the $j$th component of $U_1$, we have $$\begin{aligned}
\label{Jij}
{{\mathcal}J}_{ij}(\theta) & = & \sum_{k,\ell,p,q=1}^n
\operatorname{cov}\bigl(W_{k\ell}^i U_{1k}U_{1\ell},
W_{pq}^j U_{1p}U_{1q}\bigr)
\nonumber
\\
& = & \sum_{k\ne\ell}\bigl\{\operatorname{cov}
\bigl(W_{k\ell}^i U_{1k}U_{1\ell
},W_{k\ell}^j
U_{1k}U_{1\ell}\bigr)+\operatorname{cov}\bigl(W_{k\ell}^i
U_{1k}U_{1\ell},W_{\ell k}^j
U_{1k}U_{1\ell}\bigr)\bigr\}
\nonumber\\[-8pt]\\[-8pt]
& = & \sum_{k\ne\ell}\bigl(W_{k\ell}^i
W_{k\ell}^j + W_{k\ell}^i
W_{\ell k}^j\bigr)
\nonumber
\\
& = & \operatorname{tr}\bigl(W^i W^j\bigr) + \operatorname{tr}
\bigl\{W^i \bigl(W^j\bigr)' \bigr\} - 2\sum
_{k=1}^n W_{kk}^iW_{kk}^j.\nonumber\end{aligned}$$ As noted by @hutchinson, the terms with $k=\ell$ drop out in the second step because $U_{1j}^2=1$ with probability 1. When $K(\theta)$ is diagonal for all $\theta$, then $N=1$ gives the exact score equations, although in this case computing $\operatorname{tr}(K^{-1}K_i)$ directly would be trivial.

Writing $\kappa(\cdot)$ for the condition number of a matrix, we can bound$\operatorname{cov}_\theta\{g(\theta,N)\}$ in terms of ${{\mathcal}I}(\theta)$ and $\kappa(K)$. The proof of the following result is given in the .

\[tmain\] $$\label{Bbound}
\operatorname{cov}_\theta\bigl\{g(\theta,N)\bigr\} \preceq{{\mathcal}I}(\theta)
\biggl\{ 1 + \frac{(\kappa(K)+1)^2}{4N\kappa(K)} \biggr\}.$$

It follows from (\[Bbound\]) that $$\mathcal{E}\bigl\{g(\theta,N)\bigr\} \succeq\biggl\{ 1+ \frac{(\kappa
(K)+1)^2}{4N\kappa(K)}
\biggr\}^{-1}{{\mathcal}I}(\theta).$$ In practice, if $\frac{(\kappa(K)+1)^2}{4N\kappa(K)} < 0.01$, so that the loss of information in using (\[ascore\]) rather than (\[score\]) was at most 1%, we would generally be satisfied with using the approximate score equations and a loss of information of even 10% or larger might be acceptable when one has a massive 