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Expected value of the estimator

\begin{displaymath}S_E^{-1}(w/n) = S_E^{-1}(S_E(d)) + (w/n - S_E(d))S_E^{-1'}(S_E(d)) +
(w/n - S_E(d))^2 \frac{S_E^{-1''}(S_E(d))}{2}+ \end{displaymath}

O( (w/n - SE(d))3)

Simplifying SE-1 and its derivatives we can rewrite the above equation as

\begin{displaymath}S_E^{-1}(w/n) = d + (w/n - S_E(d)) \frac{1}{ S_E'(d)} -
...))^2 \frac{S_E^{''}(d)}{2 S_E^{'}(d)^3}+
O( (w/n - S_E(d))^3) \end{displaymath}

Taking expected values we find that

\begin{displaymath}E[d^*] = d + E[w/n - S_E(d)] \frac{1}{S_E'(d)}-
E[(w/n - S_...
...2] \frac{S_E^{''}(d)}{2 S_E^{'}(d)^3} +
O(\frac{\mu_2}{n^2}) \end{displaymath}

\begin{displaymath}= d - \frac{S_E^2(d) -S_E(d)^2}{n} \frac{S_E^{''}(d)}{2 S_E^{'}(d)^3} +
O(\frac{1}{n^2}) \end{displaymath}

since E[w/n - SE(d)] = E[w/n] - SE(d) = 0. When $n
\rightarrow \infty$ then $E[d^*] \rightarrow d$, and hence E[d*] is a consistent estimator (we could correct it to any order of n if desired).

Chantal Korostensky