\mu_{ab}^{c} + \frac{1}{2} H^{a} H^{b} H_{b}^{c} \mu_{ac}^{r}+ \\ \notag
                                  & + \frac{1}{2} H^{a} H^{b} H_{a}^{c} \mu_{cb}^{r} + \frac{1}{4} H^{a} H^{b} H^{c} \mu_{bc}^{d} \mu_{ad}^{r} + \frac{1}{4} H^{a} H^{b} H^{c} \mu_{ab}^{d} \mu_{dc}^{r} + \\ \notag
                                  & + \frac{1}{2} H^{a} H^{b} H_{ab}^{r} + \frac{1}{6} H^{a} H^{b} H^{c} \mu_{abc}^{r} +\\ \notag
                                  & + A^{a} H_{a}^{r} + \frac{1}{2} A^{a} H^{b} \mu_{ba}^{r} + \frac{1}{2} A^{a} H^{b} \mu_{ab}^{r} + A_{a}^{r} H^{a} + O_p(n^{-2}) \, ,\end{aligned}$$ where ${H^{r}_{r_1\ldots r_a}} = - {\mu^{rs}_{}} {H^{}_{sr_1\cdots r_a}}$, ${\mu^{r}_{r_1\cdots r_a}} = - {\mu^{rs}_{}} {\mu^{}_{sr_1\cdots r_a}}$, and ${A^{r}_{r_1\cdots r_a}} = - {\mu^{rs}_{}} {A^{}_{sr_1\cdots r_a}}$, with $\mu^{rs}$ denoting the matrix inverse of $\mu_{rs}$ (assumption \[invertibility\]) and $A_{r_1\cdots r_a} = \partial^{a-1} A_{r_1}({{}\theta}) / \partial
\theta^{r_2} \cdots \partial \theta^{r_a}$.

Taking expectations with respect to the underlying distribution $G$ on both sides of (\[taylor\]), assumption \[moment\_orders\] gives that the bias of $\tilde{{{}\theta}}$ is $$\expect_G{(\tilde{\theta}^{r} - \bar{\theta}^{r})} = -{\mu^{ra}_{}}
  \expect_G (A_a) + \frac{1}{2} {\mu^{ra}_{}} {\mu^{bc}_{}}
  \left({\nu^{}_{ab,c}} - {\mu^{de}_{}} {\nu^{}_{c,e}}
    {\mu^{}_{abd}} \right) + O(n^{-3/2})\, ,$$ where all terms in the right-hand side are understood as being evaluated at $\bar{{}\theta}$.

The above expansion for the bias implies that use of any adjustment ${{}A}$ with $$\label{adjustment}
  \expect_G(A_r) = \frac{1}{2} {\mu^{ab}_{}} \left( 2 {\nu^{}_{ra,b}} - {\mu^{cd}_{}} {\nu^{}_{b,d}} {\mu^{}_{rac}} \right) + O(n^{-1/2}) \, 