Why using ratio-consistent estimator instead of consistent I'm reading articles in high dimensional data. I didn't saw the definition of ratio-consistent estimator in any other field. But as in Chen an Qin's article to showing a estimator is consistent we should show: 
$$ E[\hat{\theta}] = \theta\Big(1+o(1)\Big)\quad \text{and}\quad \mathrm{var}\Big(\hat{(\theta)}\Big)= o(\theta^2).$$
Is this a weaker from of consistent estimator? or is totally different? 
any help will be appreciated.
 A: This is totally different from consistency.  A consistent estimator is one whose sequence of estimates converges in probability to the true value as the sample size $n\to\infty$.  This does not imply that the expected value of the estimator converges to the true value, or even that the expected value exists, nor does it have any implications for the variance of the estimator.  
For the former, consider an estimator $\hat{\theta}_n$ such that:
$$ \hat{\theta}_n = \begin{cases} 
   \theta &\text{ with probability }{n-1\over n} \\
   \theta+n\delta &\text{ with probability } {1\over n}
\end{cases}$$
The probability that $\hat{\theta}$ is within any fixed distance of $\theta$ goes to 1 as $n\to\infty$, but the expected value of $\hat{\theta}_n = \theta + \delta$ for all $n$ (and therefore converges to $\theta + \delta$, not $\theta$.)  In this case $\hat{\theta} = \theta(1 + O(1))$, not $\dots (1+o(1))$.  We can make it $\dots (1+O(n))$ by replacing $\delta$ with $n\delta$ above, or any order we want for that matter, which will not affect the consistency of the estimator.
For the latter, consider the sample mean as an estimator of the location of a $t_2$ distribution.  Its variance is infinite, as the variance of a $t_2$ variate is infinite, but it is a consistent estimator of the location nonetheless (and an unbiased one.)
