Variance of the estimator

The variance $\sigma^2(d^*) = E[(d^* - E[d^*])^2]$ can be derived as follows. With the asymptotic value of E[d^*] we compute the Taylor series of (d^* - E[d^*])² in powers of (w/n - S_E(d)). Then we take expected values, replacing the powers of (w/n - S_E(d))^k by the central moments described in equations 8, 9 and 10. Truncating after two terms we obtain

$$\sigma^2(d^*) = \frac{\mu_2}{S_E^{'}(d)^2 n} -\frac{S_E^{''}(... ..._3 S_E^{'}(d)^2 \right)}{S_E^{'}(d)^6 n^2} + O(\frac{1}{n^3})$$

The most effective estimator is the one among all possible estimators, which has minimal variance. The variance is

$$\sigma^2(d^*) = \frac{F(d)}{n} + O(\frac{1}{n^2})$$

where $F(d)= \frac{\mu_2}{S_E^{'}(d)^2}$. We can compute E so that it minimizes F(d). This will give us asymptotically (in n) optimal estimators. We have several choices for the minimization: we can derive the best E for

• a given distance d, e.g. $\min \limits_{E_{ij}} F(d)$.
• a norm for a range of distances, e.g. for d=0 to 200, i.e. $\min \limits_{E_{ij}} \int_0^{200} F(t) dt$.
• the minimax for a range of distances, e.g. for d=0 to 200, i.e $\min \limits_{E_{ij}} \max \limits_{d=0..200} F(d)$.

There is no hope of finding a closed formula for this optimal E, but the first and second cases can be computed numerically without much difficulty.