Suppose the MLE is difficult to compute, or that the likelihood is not log-concave, so that condition (D) fails and we cannot be sure that the MLE is consistent. If we have a reasonable starting estimate, it turns out that a single Newton-Raphson step from that estimate is enough to produce an estimator with the same limiting distribution as the MLE.
Theorem (One-step estimator): Suppose regularity conditions (A)-(C) are met and that $\btt_n$ is a $\sqrt{n}$-consistent estimator of $\bts$. Define
\[\bth_n = \btt_n + \oI_n(\btt_n)^{-1}\u(\btt_n).\]Then
\[\sqrt{n}(\bth_n - \bts) \inD \Normal(\zero, \fI(\bts)^{-1}).\]If, in addition, $\fI(\bt)$ is continuous at $\bts$, the same result holds for the Fisher scoring step $\bth_n = \btt_n + \fI_n(\btt_n)^{-1}\u(\btt_n)$.
Proof: Abbreviate $\fI(\bts)$ as $\fI$. By information continuity, $\tfrac1n\oI_n(\btt_n) \inP \fI$, which is positive definite. Multiplying both sides of the definition of $\bth_n$ by $\oI_n(\btt_n)$, we have $\oI_n(\btt_n)(\bth_n - \btt_n) = \u(\btt_n)$, and so
\[\begin{alignat*}{2} \tfrac1n\oI_n(\btt_n)\sqrt{n}(\bth_n - \bts) &= \tfrac1n\oI_n(\btt_n)\sqrt{n}(\bth_n - \btt_n) \\ &\quad + \tfrac1n\oI_n(\btt_n)\sqrt{n}(\btt_n - \bts) \\ &= \tfrac{1}{\sqrt{n}}\u(\btt_n) + \tfrac1n\oI_n(\btt_n)\sqrt{n}(\btt_n - \bts) &\hspace{4em}& \text{Def } \bth_n \\ &= \tfrac{1}{\sqrt{n}}\u(\bts) && \href{score-linearization.html}{\text{Score linearization}} \\ &\quad + \{\tfrac1n\oI_n(\btt_n) - \fI - o_p(1)\}\sqrt{n}(\btt_n - \bts) \\ &= \tfrac{1}{\sqrt{n}}\u(\bts) + o_p(1)O_p(1) && \tfrac1n\oI_n(\btt_n) \inP \fI; \href{consistent.html}{\btt_n\ \sqrt{n}\text{-consistent}} \\ &= \tfrac{1}{\sqrt{n}}\u(\bts) + o_p(1) && \href{o-notation.html}{O \text{ notation}} \end{alignat*}\]Since $\tfrac{1}{\sqrt{n}}\u(\bts) \inD \Normal(\zero, \fI)$ by the distribution of the score, the right-hand side converges to the same limit by Slutsky’s theorem. Thus, by a minor extension of Slutsky’s theorem,
\[\sqrt{n}(\bth_n - \bts) \inD \fI^{-1}\Normal(\zero, \fI) =_d \Normal(\zero, \fI^{-1}).\]For the Fisher scoring step, the only change is that $\tfrac1n\oI_n(\btt_n)$ is replaced by $\tfrac{1}{n}\fI_n(\btt_n) = \fI(\btt_n)$, which converges in probability to $\fI$ by the continuous mapping theorem, since $\fI$ is continuous at $\bts$; the rest of the proof is identical.