I'm wondering if it is always valid to use Coefficient of Variation (CV) to determine relative efficiency of parameter estimators, and to compute statistically equivalent sample sizes based on that relative efficiency? In formal terms:
Given a random variable $R > 0$ that can only have positive values, and a parameterized continuous distribution D such that $R \sim D(\theta)$, and two different methods of estimating $\theta$ based on $n$ samples: $\hat{\theta_A} = A(n), \hat{\theta_B} = B(n)$; and if method A has a smaller coefficient of variation than method B: $\forall n: \text{CV}_A(n) < \text{CV}_B(n)$; now:
I have been working under the assumption that Relative Efficiency (RE) = $\left(\frac{\text{CV}_A(n)}{\text{CV}_B(n)} \right)^2$ tells us the number of samples $n' = n \times \text{RE}$ such that $\text{CV}_A(n) = \text{CV}_B(n')$.
Is that assumption always correct? I suspect that it depends on a few more unstated conditions that I haven't violated in the course of studying common continuous random variables and estimators.
(To push this further: I have actually been using the width of the confidence intervals on the estimates, as determined by each method's sampling distribution, as an indicator of "statistical equivalence." In other words: If A and B both generate unbiased estimates of the parameter, then the sampling distributions of A(n) and B(n') are within some epsilon for all quantiles. But I assume this only happens with very nice, normalish distributions and estimators.)
