problems.

Specification of p and p under independence {#independence}
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If the assumption that ${{}Y}_1, \ldots, {{}Y}_k$ are independent under $G$ is plausible, then $$\nu_{r, s} = \sum_{i = 1}^k E_G( \psi_r^i \psi_s^i) + \sum_{i \ne j} E_G( \psi_r^i) E_G(\psi_s^j) \, .$$ Assumption \[consistency\] on the unbiasedness of the estimating function implies that the second term in the right-hand side of the above expression is zero. Hence, $l_{r|s} = \sum_{i = 1}^k \psi_r^i \psi_s^i$ is such that $\expect_G(l_{r|s}) = \nu_{r, s}$ and, under assumption \[derivative\_orders\], ${H^{}_{r | s}} = {l^{}_{r | s}} - {\nu^{}_{r, s}} = O_p(n^{1/2})$.

In matrix notation, ${{}e}({{}\theta})$ can be taken to be the $p \times p$ matrix ${{}\Psi}({{}\theta})^\top {{}\Psi}({{}\theta})$, where ${{}\Psi}({{}\theta})$ is the $n \times p$ matrix with $(i, s)$th element $\psi_s^i({{}\theta})$. In other words, the $(s, t)$th element of ${{}e}({{}\theta})$ is $$\label{e_mat}
  [{{}e}({{}\theta})]_{st} = \sum_{i = 1}^k \psi_s^i({{}\theta}) \psi_t^i({{}\theta}) \,.$$ A similar argument gives that ${{}d}_r({{}\theta})$ can be taken to be the $p \times p$ matrix $\tilde{{{}\Psi}}_r({{}\theta})^\top {{}\Psi}({{}\theta})$ where $\tilde{{{}\Psi}}_r({{}\theta})$ is the $n \times p$ matrix with $(i, s)$th element $\partial \psi_r^i / \partial \theta^s$. In other words, the $(s, t)$th element of ${{}d}_r({{}\theta})$ is $$\label{d_mat}
  [{{}d}_r({{}\theta})]_{st} = \sum_{i = 1}^k \left\{ \frac{\partial}{\partial \theta^s} \psi_r^i({{}\theta}) 