Is it necessary to exponentiate the predicted values in a log-log regression model?

For example my model is:

$\log(y) = \log(x)$

$\log(y) = -0.5141 + 0.5377 \log(x)$

if I wanted to make a prediction from this model do I need to exponentiate the result or log the x value? What is the correct way of evaluating the equation?

if $x = 100$ then y=

  1. $y = -0.5141 + 0.5377 \times 100 = 53.2559$
  2. $y = -0.5141 + 0.5377 \times \log(100) = 1.9621$
  3. $y = -0.5141 + 0.5377 \times \log(100) = 1.9621 = \exp(1.9621) = 7.114251$
  • 1
    $\begingroup$ Your model, as given ("$y=-0.5141+0.5377x$"), does not involve logarithms at all! What is really going on here? $\endgroup$ – whuber Mar 6 '15 at 21:30
  • $\begingroup$ Those are the coefficients I suppose I'm not representing the equation properly? $\endgroup$ – moku Mar 6 '15 at 21:31

Once you have an estimated log-log model, you have an equation

$$\widehat{\log(y)}=\widehat{\beta_0}+\widehat{\beta_1} \log(x).$$

Skip the hats for now to get

$$\log(y)=\beta_0+\beta_1 \log(x).$$

You want to obtain $y$. Obviously, you just have to undo the $\log$. Exponentiate both sides of the equality to get

$$\exp(\log(y))=\exp(\beta_0+\beta_1 \log(x)),$$

which is nothing more than

$$y=\exp(\beta_0+\beta_1 \log(x)).$$

You could stop here and take this as an OK solution. However, there are some subtleties that you may want to consider.

The added difficulty with hats is the following. If you ran OLS estimation, $\widehat{\log(y)}$ is the expected mean of $\log(y)$ given $x$, i.e.


Once you exponentiate, it does not hold that


That is, it des not hold that


For example, if $y \sim N(\mu,\sigma^2)$ then $\mathbb{E}(\exp(y))=\exp(\mu+\frac{1}{2} \sigma^2)$. However, this need not be a big problem in practice. Bardsen and Lutkepohl "Forecasting levels of log variables in vector autoregressions" (2011) show some examples when simple exponentiation is desirable. Dave Giles has some good discussion in his blog post "More on Prediction From Log-Linear Regressions" for alternative solutions.

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