Let the relative efficiency between two estimators $T_1$ and $T_2$ of a parameter $\theta$ be defined as:

$$e(T_1,T_2) = \frac{E[(T_2-\theta)^2]}{E[(T_1 - \theta)^2]}$$

This definition indicates that $T_1$ and $T_2$ need to be estimators of the same parameter, i.e. $\theta$.

Taking location parameters as an example - my understanding is that the sample mean and sample median are estimators for two different location parameters, namely the population mean and population median.

Is it the case then, that it is not valid to compute the relative efficiency of the sample mean and sample median on an asymmetric distribution because the parameters are different? And with that same reasoning, is it correct to say that it is valid to compute their relative efficiency on a symmetric distribution, because in that case, the mean and median happen to coincide?

  • $\begingroup$ If the mean and median of the population are different, it might be seen as perverse to consider using both the sample mean and the sample median to estimate the same location parameter $\theta$ without some kind of adjustment. At least one of them will not be a good estimator even with a large sample size. On the other hand, with an asymmetric distribution with population mean equal to median, it would not be so unreasonable to consider the two sample statistics when estimation that population parameter $\endgroup$ – Henry Aug 23 '17 at 17:52
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    $\begingroup$ The situation might not be quite as simple as it might seem: to make sense of relative efficiency it's not necessary that the $T_i$ estimate the same parameter; it's only necessary that one of the parameters be a determined function of the other. This would be the case for any one-dimensional family, for instance, regardless of the symmetry of the distributions. $\endgroup$ – whuber Aug 23 '17 at 18:20
  • $\begingroup$ @whuber , interesting! So that means that in my definition of $e(T_1,T_2)$, $\theta$ in the numerator can in fact be something other than the $\theta$ in the denominator (say $\theta'$), as long as $\theta'$ is a function of $\theta$? I've not heard the term "determined" function before, so not sure what that implies additionally. If taking exponential distribution as an example (asymmetric, and one-dimensional(?) with parameter $\lambda >0$), the median can be expressed as $\mu \ln(2)$ with mean $\mu$. Therefore, relative efficiency on mean and median on exponential distributed data is valid? $\endgroup$ – Patrick Aug 23 '17 at 22:07
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    $\begingroup$ By "determined" I meant it does not depend on the data. The proper expression for relative efficiency would be $$\frac{E((T_1-\theta)^2)}{E((h(T_2) - \theta)^2)}$$ where $T_2$ estimates a parameter $\phi$ and $h(\phi)=\theta$. The Exponential (or, say, any scale family of Gammas with a fixed shape parameter) is a good example to consider. $\endgroup$ – whuber Aug 23 '17 at 22:13

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