Previously, we employed a (natural) log transformation to make right-skewed data appear more symmetric, but this is not all we can do with logs. We use logs when we want to consider a proportional change in one or more variables instead of a level change. We call the proportional change in Y for a proportional change in X the elasticity of Y with respect to X. We call the proportional change in Y for a level change in X the semi-elasticity of Y with respect to X.
For example, we may care more about the % change in demand rather than the unit amount change. The % change in demand for a 1% change in price would be the price elasticity of demand. The % change in demand for a $1 change in price would be the price semi-elasticity of demand.
Natural Log: Approximating Proportional Change
A proportional change in X is given by \(\frac{x_1-x_0}{x_0}=\frac{\Delta x}{x}\). We can use the natural log to approximate a proportional change in X due to the function’s derivative when \(dx\) is small:
\[\begin{aligned}
\frac{d\ln x}{dx}&=\frac1x \\
\\
d\ln x &= \frac{dx}{x} \\
\\
\Rightarrow \Delta\ln x &\approx \frac{\Delta x}{x}
\end{aligned}\]
Thus, a 0.01 change in the log of X is roughly equivalent to a 0.01 proportional change in X: a 1% change in X.
Natural Log: Elasticity and Semi-Elasticity
As we said before, we can compute elasticities and semi-elasticities between two variables. With our proof that \(\Delta\ln x\approx\frac{\Delta x}{x}\) above:
\[\text{semi-elasticity: } \frac{\Delta\ln y}{\Delta x}\text{ or }\frac{\Delta y}{\Delta\ln x}\]
We can estimate these [semi-]elasticities with bivariate OLS after log transforming one or both of our variables.
Natural Log: Log-Linear, Linear-Log, and Log-Log Models
We have three types of bivariate OLS models that incorporate logs. We name the model in order of dependent then independent variable transformation:
\[\text{log-linear: } \widehat{\ln y}=b_1+b_2x\]
\[\text{linear-log: } \hat{y}=b_1+b_2\ln x\]
\[\text{log-log: } \widehat{\ln y}=b_1+b_2\ln x\]
Obviously, we cannot apply log transformations to negative numbers. Thus, any variable must not contain any negative values to be eligible for log transformation.
Natural Log: Slope Interpretation
In each case, the interpretation of \(b_2\) is different:
log-linear model: a 1-unit change in X is associated with a \((b_2\times100)\)% change in Y
log-log model: a 1% change in X is associated with a \(b_2\)% change in Y
linear-log model: a 1% change in X is associated with a \((b_2/100)\)-unit change in Y
Natural Log: Practice Problems
How would we interpret the following results:
\(b_2=4\) from a log-log regression
\(b_2=0.2\) from a log-linear regression
\(b_2=0.04\) from a log-log regression
\(b_2=3\) from a linear-log regression
\(b_2=40\) from a log-linear regression
“A 1-[something] change in X is associated with a …”
Natural Log: Stata Example
sysuse auto, clear// use system datasetgen ln_price = ln(price) // log transform price reg price mpg // linear regressionreg ln_price mpg // log-linear regression
We are first log-transforming the price variable and then running two regressions. The first regression is a linear model, while the second is a log-linear model. The first regression estimates the change in price for a 1-unit change in miles per gallon (mpg), while the second regression estimates the % change in price for a 1-unit change in mpg.
Natural Log: Exponential Growth
We may also want to use the natural log to linearize economic quantities that grow based on a power rule or exponentially, such as returns on financial assets.
If we think our returns on a principal investment \(x_0\) grow exponentially each period \(t\) based on some positive interest rate \(r\), we can model these returns for any period by: \[x_t=x_0(1+r)^t\]
We may have data on the asset value each period, but not know the principal \(x_0\) or interest rate \(r\). How could we use log transformations to estimate these?
Natural Log: Deriving the Log-Linear Time Model
We start with our asset value for any period \(x_t=x_0(1+r)^t\) and take the natural log of both sides:
\[\begin{aligned}
\ln x_t &= \ln(x_0(1+r)^t) \\
&= \ln x_0 + \ln((1+r)^t)\text{ (because }\ln(ab)=\ln a+\ln b)\\
&= \ln x_0 + \ln(1+r)\times t \text{ (because }\ln(a^b)=b\times\ln a)\\
&= \ln x_0 + r\times t \text{ (because }\ln(1+r)\approx r\text{ for small }r)
\end{aligned}\]
With our new equation \(\boldsymbol{\ln x_t = \ln x_0+r\times t}\), we can log-transform our outcome variable and estimate \(x_0\) and \(r\) using OLS:
reg ln_x t\(\rightarrow r=b_2, x_0=\exp(b_1)\).
Natural Log: Why Base e?
Why use natural log?
We could choose to use log base 10, base 5, or any other base, but we almost always use the natural log or base \(e\). This is because \(\frac{d}{dx}e^x=e^x\) (the exponential1) and \(\ln(e)=1\). Thus, when we take \[\frac{d}{dx}\log_a(x)\left(=\frac{1}{x\ln(a)}\right)\text{ with } a=e\] we get a clean \(\frac{dx}{x}\approx\frac{\Delta x}{x}\).
End of Lecture Material
Knowledge Check 9
Describe what we are doing here. What kind of regression is this and how would you write it? What do these results say? Interpret the coefficient, t-stat, p-value, and confidence interval for \(b_2\).
Source | SS df MS Number of obs = 248
-------------+---------------------------------- F(1, 246) = 291.11
Model | .715537701 1 .715537701 Prob > F = 0.0000
Residual | .604652438 246 .002457937 R-squared = 0.5420
-------------+---------------------------------- Adj R-squared = 0.5401
Total | 1.32019014 247 .005344899 Root MSE = .04958
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ln_close | Coefficient Std. err. t P>|t| [95% conf. interval]
-------------+----------------------------------------------------------------
month | -.0156221 .0009156 -17.06 0.000 -.0174255 -.0138187
_cons | 7.183856 .0067197 1069.07 0.000 7.170621 7.197092
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