• Not exactly. A jump from 0.001 to 0.3 would be associated by an average increase of $\log(y)$ by $\hat \beta_1 \cdot (0.3-0.001) =0.4485$. There is no need to talk of relative changes on the log-scale. – Michael M Nov 6 '13 at 15:05

The maximum by which X1 can increase is 0.30-0.001. Following your explanation, if I want to compare the max. value with the lowest value, does this mean that the average log(y) increases by 1.5/100 * 0.3-0.001, i.e. around 45%?

No, because the simple way of multiplying the coefficient by 100 and then interpreting as a factorial change only works when both the coefficient and the change in the independent variable are relatively small.

Consider the equation:

$\ln(y) = \alpha + 1.5X_1 + 0.8 X_2$

With one percent point increase:

$\ln(y') = \alpha + 1.5(X_1+0.01) + 0.8 X_2$

Subtracting for the difference:

$\ln(y') - \ln(y) = 1.5 \times 0.01$

$\ln(y'/y) = 0.015$

The factorial change is then:

$y'/y = 1.015113$,

which is about 1.5%. So, that approximation shortcut works.

When the change is large, like your 0.299, the approximation can deviate:

$\ln(y'/y) = 1.5 \times 0.299$

The factorial change is then:

$y'/y = 1.565961$,