Logarithmic regression with individual errors If I have a dataset of $N$ pairs ($x_i$,$y_i$) where each $y_i$ has an individual error $\sigma_{y_i}$, I can incorporate this into regression by using the inverse of this as weights.
If I now do logarithmic regression $\ln y_i = b_0 + b_1\,\ln x_i + \epsilon_i$, how do I incorporate now the errors? Are the weights just plain the same as in the linear case or do I have to convert them using the lognormal distribution: $Var = \mathrm{e}^{2\mu+\sigma^{2}}(\mathrm{e}^{\sigma^{2}}-1)$ 
Thanks for your help.
 A: It is good to remember what is the purpose of the weighting. Suppose we have linear regression
\begin{align}
\mathbf{y}=\mathbf{X}\mathbf{\beta}+\mathbf{\varepsilon}
\end{align}
In this case $\sigma_{y_i}=\sigma_{\varepsilon_i}=\sigma_i$ and we can write that 
$$\Omega=E(\varepsilon\varepsilon')=diag(\sigma_i)$$
Now weighted regression with weights inverse to $\sigma_i$ is generalised least squares estimate with $\Omega$. Generalised least squares estimates are best linear unbiased estimates if variance of errors is $\Omega$. 
So you need to know the variance of errors, not the $y$. If you fit logarithmic regression and you know that $y$ is log normal, you need to weight with the variance of $\log y$, which in your case will correspond to variance of $\epsilon$. Then your weighted least squares estimates will be best linear unbiased estimates, and you will retain the same property as in linear case.
A: If the variance of $y_i$ is $\sigma_i^2,$ then, by the delta method, the variance of $\log(y_i)$ is approximately $\sigma_i^2 / y_i^2 .$ So if you're doing weighted least squares using inverse-variance weights, instead of weighting by $1/\sigma_i^2 ,$ you need to weight by $y_i^2 / \sigma_i^2 .$
