Log Approximation vs Non-log Approximation for Growth Rate

[Jared Greathouse]

Consider

* Example generated by -dataex-. For more info, type help dataex
clear
input byte id float(realimp date)
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95 21395792 684
end
format %tm date
label values id id
label def id 95 "Luxury Watches", modify


cls

tsset date, m

g logimp = ln(realimp)

g grow2 = D.logimp

g grow = D.realimp / L.realimp

br

Here, we have real-imports of Luxury Watches, put together from Python which I've discussed here before. Precisely, we have interpolated versions of real import data, which the original authors don't mention. Anyways, I wanted to make a growth rate outcome variable. So, I followed the formulae from Carlo Lazzaro and Sebastian Kripfganz and Clyde Schechter who respectively give two formulae to calculate this here and here.

i get different results when I use these though. Not incremental differences of a few thousandths, but by the hundredths, which in the aggregate can be pretty meaningful. I guess my question is, why do these two give different results? One is a log approximation, whereas the other one is more precise (presumably)? Why is this, and are there any advantages from one formula to the next?

[Andrew Musau]

\(ln(x+ \Delta x) - ln(x)\) is approximately equal to \(\frac{\Delta x}{x}\) for small \(\frac{\Delta x}{x}\). Consider 3 cases:

1. \(\Delta x= 10\), \(x= 100\)
2. \(\Delta x= 1\), \(x= 100\)
3. \(\Delta x= 0.1\), \(x= 100\)

then

1. \(ln(x+ \Delta x) - ln(x) = ln(110) - ln(100)\)

Res.:

. di 10/100
.1

. di ln(110)-ln(100)
.09531018

2. \(ln(x+ \Delta x) - ln(x) = ln(101) - ln(100)\)

Res.:

. di 1/100
.01

. di ln(101)-ln(100)
.00995033

3. \(ln(x+ \Delta x) - ln(x) = ln(100.1) - ln(100)\)

Res.:

. di .1/100
.001

. di ln(100.1)-ln(100)
.0009995

So you see the approximation gets better the smaller the relative change is. Lagging and taking differences in time-series data is easy, so that's one advantage for you.