I'm getting in a real mess at the moment over something I think is very simple, as well as the wording/terminology.

I have a model - $\ln(Y(x))=a+b\ln(x)+\epsilon, \quad\epsilon\sim\mathrm{N}(0,\sigma^2)$.

Am I right in saying that this is equivalent to $Y(x)=\exp(a)x^b+\tilde{\epsilon}$ where $\tilde{\epsilon}$ is log-normally distributed?

Also from this, is it ok to say that an equation for $y(x)$ is $y(x)=\exp(a)x^b$?




In your notation, $\ln(Y)$ has a normal distribution -- $\ln(Y) \sim N(\mu,\sigma^2)$ -- where

$$E[\ln(Y)]= \mu = a + b\ln(x).$$

Then $Y=\exp{[\ln(Y)]}$ has a lognormal distribution.

We can represent $Y$ as

$$Y= \exp{[a+b\ln(x)]}\exp{(\epsilon)}=\exp(a)x^b\exp{(\epsilon)}$$

where $\epsilon \sim N(0,\sigma^2).$

Note that

$$E(Y) \neq \exp(a)x^b,$$


$$E[\exp(\epsilon)]= \exp(\frac1{2}\sigma^2).$$

Your last equation is missing a random factor.

  • $\begingroup$ That's great, thanks a lot. $\endgroup$
    – user135174
    Jun 6 '14 at 1:06
  • $\begingroup$ You're welcome. You would also run into trouble trying to find a representation of $Y$ as a sum of $\exp(a)x^b$ and a log-normal RV. As long as you express it as above it will work out. $\endgroup$
    – RRL
    Jun 6 '14 at 3:29

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