2
$\begingroup$

In Wooldridge's "Introductory Econometrics: A Modern Approach", 5th edition, p. 212-215, the author describes the procedure for obtaining predictions from an OLS regression when the regressand is in log form.

In short, due to the non-linearity of the log/exp transformations, we can't simply use $exp(\hat{log(y_0)})$, where $\hat{log(y_0)}=\hat{\beta}*x_0$, as a predictor for $y$ (note that the "hat" in $\hat{log(y_0)}$ applies to the log(y), not only to the y). Rather, one has to take into account that $E(y|x)=exp(\sigma^2/2)*exp(\hat{\beta}*x_0)$; Wooldridge then presents several possible estimators for $exp(\sigma^2/2)$ (no need to talk about them for this particular question).

So far, so good. However, I do not understand how the prediction interval is derived. Wooldridge doesn't go into detail about the theory behind it and just mentions in example 6.8 that one can simply apply the exponential function to the lower and upper bounds of the prediction interval for $log(y_0)$ to get the lower and upper bounds of the prediction interval for $y_0$ "because the exponential function is strictly increasing". Why is this the case?

References welcome.

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Hint: this is simpler than you think. If you determine there is (say) a 95% chance that $\log(y)$ will lie between $a$ and $b$, what does that tell you about where $y$ is likely to lie? $\endgroup$
    – whuber ♦
    Jan 2, 2018 at 22:44
  • $\begingroup$ Thanks @whuber, that makes intuitive sense. However, one thing that I still don't understand is the relationship between the expected value and the prediction interval - the former must be adjusted by the factor $exp(\sigma^2/2)$, whilst the latter is not. Hence it would actually be possible, in a case where (i) $exp(\sigma^2/2)$ is substantially larger than unity and where (ii) the critical values of the appropriate t-distribution are sufficiently small, that the expected value could lie outside the prediction interval! Can that be and what would it mean? Or am I missing something here? $\endgroup$
    – Candamir
    Jan 3, 2018 at 16:44
  • $\begingroup$ (Please let me know if you think the above would make more sense as a separate question.) $\endgroup$
    – Candamir
    Jan 3, 2018 at 16:46
  • 1
    $\begingroup$ That's a good observation, but it's not paradoxical that the prediction interval would fail to include the estimated mean, because for lognormal distributions with large geometric means ($\sigma$), the estimated mean can be way out into the right tail: that is, most future observations will be less than an estimate of the mean based on a sample. $\endgroup$
    – whuber ♦
    Jan 3, 2018 at 20:19
  • $\begingroup$ @whuber Thanks, that makes sense. I think that another way to describe it is that the prediction interval is more „related“ to the median than to the mean and that in many non-normal distributions, including the log-normal distribution, these are not identical. $\endgroup$
    – Candamir
    Jan 3, 2018 at 22:53

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.