23
$\begingroup$

Often I see authors estimate a "log difference" model, e.g.

$\log (y_t)-\log(y_{t-1}) = \log(y_t/y_{t-1}) = \alpha + \beta x_t$

I agree this is appropriate to relate $x_t$ to a percentage change in $y_t$ while $\log (y_t)$ is $I(1)$.

But the log difference is an approximation, and it seems one could just as well estimate a model without the log transformation, e.g.

$y_t/y_{t-1} -1 = (y_t - y_{t-1}) / y_{t-1}=\alpha+\beta x_t$

Moreover the growth rate would precisely describe the percent change, while the log difference would only approximate the percent change.

However, I've found the log difference approach is used much more often. In fact, using the growth rate $y_t/y_{t-1}$ seems just as appropriate to address stationarity as taking the first difference. In fact, I have found that forecasting becomes biased (sometimes called the retransformation problem in the literature) when transforming the log variable back to the level data.

What are the benefits to using the log difference compared to the growth rate? Are there any inherent problems with the growth rate transformation? I'm guessing I am missing something, otherwise it would seem obvious to use that approach more often.

$\endgroup$
2
  • 1
    $\begingroup$ Thank you for your comments. I agree the symmetry and bounding is a significant advantage. It seems the bounding would help control heteroskedasticity and the symmetry would help hold the mean constant. $\endgroup$
    – A. Smith
    Commented Jul 2, 2016 at 15:46
  • 2
    $\begingroup$ The log-difference is not an approximation. It is a continuously compounded or exponential growth rate, as opposed to a period-over-period rate. They are different things. Laypersons understand the second one better, but the first one has cleaner mathematical properties (e.g. average growth is just the mean of the growth rates, growth rate of product is the sum of the rates, etc). The bit about forecasting is either unnecessary transformation leading to explosive forecasts, or median-unbiased but not mean-unbiased, which is fine. It has nothing to do with continuous vs. period rates. $\endgroup$
    – Chris Haug
    Commented Jul 22, 2017 at 15:58

2 Answers 2

22
$\begingroup$

One major advantage of log-differences is symmetry: if you have a log difference of $0.1$ today and one of $-0.1$ tomorrow, you are back from where you started. In contrast, 10% growth today and 10% decline tomorrow will not bring you back to the initial value.

$\endgroup$
4
  • 10
    $\begingroup$ The symmetry / bounding is the main advantage I see. Going from 100 to 10 is a log10 difference of -1, but -90%. Going from 100 to 1000 is also a log difference of 1, but 900%. A linear model is going to pay inordinate attention to that 900% observation. $\endgroup$
    – zbicyclist
    Commented Jul 2, 2016 at 14:40
  • $\begingroup$ 7 years later but, a paper I'm reading points out that log-transforming the dependent variable is not innocuous. Having $ln(y)=x$ instead of $y=x$ implies an exponential relationship $y=exp(x)$. With, for example, $ln(income)=\beta_0 + \beta_1 education + u$ the transformation is reasonable because income increases nonlinearly and relative to the current level of education (e.g., 3 years of PhD are more valuable than 3 years of middle school). The paper argues we should only log-transform if an exponential relationship makes sense, and not out of convenience. $\endgroup$
    – suckrates
    Commented Jan 10 at 19:06
  • $\begingroup$ I'd be interested to hear your thoughts on that @Christoph Hanck. It's "Guideline 1" in this paper: journals.sagepub.com/doi/full/10.1177/1094428121991907. $\endgroup$
    – suckrates
    Commented Jan 10 at 19:07
  • $\begingroup$ I agree with the statement. I also think it is perfectly compatible with this post, which "only" addresses two alternative ways of computing such percentage changes. $\endgroup$ Commented Jan 11 at 6:35
7
$\begingroup$

Many macroeconomic indicators are tied to population growth, which is exponential, and thus have an exponential trend themselves. So the process before modelling with ARIMA, VAR or other linear methods is usually:

  • Take logs to get a series with a linear trend
  • Then difference to get a stationary series
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.