This is a regression problem. Let $r_i=f_iu_i+e_i,i=1,...,n$, where $u_i\sim^{i.i.d}N(0,1)$, and $e_i=O_p(\alpha_n),\alpha_n\rightarrow 0$ as $n\rightarrow 0$. When I take the log transformation of $r_i$ and get \begin{align} \log|r_i|&=\log|f_iu_i+e_i|\\ &=\log|f_i|+\log|u_i|+\log|1+\frac{e_i}{f_iu_i}|. \end{align} The third term is $O_p(\alpha_n)$ since $\log(1+x)\sim x$ as $x\rightarrow0$. My question is that, if I want to model $\log|f_i|=x_i'\beta$, and denote by $\epsilon_i=\log|u_i|+\log|1+\frac{e_i}{f_iu_i}|$ the remainder. In regression problem, we need basic assumption that $E\epsilon_i=0$, but how can I achieve this? We can calculate the mean of $\log|u_i|$, denote by $c$, and hence \begin{align} \log|r_i|-c&=x_i'\beta+(\log|u_i|-c)+\log|1+\frac{e_i}{f_iu_i}|, \end{align} but what is the mean of $\log|1+\frac{e_i}{f_iu_i}|$? In this case, how to derive the convergence rate of $\hat{\beta}$?

  • $\begingroup$ Maybe I can derive the result directly as usual, since the mean of $\epsilon_i^\star=(\log|u_i|-c)$ is 0 and the variance is finite. Hence $\epsilon_i^\star=O_p(1)$. The term $\log|1+\frac{e_i}{f_iu_i}|=O_p(\alpha_n)$ can be ignored since it does not affect the asymptoic result. $\endgroup$ Sep 5 '21 at 2:13

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