Expected value of the estimator

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

O( (w/n - S_E(d))³)

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

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

Taking expected values we find that

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

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

since E[w/n - S_E(d)] = E[w/n] - S_E(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).