lugsail ABM estimator: $$\hat{\Sigma}_{A,L} = \dfrac{1}{m}\sum_{k=1}^{m}\hat{\Sigma}_{k,L}\,.$$ Since the $m$ Markov chains are independent, large sample properties of $\hat{\Sigma}_{k,L}$ are shared by $\hat{\Sigma}_{A,L}$. This includes strong consistency, which was discussed in [@vats:fleg:jon:2019], a CLT analogue for the univariate estimator [@chak:khare:2019], and large-sample bias and variance results. To be explicit, the following corollary is a consequence of [@vats:flegal:2018].

\[thm:abm\_bias\] Under Assumption \[ass:mixing\] $$\text{Bias} \left(\hat{\Sigma}_{A,L} \right) = \dfrac{\Gamma}{b}\left(\dfrac{1 - rc}{1 - c}\right) + o\left(\dfrac{1}{b}\right) \,.$$ Further, if Assumption \[ass:bn\] holds, ${\text{E}}D^{4} < \infty$, $E_{F}\|Y_{1,1}\|^{4} < \infty$, and $b^{-1}\psi^{2}(n)\log n \to 0$ as $n \to \infty$, $$\dfrac{n}{b}Var\left(\hat{\Sigma}^{ij}_{A,L}\right) =  \dfrac{1}{m}\left[\dfrac{1}{r} + \dfrac{r - 1}{r(1 - c)^{2}}\right]\left(\Sigma^{2}_{ij} + \Sigma_{ii}\Sigma_{jj}\right) + o(1)  \; .$$

For large $n$, there is essentially no difference between ABM and RBM; this is also verified in all of our examples. However, and most critically, for small $n$ and particularly when any of the $m$ Markov chains has not sufficiently explored the state space, the RBM estimator exhibits far superior finite sample properties. This is illustrated in all of our examples in Section \[sec:examples\].

### Naive estimator

The naive estimator of $\Sigma$ used in the convergence diagnostic of [@bro:gel:1998] is obtained by taking the sample variance of the $m$ sample means $\hat{\mu}_k$. That 