$t_{2}$ only through $t_{1}^{\alpha}t_{2}^{\beta}$ for some $\alpha,\beta\in\mathbb{R}$ for each $1\leq j,l\leq\lambda$, then the bias is $$\begin{aligned}
 & \lefteqn{\bias\left[\g{h_{X},h_{X|Y}}\right]}\nonumber \\
 & =\sum_{\substack{m,n=0\\
i+j+m+n\neq0
}
}^{\left\lfloor \lambda/2\right\rfloor }\sum_{i,j=0}^{r}c_{14,j,i,m,n}\frac{h_{X}^{i}l^{j}N^{-j\beta}}{\left(Nh_{X}^{d_{X}}\right)^{m}\left(l^{d_{X}}N^{1-\beta d_{X}}\right)^{n}}\nonumber \\
 & +O\left(h_{X}^{s}+N^{-s\beta}+\frac{1}{\left(Nh_{X}^{d_{X}}\right)^{\lambda/2}}+\frac{1}{\left(N^{1-\beta d_{X}}\right)^{\lambda/2}}\right).\label{eq:bias_mixed2}\end{aligned}$$

We focus on (\[eq:bias\_mixed1\]) as (\[eq:bias\_mixed2\]) follows similarly. It can be shown that $$\bias\left[\g{h_{X},h_{X|Y}}\right]=\bE\left[\sum_{y\in\mathcal{S}_{Y}}\frac{\mathbf{N}_{y}}{N}\bias\left[\left.\g{h_{X},h_{X|y}}\right|\mathbf{Y}_{1},\dots,\mathbf{Y}_{N}\right]\right].$$ The conditional bias of $\g{h_{X},h_{X|y}}$ given $\mathbf{Y}_{1},\dots,\mathbf{Y}_{N}$ can then be obtained from Theorem \[thm:bias\] as $$\begin{aligned}
 & \lefteqn{\bias\left[\left.\g{h_{X},h_{X|y}}\right|\mathbf{Y}_{1},\dots,\mathbf{Y}_{N}\right]}\\
 & =\sum_{\substack{i,j=0\\
i+j\neq0
}
}^{r}c_{10,i,j}h_{X}^{i}\mathbf{h}_{X|y}^{j}\\
 & +O\left(h_{X}^{s}+\mathbf{h}_{X|y}^{s}+\frac{1}{\mathbf{N}_{y}h_{X}^{d_{X}}}+\frac{1}{\mathbf{N}_{y}\mathbf{h}_{X|y}^{d_{X}}}\right)\end{aligned}$$ Then given that $\mathbf{h}_{X|y}\propto\mathbf{N}_{y}^{-\beta}$, (\[eq:mixed\_est\]) gives terms of the form of $\mathbf{N}_{y}^{1-\gamma}$ with $\gamma>0$. $\mathbf{N}_{y}$ is a binomial random variable with parameter $f_{Y}(y)$, $N$ trials, and mean $Nf_{Y}(y)$. Thus we need to compute the fractional moments of a binomial random variable. By the generalized binomial theorem, we have that $$\begin{aligned}
\mathbf{N}_{y}^{\alpha} & =\left(\mathbf{N}_{y}-Nf_{Y}(y)+Nf_{Y}(y)\right)^{\alpha}\nonumber \\
 & =\sum_{i=0}^{\infty}\left(\begin{array}{c}
\alpha\\
i
\end{array}\right)\left(Nf_{Y}(y)\right)^{\alpha-i}\left(\mathbf{N}_{y}-Nf_{Y}(y)\right)^{i},\nonumber \\
 & \lefteqn{\implies\bE\left[\mathbf{N}_{y}^{\alpha}\right]}\nonumber \\
 & =\sum_{i=0}^{\infty}\left(\begin{array}{c}
\alpha\\
i
\end{array}\right)\left(Nf_{Y}(y)\right)^{\alpha-i}\bE\left[\left(\mathbf{N}_{y}-Nf_{Y}(y)\right)^{i}\right].\label{eq:fractional_moment}\end{aligned}$$ From [@riordan1937moment], the $i$-th central moment of $\mathbf{N}_{y}$ has the form of $$\bE\left[\left(\mathbf{N}_{Y}-Nf_{Y}(y)\right)^{i}\right]=\sum_{n=0}^{\left\lfloor i/2\right\rfloor }c_{n,i}(f_{Y}(y))N^{n}.$$ Thus $\bE\left[\mathbf{N}_{y}^{1-\gamma}\right]$ has terms proportional to $N^{1-\gamma-i+n}\leq N^{1-\gamma-\left\lfloor i/2\right\rfloor }$ for $i=0,1,\dots$ since $n\leq\left\lfloor i/2\right\rfloor $. Then since there is an $N$ in the denominator of (\[eq:mixed\_est\]), this leaves terms of the form of $N^{-\gamma}$ when $i=0,1$ and $N^{-1}$ for $i\geq2$. This completes the proof for the 