if $M$ is block diagonal (i.e., $M^b_{k\ell}$ is a matrix of zeroes for all $k\ne\ell$), (\[UjMUj\]) yields $\operatorname{tr}(M)$ without error. This result is an extension of the result that independent $U_j$’s give $\operatorname{tr}(M)$ exactly for diagonal $M$. Furthermore, it turns out that, at least in terms of the variance of $\frac{1}{N}\sum_{j=1}^N U_j'MU_j$, for the elements of $M$ off the block diagonal, we do exactly the same as we do when the $U_j$’s are independent. Write $B(\theta)$ for $\operatorname{cov}\{g(\theta,N)\}$ with $g(\theta,N)$ defined as in (\[ascore\]) with independent $U_j$’s. Define $g^d(\theta,N)=0$ for the unbiased estimating equations defined by (\[ascore\]) with dependent $U_j$’s defined by (\[Uj\]) and $B^d(\theta)$ to be the covariance matrix of $g^d(\theta,N)$. Take $T(N,n)$ to be the set of pairs of positive integers $(k,\ell)$ with $1\le\ell<k \le n$ for which $\lfloor k/N\rfloor= \lfloor\ell
/N\rfloor$. We have the following result, whose proof is given in the .

\[tdependent\] For any vector $v=(v_1,\ldots,v_p)'$, $$\label{improve}
v'B(\theta)v - v'B^d(\theta)v =
\frac{2}{N} \sum_{(k,\ell)\in T(N,n)} \Biggl\{ \sum
_{i=1}^p v_i \bigl( W_{k\ell}^i+W_{\ell k}^i
\bigr) \Biggr\}^2.$$

Thus, $B(\theta) \succeq B^d(\theta)$. Since $E_\theta\dot{g}
(\theta,N) = E_\theta\dot{g}^d(\theta,N)=-\mathcal{I}(\theta)$, it follows that $\mathcal{E}\{g^d(\theta,N)\}\succeq\mathcal{E}\{g(\theta,N)\}$.

How much of an improvement will result from using dependent $U_j$’s depends on the size of the $W_{k\ell}^i$’s within 