1
$\begingroup$

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 $y(x)=\exp(a)x^b$?

Thanks

James

$\endgroup$
1
$\begingroup$

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

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

because

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

Your last equation is missing a random factor.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

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

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