bias. See Appendix \[sec:MixedProofs\] for more details.

\[thm:var\_mixed\]If the functional $g$ is Lipschitz continuous in both of its arguments and $\mathcal{S}_{Y}$ is finite, then the variance of $\g{h_{X},h_{X|Y}}$ is $O(1/N)$.

By the law of total variance, we have $$\begin{aligned}
\var\left[\g{h_{X},h_{X|Y}}\right] & =\bE\left[\var\left[\left.\g{h_{X},h_{X|Y}}\right|\mathbf{Y}_{1},\dots,\mathbf{Y}_{N}\right]\right]\\
 & +\var\left[\bE\left[\left.\g{h_{X},h_{X|Y}}\right|\mathbf{Y}_{1},\dots,\mathbf{Y}_{N}\right]\right].\end{aligned}$$ Given all of the $\mathbf{Y}_{i}$’s, the estimators $\g{h_{X},h_{X|y}}$ are all independent since they use different sets of $\mathbf{X}_{i}$’s for each $y$. From Theorem \[thm:variance\], we know that $\var\left[\left.\g{h_{X},h_{X|Y}}\right|\mathbf{Y}_{1},\dots,\mathbf{Y}_{N}\right]=O\left(\sum_{y\in\mathcal{S}_{Y}}\mathbf{N}_{y}/N^{2}\right)$. Taking the expectation then yields $O(1/N)$.

For the second term, we know from the proof of Theorem \[thm:bias\_mixed\] that $\bE\left[\left.\g{h_{X},h_{X|Y}}\right|\mathbf{Y}_{1},\dots,\mathbf{Y}_{N}\right]$ yields a sum of terms of the form of $\mathbf{N}_{y}^{\gamma}/N$ for $0<\gamma\leq1$. Taking the variance of the sum of these terms yields a sum of terms of the form $\var\left[\mathbf{N}_{y}^{\gamma}\right]/N^{2}$ (the covariance terms can be bounded by the Cauchy-Schwarz inequality to yield similar terms). Then $\var\left[\mathbf{N}_{y}^{\gamma}\right]$ can be bounded by taking a Taylor series expansion of the functions $\mathbf{N}_{y}^{\gamma}$ and $\mathbf{N}_{y}^{2\gamma}$ at the point $Nf_{Y}(y)$ which yields an expression that depends on the central moments of $\mathbf{N}_{y}$. From this, we obtain $\var\left[\mathbf{N}_{y}^{\gamma}\right]=O(N)$ which completes the proof. See Appendix \[sec:MixedProofs\] for details.

Theorems \[thm:bias\_mixed\] and \[thm:var\_mixed\] provide exact expressions for the bias and bounds on the variance of the 