# Logs as percentage changes in a log-level-model

In the Econometrics class I taught last winter term, I explained why the coefficient of a regressor in levels can be interpreted as approximately the percentage increase of the dependent variable in logs. I have searched the web and textbooks for a concise, straightforward derivation but found none, so I made my own. I guess it’s worth sharing. If this helps you, let me know in the comments.

With Taylor’s theorem (see also Taylor series) we can approximate the natural logarithm around some $$a>0$$ by:
[ ln(t) approx ln(a) + ln'(a) cdot (t-a) ]
Recall that the first derivative of the natural logarithm is $$ln'(t)=frac{1}{t}$$ with $$t>0$$.
Then the above approximation becomes:
[ln(t) approx ln(a) + frac{1}{a} cdot (t-a)]
Note that this approximation becomes the worse, the larger the difference of $$t-a$$, i.e. the farther one is away from the expansion point $$a$$.

We are interested in an approximation for the log of a percent increase $$ln(1+p)$$, so let $$t=1+p$$:
[ln(1+p) approx ln(a) + frac{1}{a} cdot (1+p-a)]
Further, let $$a=1$$:
[ln(1+p) approx ln(1) + frac{1}{1} cdot (1+p-1)\ln(1+p) approx 0 + 1 cdot p\ln(1+p)approx p]

The percentage change $$p$$ is given by the new  value minus the initial value all divided by the initial value.
[p=frac{(z+Delta z) – z}{z} = frac{z+Delta z}{z} – frac{z}{z} = frac{z+Delta z}{z} – 1 = frac{Delta z}{z}]

Now consider the log-level model:
[ln(hat{y}) = hat{beta}_0 + hat{beta}_1 x]

Increasing $$x$$ by one:
[ln(hat{y}_text{new}) = hat{beta}_0 + hat{beta}_1 (x+1)]

The difference is then:
[ ln(hat{y}_text{new}) – ln(hat{y}) = hat{beta}_0 + hat{beta}_1 (x+1) – hat{beta}_0 – hat{beta}_1 x \ ln(frac{hat{y}_text{new}}{hat{y}}) = hat{beta}_1 \ ln(frac{ hat{y} + Delta hat{y}}{hat{y}}) = hat{beta}_1 \ ln(1 + frac{Deltahat{y}}{hat{y}}) = hat{beta}_1 \ ln(1+p) = hat{beta}_1\ p approx hat{beta}_1 ]

From this, we can also see that the exact percent change is:
[ ln(1+p) = hat{beta}_1 \ p = e^{hat{beta}_1} – 1 ]

A quick comparison of the approximation and the exact value shows that the approximation is less than 5% off of the exact value if $$|p|<0.1$$, that is, if the change is less than $$pm 10%$$.