text:
\begin{document}

\maketitle

\section{\nopunct}

In this section, we prove Proposition 3.1.
The proof is based on the Taylor formula and well-known results
on sums of independent random variables.

\begin{proof}[Proof of Proposition 3.1]
Let $\xi'(dz)$, $\xi''(dz)$ be any probability measures on ${\cal Z}$.
As a direct consequence of (3.2), we get
\begin{align*}
	g(z')\leq \beta g(z''),
	\;\;\;\;\;
	f(z') g(z'')
	\leq
	(f(z'')+\alpha g(z'') ) g(z')
\end{align*}
for $z',z''\in{\cal Z}$.
Then, we have
\begin{align*}
	\xi'(g)
	=
	\int\int g(z') \xi'(dz')\xi''(dz'')
	\leq
	\beta \int\int g(z'') \xi'(dz')\xi''(dz'')
	=
	\beta \xi''(g).
\end{align*}
We also have
\begin{align*}
	\xi'(f)\xi''(g)
	\!=\!
	\int\int f(z')g(z'') \xi'(dz')\xi''(dz'')
	\!\leq\! &
	\int\int (f(z'')+\alpha g(z'') ) g(z') \xi'(dz')\xi''(dz'')
	\\
	\!=\! &
	(\xi''(f)+\alpha\xi''(g) ) \xi'(g).
\end{align*}
Consequently, we get
\begin{align}\label{p2.1.1}
	\frac{\xi'(g)}{\xi''(g)}
	\leq
	\beta,
	\;\;\;\;\;
	\frac{\xi'(f)}{\xi'(g)}
	\leq
	\frac{\xi''(f)}{\xi''(g)}
	+
	\alpha.
\end{align}
Reverting the roles of $\xi'(dz)$, $\xi''(dz)$,
we conclude $\xi''(f)/\xi''(g)\leq\xi'(f)/\xi'(g)+\alpha$.
Since $\beta\geq 1$, we then deduce
\begin{align}\label{p2.1.3}
	\left|
	\frac{\xi'(f)}{\xi'(g)} - \frac{\xi''(f)}{\xi''(g)}
	\right|
	\leq
	\alpha,
	\;\;\;\;\;
	\left|
	\frac{\xi'(g)}{\xi''(g)} - 1
	\right|
	\leq
	\frac{\xi'(g)}{\xi''(g)} + 1
	\leq
	2\beta.
\end{align}

Let $k\geq 1$ be any integer. It is straightforward to verify
\begin{align*}
	\frac{\xi_{k}(f) }{\xi_{k}(g) }
	=&
	\frac{\xi_{k}(f) }{\xi(g) }
	-
	\frac{\xi_{k}(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)
	+
	\frac{\xi(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\\
	&+
	\left(\frac{\xi_{k}(f) }{\xi_{k}(g) } - \frac{\xi(f) }{\xi(g) } \right)
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}.
\end{align*}
Consequently, we get
\begin{align*}
	E\left(\frac{\xi_{k}(f) }{\xi_{k}(g) } \right)
	-
	\frac{\xi(f) }{\xi(g) }
	=&
	-
	E\left(
	\frac{\xi_{k}(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)
	\right)
	+
	E\left(
	\frac{\xi(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right)
	\\ \nonumber
	&+
	E\left(
	\left(\frac{\xi_{k}(f) }{\xi_{k}(g) } - \frac{\xi(f) }{\xi(g) } \right)
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right).
\end{align*}
Then, owing to (\ref{p2.1.3}), we have
\begin{align}\label{p2.1.9}
	\left|
	E\left(\frac{\xi_{k}(f) }{\xi_{k}(g) } \right)
	-
	\frac{\xi(f) }{\xi(g) }
	\right|
	\leq &
	\left|
	E\left(
	\frac{\xi_{k}(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)
	\right)
	-
	E\left(
	\frac{\xi(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right)
	\right|
	\\ \nonumber
	&+
	E\left(
	\left|\frac{\xi_{k}(f) }{\xi_{k}(g) } - \frac{\xi(f) }{\xi(g) } \right|
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right)
	\\ \nonumber
	\leq &
	\left|
	E\left(
	\frac{\xi_{k}(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)
	\right)
	-
	E\left(
	\frac{\xi(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right)
	\right|
	\\ \nonumber
	&+
	\alpha
	E\left(
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right).
\end{align}
As $\{Z_{k} \}_{k\geq 1}$ are distributed according to $\xi(dz)$, we also have
\begin{align}\label{p2.1.21}
	E\left(\frac{g(Z_{k} ) }{\xi(g) } - 1 \right) = 0.
\end{align}
Since $\{Z_{k} \}_{k\geq 1}$ are independent, (\ref{p2.1.21}) implies
\begin{align}\label{p2.1.23}
	E\left(
	\frac{\xi_{k}(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)
	\right)
	=&
	\frac{1}{k^{2} } \sum_{i=1}^{k}
	E\left(\frac{f(Z_{i} ) }{\xi(g) }
	\left(\frac{g(Z_{i} ) }{\xi(g) } - 1 \right)
	\right)
	\\ \nonumber
	&+
	\frac{1}{k^{2} } \sum_{\stackrel{\scriptstyle 1\leq i,j \leq k}{i\neq j} }
	E\left(\frac{f(Z_{i} ) }{\xi(g) } \right)
	E\left(\frac{g(Z_{j} ) }{\xi(g) } - 1 \right)
	\\ \nonumber
	=&
	\frac{1}{k}
	\int \frac{f(z)}{\xi(g) } \left(\frac{g(z)}{\xi(g)} - 1 \right)
	\xi(dz)
	\\ \nonumber
	=&
	\frac{1}{k}
	\left(
	\int \frac{f(z) g(z) }{\xi^{2}(g) } \xi(dz)
	-
	\frac{\xi(f)}{\xi(g) }
	\right).
\end{align}
For the same reason, (\ref{p2.1.21}) yields
\begin{align}\label{p2.1.25}
	E\left(
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right)
	=&
	\frac{1}{k^{2} } \sum_{i=1}^{k}
	E\left(
	\left(\frac{g(Z_{i} ) }{\xi(g) } - 1 \right)^{2}
	\right)
	\\ \nonumber
	&+
	\frac{1}{k^{2} } \sum_{\stackrel{\scriptstyle 1\leq i,j \leq k}{i\neq j} }
	E\left(\frac{g(Z_{i} ) }{\xi(g) } - 1 \right)
	E\left(\frac{g(Z_{j} ) }{\xi(g) } - 1 \right)
	\\ \nonumber
	=&
	\frac{1}{k}
	\int \left(\frac{g(z)}{\xi(g)} - 1 \right)^{2}
	\xi(dz)
	\\ \nonumber
	=&
	\frac{1}{k}
	\left(
	\int \frac{g^{2}(z) }{\xi^{2}(g) } \xi(dz)
	-
	1
	\right).
\end{align}
Using (\ref{p2.1.23}), (\ref{p2.1.25}), we deduce
\begin{align*}
	&
	E\left(
	\frac{\xi_{k}(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)
	\right)
	-
	E\left(
	\frac{\xi(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right)
	\\
	&=
	\frac{1}{k}
	\int \frac{g^{2}(z) }{\xi^{2}(g) }
	\left(\frac{f(z) }{g(z) } - \frac{\xi(f) }{\xi(g) } \right)
	\xi(dz).
\end{align*}
Combining this with (\ref{p2.1.1}), (\ref{p2.1.3}), we get
\begin{align}\label{p2.1.27}
	&
	\left|
	E\left(
	\frac{\xi_{k}(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)
	\right)
	-
	E\left(
	\frac{\xi(f) }{\xi(g) }
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right)
	\right|
	\\ \nonumber
	&\leq
	\frac{1}{k}
	\int \frac{g^{2}(z) }{\xi^{2}(g) }
	\left|\frac{f(z) }{g(z) } - \frac{\xi(f) }{\xi(g) } \right|
	\xi(dz)
	\leq
	\frac{\alpha\beta^{2} }{k}.
\end{align}
Similarly, combining (\ref{p2.1.1}), (\ref{p2.1.25}), we get
\begin{align}\label{p2.1.29}
	E\left(
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right)^{2}
	\right)
	\leq
	\frac{1}{k}
	\int \frac{g^{2}(z) }{\xi^{2}(g) } \xi(dz)
	\leq
	\frac{\beta^{2} }{k}.
\end{align}
Relying on (\ref{p2.1.9}), (\ref{p2.1.27}), (\ref{p2.1.29}),
we conclude that the first part of (3.1) is true.

It is straightforward to verify
\begin{align*}
	\frac{\xi_{k}(f) }{\xi_{k}(g) }
	=
	\frac{\xi(f) }{\xi(g) }
	+
	\left(
	\frac{\xi_{k}(f) }{\xi(g) } - \frac{\xi(f)\xi_{k}(g) }{\xi^{2}(g) }
	\right)
	-
	\left(\frac{\xi_{k}(f) }{\xi_{k}(g) } - \frac{\xi(f) }{\xi(g) } \right)
	\left( \frac{\xi_{k}(g) }{\xi(g) } - 1 \right).
\end{align*}
Then, Minkowski inequality and (\ref{p2.1.3}) imply
\begin{align}\label{p2.1.31}
	\left(
	E\left(
	\left|
	\frac{\xi_{k}(f) }{\xi_{k}(g) } \!- \frac{\xi(f) }{\xi(g) }
	\right|^{2}
	\right)
	\right)^{\!\!1/2}
	\!\leq &
	\left(
	E\left(
	\left| \frac{\xi_{k}(f) }{\xi(g) } - \frac{\xi(f)\xi_{k}(g) }{\xi^{2}(g) } \right|^{2}
	\right)
	\right)^{\!\!1/2}
	\\ \nonumber
	&+
	\left(
	E\left(
	\left|\frac{\xi_{k}(f) }{\xi_{k}(g) } - \frac{\xi(f) }{\xi(g) } \right|^{2}
	\left| \frac{\xi_{k}(g) }{\xi(g) } - 1 \right|^{2}
	\right)
	\right)^{\!\!1/2}
	\\ \nonumber
	\!\leq &
	\left(
	E\left(
	\left| \frac{\xi_{k}(f) }{\xi(g) } - \frac{\xi(f)\xi_{k}(g) }{\xi^{2}(g) } \right|^{2}
	\right)
	\right)^{\!1/2}
	\\ \nonumber
	&+
	\alpha
	\left(
	E\left(
	\left| \frac{\xi_{k}(g) }{\xi(g) } - 1 \right|^{2}
	\right)
	\right)^{\!1/2}.
\end{align}
As $\{Z_{k} \}_{k\geq 1}$ are distributed according to $\xi(dz)$, we have
\begin{align}\label{p2.1.33}
	E\left(
	\frac{f(Z_{k} ) }{\xi(g) } - \frac{\xi(f)g(Z_{k} ) }{\xi^{2}(g) }
	\right)
	=
	0.
\end{align}
Since $\{Z_{k} \}_{k\geq 1}$ are independent,
(\ref{p2.1.1}), (\ref{p2.1.3}), (\ref{p2.1.33}) yield
\begin{align}\label{p2.1.35}
	E\left(
	\left| \frac{\xi_{k}(f) }{\xi(g) } - \frac{\xi(f)\xi_{k}(g) }{\xi^{2}(g) } \right|^{2}
	\right)
	=&
	\frac{1}{k^{2} }
	\sum_{i=1}^{k}
	E\left(
	\left|\frac{f(Z_{i} ) }{\xi(g) } - \frac{\xi(f)g(Z_{i} ) }{\xi^{2}(g) } \right|^{2}
	\right)
	\\ \nonumber
	&
	\begin{aligned}[t]
	+
	\frac{1}{k^{2} } \sum_{\stackrel{\scriptstyle 1\leq i,j \leq k}{i\neq j} }
	&
	E\left(
	\frac{f(Z_{i} ) }{\xi(g) } - \frac{\xi(f)g(Z_{i} ) }{\xi^{2}(g) }
	\right)
	\\
	&\cdot
	E\left(
	\frac{f(Z_{j} ) }{\xi(g) } - \frac{\xi(f)g(Z_{j} ) }{\xi^{2}(g) }
	\right)
	\end{aligned}
	\\ \nonumber
	= &
	\frac{1}{k}
	\int \left|\frac{g(z)}{\xi(g)} \right|^{2}
	\left| \frac{f(z)}{g(z)} - \frac{\xi(f) }{\xi(g) } \right|^{2} \xi(dz)
	\\ \nonumber
	\leq &
	\frac{\alpha^{2}\beta^{2} }{k}.
\end{align}
Relying on (\ref{p2.1.29}), (\ref{p2.1.31}), (\ref{p2.1.35}),
we conclude that the second part of (3.1) is true.
\end{proof}

\section{\nopunct}

In this section, we prove Proposition 4.1.
The proposition is based on the results of \cite{legland&oudjane} -- \cite{tadic&doucet3}.
To show Proposition 4.1,
we use the following additional notation.
$\tilde{r}_{\theta,\boldsymbol y}^{n}\!(x'|x)$ is the function defined by
\begin{align}\label{3.1}
	\tilde{r}_{\theta,\boldsymbol y}^{n}(x'|x)
	=
	q_{\theta}(y_{n}|x') p_{\theta}(x'|x)
\end{align}
for $\theta\in\Theta$, $x,x'\in{\cal X}$, $n\geq 0$
and a sequence $\boldsymbol y = \{y_{n} \}_{n\geq 0}$ in ${\cal Y}$.
$\tilde{r}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )$ and
$\tilde{s}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )$
are the functions defined by
\begin{align}
	\label{3.3}
	\tilde{r}_{\theta,\boldsymbol y}^{m:m}(x_{m:m} ) = 1,
	&\;\;\;\;\;
	\tilde{r}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	=
	\prod_{k=m+1}^{n}
	\tilde{r}_{\theta,\boldsymbol y}^{k}(x_{k}|x_{k-1} ),
	\\
	\label{3.5}
	\tilde{s}_{\theta,\boldsymbol y}^{m:m}(x_{m:m} ) = 0,
	&\;\;\;\;\;
	\tilde{s}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	=
	\nabla_{\theta} \tilde{r}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
\end{align}
for $x_{m},\dots,x_{n}\in{\cal X}$, $n>m\geq 0$.
$\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)$ and
$\tilde{S}_{\theta,\boldsymbol y}^{m:n}(dx'|x)$
are the elements of ${\cal M}_{p}({\cal X})$ and
${\cal M}_{s}^{d}({\cal X})$ (respectively) defined by
\begin{align}
	\label{3.7}
	\tilde{R}_{\theta,\boldsymbol y}^{m:m}(B|x)
	=
	\delta_{x}(B),
	&\;\;\;\;\;
	\begin{aligned}[t]
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(B|x)
	=&
	\int\cdots\int\int
	I_{B}(x_{n} )	\tilde{r}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\\
	&\cdot
	\mu(dx_{n})\cdots\mu(dx_{m+1})\delta_{x}(dx_{m}),
	\end{aligned}
	\\
	\label{3.9}
	\tilde{S}_{\theta,\boldsymbol y}^{m:m}(B|x)
	=
	0,
	&\;\;\;\;\;
	\begin{aligned}[t]
	\tilde{S}_{\theta,\boldsymbol y}^{m:n}(B|x)
	=&
	\int\cdots\int\int
	I_{B}(x_{n} )	\tilde{s}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\\
	&\cdot
	\mu(dx_{n})\cdots\mu(dx_{m+1})\delta_{x}(dx_{m})
	\end{aligned}
\end{align}
for $B\in{\cal B}({\cal X})$.
$\tilde{F}_{\theta,\boldsymbol y}^{m:n}(dx|\xi)$,
$\tilde{G}_{\theta,\boldsymbol y}^{m:n}(dx|\xi,\zeta)$ and
$\tilde{H}_{\theta,\boldsymbol y}^{m:n}(dx|\xi,\zeta)$
are the elements of ${\cal P}({\cal X})$,
${\cal M}_{s}^{d}({\cal X})$ and ${\cal M}_{s}^{d}({\cal X})$ (respectively)
defined by
\begin{align}\label{3.21}
	\tilde{F}_{\theta,\boldsymbol y}^{m:m}(B|\xi)
	=
	\xi(B),
	&\;\;\;
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}(B|\xi)
	=
	\frac{\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}(B|x) \xi(dx)}
	{\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}({\cal X}|x) \xi(dx)},
	\\
	\label{3.23}
	\tilde{H}_{\theta,\boldsymbol y}^{m:m}(B|\xi,\zeta)
	\!=\!
	\zeta(B),
	&\;\;\;
	\tilde{H}_{\theta,\boldsymbol y}^{m:n}(B|\xi,\zeta)
	\!=\!
	\frac{\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}(B|x) \zeta(dx)
	\!+\!
	\int \tilde{S}_{\theta,\boldsymbol y}^{m:n}(B|x) \xi(dx)}
	{\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}({\cal X}|x) \xi(dx)},
	\\
	\label{3.25}
	\tilde{G}_{\theta,\boldsymbol y}^{m:m}(B|\xi,\zeta)
	\!=\!
	\zeta(B),
	&\;\;\;
	\tilde{G}_{\theta,\boldsymbol y}^{m:n}(B|\xi,\zeta)
	\!=\!
	\tilde{H}_{\theta,\boldsymbol y}^{m:n}(B|\xi,\zeta)
	\!-\!
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}(B|\xi)
	\tilde{H}_{\theta,\boldsymbol y}^{m:n}({\cal X}|\xi,\zeta)
\end{align}
for $\xi\!\in{\cal P}({\cal X})$, $\zeta\!\in{\cal M}_{s}^{d}({\cal X})$.
$\tilde{\alpha}_{\theta,\boldsymbol y}^{n}(dx|\xi)$ and
$\tilde{\beta}_{\theta,\boldsymbol y}^{n}(dx|\xi,\zeta)$
are the elements of ${\cal P}({\cal X})$ and ${\cal M}_{s}^{d}({\cal X})$ (respectively)
defined by
\begin{align}
	&\label{3.27}
	\tilde{\alpha}_{\theta,\boldsymbol y}^{n}(B|\xi)
	=
	\frac{\int_{B} q_{\theta}(y_{n}|x) \xi(dx) }{\int q_{\theta}(y_{n}|x) \xi(dx) },
	\\
	&\label{3.29}
	\tilde{\beta}_{\theta,\boldsymbol y}^{n}(B|\xi,\zeta)
	=
	\frac{\int_{B} q_{\theta}(y_{n}|x) \zeta(dx)
	+ \int_{B} \nabla_{\theta} q_{\theta}(y_{n}|x) \xi(dx)  }
	{\int q_{\theta}(y_{n}|x) \xi(dx) }
\end{align}
for $n\geq 0$.
$\tilde{F}_{\theta,\boldsymbol y}^{m:n}(\xi)$, $\tilde{G}_{\theta,\boldsymbol y}^{m:n}(\xi,\zeta)$
and $\tilde{H}_{\theta,\boldsymbol y}^{m:n}(dx|\xi,\zeta)$
are a `short-hand' notation for $\tilde{F}_{\theta,\boldsymbol y}^{m:n}(dx|\xi)$, $\tilde{G}_{\theta,\boldsymbol y}^{m:n}(dx|\xi,\zeta)$
and $\tilde{H}_{\theta,\boldsymbol y}^{m:n}(dx|\xi,\zeta)$ (respectively).
Similarly, $\tilde{\alpha}_{\theta,\boldsymbol y}^{n}(\xi)$ and $\tilde{\beta}_{\theta,\boldsymbol y}^{n}(\xi,\zeta)$
are a `short-hand' notation for
$\tilde{\alpha}_{\theta,\boldsymbol y}^{n}(dx|\xi)$ and $\tilde{\beta}_{\theta,\boldsymbol y}^{n}(dx|\xi,\zeta)$ (respectively).

\begin{vremark}
Let Assumptions 2.1 -- 2.3 hold.
Moreover, let $\zeta_{\theta}(dx)$ and $w_{\theta}(x)$ be of the form (2.4).
Then, it is straightforward to verify
\begin{align*}
	&
	\tilde{F}_{\theta,\boldsymbol y}^{0:n}(B|\xi_{\theta} )
	=
	P\left(\left.
	X_{n}^{\theta}\in B
	\right|
	Y_{1:n}^{\theta} = y_{1:n}
	\right),
	\;\;\;\;\;
	\tilde{G}_{\theta,\boldsymbol y}^{0:n}(B|\xi_{\theta},\zeta_{\theta} )
	=
	\nabla_{\theta}
	\tilde{F}_{\theta,\boldsymbol y}^{0:n}(B|\xi_{\theta} )
\end{align*}
for all $\theta\in\Theta$, $B\in{\cal B}({\cal X})$,
$n\geq 1$ and any sequence $\boldsymbol y = \{y_{n} \}_{n\geq 0}$ in ${\cal Y}$.
Hence, $\tilde{F}_{\theta,\boldsymbol y}^{m:n}(dx|\xi)$ and
$\tilde{G}_{\theta,\boldsymbol y}^{m:n}(dx|\xi,\zeta)$
can be considered as (a generalization of)
the optimal filter and its gradient.
\end{vremark}

\begin{lemma}\label{lemma3.1}
(i) Let Assumption 2.1 hold. Then, we
have
\begin{align}
	\label{l3.1.3*}
	F_{\theta,\boldsymbol y}^{m:n+1}(B|\xi)
	=&
	\int\int I_{B}(x'') p_{\theta}(x''|x') \mu(dx'')
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}\big(dx'|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi) \big)
\end{align}
for all $\theta\in\Theta$, $B\in{\cal B}({\cal X})$, $\xi\in{\cal P}({\cal X})$,
$n\geq m\geq 0$ and any sequence $\boldsymbol y = \{y_{n} \}_{n\geq 0}$ in ${\cal Y}$.

(ii) Let Assumptions 2.1 and 2.2 hold. Then, we
have
\begin{align}
	\label{l3.1.7*}
	G_{\theta,\boldsymbol y}^{m:n+1}(B|\xi,\zeta)
	=&
	\int\int I_{B}(x'') \nabla_{\theta} p_{\theta}(x''|x') \mu(dx'')
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}
	\big(dx'|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi) \big)
	\\ \nonumber
	&+
	\int\int I_{B}(x'') p_{\theta}(x''|x') \mu(dx'')
	\tilde{G}_{\theta,\boldsymbol y}^{m:n}
	\big(dx'|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi), \tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi,\zeta) \big)
\end{align}
for all $\theta\in\Theta$, $B\in{\cal B}({\cal X})$,
$\xi\in{\cal P}({\cal X})$,
$\zeta\in{\cal M}_{s}^{d}({\cal X})$,
$n\geq m\geq 0$ and any sequence $\boldsymbol y = \{y_{n} \}_{n\geq 0}$ in ${\cal Y}$.
\end{lemma}

\begin{proof}
(i) Throughout this part of the proof,
the following notation is used.
$\theta$ is any element of $\Theta$,
while $x$ is any element of ${\cal X}$.
$B$, $\xi$ are any elements of ${\cal B}({\cal X} )$, ${\cal P}({\cal X} )$ (respectively),
while $\boldsymbol y = \{y_{n} \}_{n\geq 0}$ is any sequence in ${\cal Y}$.
$n$, $m$ are any integers satisfying $n\geq m\geq 0$.

Using (2.2), (\ref{3.3}), it is straightforward to verify
\begin{align}\label{l3.1.1}
	r_{\theta,\boldsymbol y}^{m:n+1}(x_{m:n+1} )
	=
	p_{\theta}(x_{n+1}|x_{n} ) q_{\theta}(y_{m}|x_{m} )
	\tilde{r}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
\end{align}
for $x_{m},\dots,x_{n+1}\in{\cal X}$.
Therefore, (4.2), (\ref{3.7}) imply
\begin{align}\label{l3.1.3}
	R_{\theta,\boldsymbol y}^{m:n+1}(B|x)
	=&
	\begin{aligned}[t]
	&\int\cdots\int\int
	I_{B}(x_{n+1} ) p_{\theta}(x_{n+1}|x_{n} ) q_{\theta}(y_{m}|x_{m} )
	\tilde{r}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\\
	&\cdot
	\mu(dx_{n+1} )\cdots\mu(dx_{m+1})\delta_{x}(dx_{m} )
	\end{aligned}
	\\ \nonumber
	=&
	\int\int I_{B}(x'') p_{\theta}(x''|x')q_{\theta}(y_{m}|x')
	\mu(dx'')\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x).
\end{align}
Then, using (\ref{3.27}), we conclude
\begin{align}\label{l3.1.5}
	\frac{\int R_{\theta,\boldsymbol y}^{m:n+1}(B|x)\xi(dx) }
	{\int q_{\theta}(y_{m}|x)\xi(dx) }
	=&
	\frac{\int\int\int I_{B}(x'') p_{\theta}(x''|x') q_{\theta}(y_{m}|x) \mu(dx'')
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x) \xi(dx) }
	{\int q_{\theta}(y_{m}|x)\xi(dx) }
	\\ \nonumber
	=&
	\int\int\int I_{B}(x'') p_{\theta}(x''|x') \mu(dx'')
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi).
\end{align}
Thus, we have
\begin{align}\label{l3.1.5'}
	\frac{\int R_{\theta,\boldsymbol y}^{m:n+1}({\cal X}|x)\xi(dx) }
	{\int q_{\theta}(y_{m}|x)\xi(dx) }
	=
	\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}({\cal X}|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi).
\end{align}
Combining this with (4.4), (\ref{3.21}), we get
\begin{align*}
	F_{\theta,\boldsymbol y}^{m:n+1}(B|\xi)
	=&
	\frac{\int\int\int I_{B}(x'') p_{\theta}(x''|x') \mu(dx'')
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi)}
	{\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}({\cal X}|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi)}
	\\
	=&
	\int\int I_{B}(x'') p_{\theta}(x''|x') \mu(dx'')
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}
	\big(dx'|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi) \big).
\end{align*}

(ii) Let $\theta$, $x$, $B$, $\xi$, $\boldsymbol y$, $n$, $m$ have the same meaning as in (i).
Moreover, let $\zeta$ be any element of
${\cal M}_{s}^{d}({\cal X} )$.

Differentiating (\ref{l3.1.1}) in $\theta$ and using (4.1), (\ref{3.5}),
we get
\begin{align*}
	s_{\theta,\boldsymbol y}^{m:n+1}(x_{m:n+1} )
	=&
	\nabla_{\theta}\big( p_{\theta}(x_{n+1}|x_{n} )q_{\theta}(y_{m}|x_{m} ) \big)
	\tilde{r}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\\
	&+
	p_{\theta}(x_{n+1}|x_{n} )q_{\theta}(y_{m}|x_{m} )
	\tilde{s}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
\end{align*}
for $x_{m},\dots,x_{n+1}\in{\cal X}$.
Therefore, (4.3), (\ref{3.7}), (\ref{3.9}) imply
\begin{align*}
	S_{\theta,\boldsymbol y}^{m:n+1}(B|x)
	=&
	\begin{aligned}[t]
	&\int\cdots\int\int
	I_{B}(x_{n+1} ) \nabla_{\theta}\big( p_{\theta}(x_{n+1}|x_{n} )q_{\theta}(y_{m}|x_{m} ) \big)
	\tilde{r}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\\
	&\cdot
	\mu(dx_{n+1} )\cdots\mu(dx_{m+1})\delta_{x}(dx_{m} )
	\end{aligned}
	\\
	&+
	\begin{aligned}[t]
	&\int\cdots\int\int
	I_{B}(x_{n+1} ) p_{\theta}(x_{n+1}|x_{n} )q_{\theta}(y_{m}|x_{m} )
	\tilde{s}_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\\
	&\cdot
	\mu(dx_{n+1} )\cdots\mu(dx_{m+1})\delta_{x}(dx_{m} )
	\end{aligned}
	\\
	=&
	\int\int I_{B}(x'') \nabla_{\theta}\big( p_{\theta}(x''|x')q_{\theta}(y_{m}|x') \big)
	\mu(dx'') \tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\\
	&+
	\int\int I_{B}(x'') p_{\theta}(x''|x')q_{\theta}(y_{m}|x')
	\mu(dx'') \tilde{S}_{\theta,\boldsymbol y}^{m:n}(dx'|x).
\end{align*}
Then, using (\ref{3.27}), (\ref{3.29}), (\ref{l3.1.3}),
we conclude
\begin{align*}
	&
	\frac{\int R_{\theta,\boldsymbol y}^{m:n+1}(B|x)\zeta(dx)
	+ \int S_{\theta,\boldsymbol y}^{m:n+1}(B|x)\xi(dx) }
	{\int q_{\theta}(y_{m}|x) \xi(dx) }
	\\
	&
	=
	\begin{aligned}[t]
	&
	\frac{\int\int\int I_{B}(x'')
	\nabla_{\theta}\big( p_{\theta}(x''|x') q_{\theta}(y_{m}|x) \big)
	\mu(dx'')\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x) \xi(dx) }
	{\int q_{\theta}(y_{m}|x)\xi(dx) }
	\\
	&+
	\frac{\int\int\int I_{B}(x'')
	p_{\theta}(x''|x') q_{\theta}(y_{m}|x)
	\mu(dx'')
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)\zeta(dx) }
	{\int q_{\theta}(y_{m}|x)\xi(dx) }
	\\
	&+
	\frac{\int\int\int I_{B}(x'')
	p_{\theta}(x''|x') q_{\theta}(y_{m}|x)
	\mu(dx'')
	\tilde{S}_{\theta,\boldsymbol y}^{m:n}(dx'|x)\xi(dx) }
	{\int q_{\theta}(y_{m}|x)\xi(dx) }
	\end{aligned}
	\\
	&
	\begin{aligned}[t]
	=&
	\int\int\int I_{B}(x'')
	p_{\theta}(x''|x') \mu(dx'')
	\tilde{S}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi)
	\\
	&+
	\int\int\int I_{B}(x'')
	\nabla_{\theta} p_{\theta}(x''|x') \mu(dx'')
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi)
	\\
	&+
	\int\int\int I_{B}(x'')
	p_{\theta}(x''|x') \mu(dx'')
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(dx|\xi,\zeta).
	\end{aligned}
\end{align*}
Combining this with (4.5), (\ref{3.21}), (\ref{3.23}), (\ref{l3.1.5'}),
we get
\begin{align}\label{l3.1.21}
	H_{\theta,\boldsymbol y}^{m:n+1}(B|\xi,\zeta)
	=&
	\frac{\int\int\int I_{B}(x'')
	\nabla_{\theta} p_{\theta}(x''|x') \mu(dx'')
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi) }
	{\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}({\cal X}|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi) }
	\\ \nonumber
	&+
	\frac{\int\int\int I_{B}(x'')
	p_{\theta}(x''|x') \mu(dx'')
	\tilde{R}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(dx|\xi,\zeta) }
	{\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}({\cal X}|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi) }
	\\ \nonumber
	&+
	\frac{\int\int\int I_{B}(x'')
	p_{\theta}(x''|x') \mu(dx'')
	\tilde{S}_{\theta,\boldsymbol y}^{m:n}(dx'|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi) }
	{\int \tilde{R}_{\theta,\boldsymbol y}^{m:n}({\cal X}|x)
	\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(dx|\xi) }
	\\ \nonumber
	=&
	\int\int I_{B}(x') \nabla_{\theta} p_{\theta}(x'|x) \mu(dx')
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}
	\big(dx|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi) \big)
	\\ \nonumber
	&+
	\int\int I_{B}(x') p_{\theta}(x'|x) \mu(dx')
	\tilde{H}_{\theta,\boldsymbol y}^{m:n}
	\big(dx|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi,\zeta) \big).
\end{align}
Since $\int \nabla_{\theta} p_{\theta}(x'|x) \mu(dx') = 0$,
we then have
\begin{align}\label{l3.1.23}
	H_{\theta,\boldsymbol y}^{m:n+1}({\cal X}|\xi,\zeta)
	=
	\tilde{H}_{\theta,\boldsymbol y}^{m:n}
	\big({\cal X}|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi,\zeta) \big).
\end{align}

Using (4.6), (\ref{3.25}), (\ref{l3.1.21}), (\ref{l3.1.23}), we conclude
\begin{align*}
	G_{\theta,\boldsymbol y}^{m:n+1}(B|\xi,\zeta)
	=&
	H_{\theta,\boldsymbol y}^{m:n+1}(B|\xi,\zeta)
	-
	F_{\theta,\boldsymbol y}^{m:n+1}(B|\xi)
	H_{\theta,\boldsymbol y}^{m:n+1}({\cal X}|\xi,\zeta)
	\\
	=&
	\int\int I_{B}(x') \nabla_{\theta} p_{\theta}(x'|x) \mu(dx')
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}
	\big(dx|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi) \big)
	\\
	&
	\begin{aligned}[t]
	+&
	\int\int I_{B}(x') p_{\theta}(x'|x) \mu(dx')
	\tilde{H}_{\theta,\boldsymbol y}^{m:n}
	\big(dx|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi,\zeta) \big)
	\\
	-&
	\tilde{H}_{\theta,\boldsymbol y}^{m:n}
	\big({\cal X}|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi,\zeta) \big)
	\\
	\cdot&
	\int\int I_{B}(x') p_{\theta}(x'|x) \mu(dx')
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}
	\big(dx|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi) \big)
	\end{aligned}
	\\
	=&
	\int\int I_{B}(x') \nabla_{\theta} p_{\theta}(x'|x) \mu(dx')
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}
	\big(dx|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi) \big)
	\\
	&+
	\int\int I_{B}(x') p_{\theta}(x'|x) \mu(dx')
	\tilde{G}_{\theta,\boldsymbol y}^{m:n}
	\big(dx|\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi,\zeta) \big).
\end{align*}
\end{proof}

\begin{proof}[Proof of Proposition 4.1]
(i) Using \cite[Theorem 4.1]{{legland&oudjane}} (or \cite[Theorem 3.1]{tadic&doucet1}), we conclude that
there exist real numbers $\rho_{1}\in(0,1)$, $\tilde{C}_{1}\in[1,\infty)$
(depending only on $\varepsilon$) such that
\begin{align*}
	\left\| \tilde{F}_{\theta,\boldsymbol y}^{m:n}(\xi')
	-
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}(\xi'') \right\|
	\leq
	\tilde{C}_{1}\rho_{1}^{n-m}\|\xi'-\xi''\|
\end{align*}
for all $\theta\in\Theta$, $\xi',\xi''\in{\cal P}({\cal X})$,
$n\geq m\geq 0$ and any sequence $\boldsymbol y = \{y_{n} \}_{n\geq 0}$ in ${\cal Y}$.
As $\|\xi'-\xi''\|\leq 1$, we have
\begin{align}\label{l3.2.501}
	\left\| \tilde{F}_{\theta,\boldsymbol y}^{m:n}(\xi')
	-
	\tilde{F}_{\theta,\boldsymbol y}^{m:n}(\xi'') \right\|
	\leq
	\tilde{C}_{1}\rho_{1}^{n-m}.
\end{align}
Throughout this part of the proof, the following notation is used.
$\theta$ is any element of $\Theta$, while $x$, $x'$ are any elements of ${\cal X}$.
$B$ is any element of ${\cal B}({\cal X} )$, while
$\xi$, $\xi'$, $\xi''$ are any elements of ${\cal P}({\cal X} )$.
$\boldsymbol y = \{y_{n} \}_{n\geq 0}$ is any sequence in ${\cal Y}$.
$C_{1}$ is the real number defined by
$C_{1}=\tilde{C}_{1}\varepsilon^{-4}\rho_{1}^{-1}$
($\varepsilon$ is specified in Assumption 2.1).

When, $m=n$, the left-hand sides of inequalities (4.8) reduce to
$\|\xi'-\xi''\|$ and $1$ (respectively).
Hence, it is sufficient to show (4.8) for $n>m\geq 0$.
In what follows in this part of the proof,
$n$, $m$ are any integers satisfying $n>m\geq 0$.

Relying on (\ref{l3.2.501}), we conclude
\begin{align}\label{l3.2.3}
	\left\|
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}
	\big(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi') \big)
	-
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}
	\big(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi'') \big)
	\right\|
	\leq &
	\tilde{C}_{1}\rho_{1}^{n-m-1}
	\leq
	C_{1}\rho_{1}^{n-m}.
\end{align}
Then, using Lemma \ref{lemma3.1}, we deduce
\begin{align*}
	&
	\left|
	F_{\theta,\boldsymbol y}^{m:n}(B|\xi')
	-
	F_{\theta,\boldsymbol y}^{m:n}(B|\xi'')
	\right|
	\\
	&\leq
	\int\int I_{B}(x') p_{\theta}(x'|x) \mu(dx')
	\left|
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}
	(dx| \tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi') )
	-
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}
	(dx| \tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi'') )
	\right|
	\\ \nonumber
	&\leq
	\left\|
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}
	(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi') )
	-
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}
	(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi'') )
	\right\|
	\leq
	C_{1}\rho_{1}^{n-m}.
\end{align*}
Thus, the first part of (4.8) holds.

Owing to Assumption 2.1 and (2.1), we have
\begin{align}\label{l3.2.7}
	\varepsilon^{2} \leq r_{\theta,\boldsymbol y}^{n}(x'|x) \leq \frac{1}{\varepsilon^{2} }.
\end{align}
for $n\geq 1$.
Moreover, due to (2.2), we have
\begin{align*}
	r_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	=
	r_{\theta,\boldsymbol y}^{m+1:n}(x_{m+1:n} ) r_{\theta,\boldsymbol y}^{m+1}(x_{m+1}|x_{m} ).
\end{align*}
Hence, we get
\begin{align}\label{l3.2.303}
	\varepsilon^{2} r_{\theta,\boldsymbol y}^{m+1:n}(x_{m+1:n} )
	\leq
	r_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\leq
	\frac{1}{\varepsilon^{2} } r_{\theta,\boldsymbol y}^{m+1:n}(x_{m+1:n} ).
\end{align}
Consequently, (4.2) implies
\begin{align}
	&\label{l3.2.307}
	\begin{aligned}
	\int R_{\theta,\boldsymbol y}^{m:n}({\cal X}|x)\xi(dx)
	\geq &
	\varepsilon^{2}
	\int\cdots\int\int r_{\theta,\boldsymbol y}^{m+1:n}(x_{m:n} )
	\mu(dx_{n} )\cdots\mu(dx_{m+1} )\xi(dx_{m} )
	\\
	=&
	\varepsilon^{2}
	\int R_{\theta,\boldsymbol y}^{m+1:n}({\cal X}|x)\mu(dx),
	\end{aligned}
	\\ \nonumber
	&
	\begin{aligned}
	\int R_{\theta,\boldsymbol y}^{m:n}({\cal X}|x)\xi(dx)
	\leq &
	\frac{1}{\varepsilon^{2} }
	\int\cdots\int\int r_{\theta,\boldsymbol y}^{m+1:n}(x_{m:n} )
	\mu(dx_{n} )\cdots\mu(dx_{m+1} )\xi(dx_{m} )
	\\
	=&
	\frac{1}{\varepsilon^{2} }
	\int R_{\theta,\boldsymbol y}^{m+1:n}({\cal X}|x)\mu(dx).
	\end{aligned}
\end{align}
Therefore, we have
\begin{align*}
	\frac{\int R_{\theta,\boldsymbol y}^{m:n}({\cal X}|x) \xi'(dx) }
	{\int R_{\theta,\boldsymbol y}^{m:n}({\cal X}|x) \xi''(dx) }
	\leq
	\frac{1}{\varepsilon^{4}}
	\leq
	C_{1}.
\end{align*}

(ii) Relying on \cite[Theorem 3.2]{tadic&doucet1}
(or \cite[Theorem 2.2]{tadic&doucet3}), we deduce that
there exist real numbers $\rho_{2}\in[\rho_{1},1)$, $\tilde{C}_{2}\in[C_{1},\infty)$
(depending only on $\varepsilon$, $d$, $K$) such that
\begin{align*}
	\left\| \tilde{G}_{\theta,\boldsymbol y}^{m:n}(\xi',\zeta')
	-
	\tilde{G}_{\theta,\boldsymbol y}^{m:n}(\xi'',\zeta'') \right\|
	\leq &
	\tilde{C}_{2}\rho_{2}^{n-m}\|\xi'-\xi''\|(1 + \|\zeta'\| + \|\zeta''\| )
	\\
	&+
	\tilde{C}_{2}\rho_{2}^{n-m}\|\zeta'-\zeta''\|
\end{align*}
for all $\theta\in\Theta$, $\xi',\xi''\in{\cal P}({\cal X})$,
$\zeta',\zeta''\in{\cal M}_{s}^{d}({\cal X})$,
$n\geq m\geq 0$ and any sequence $\boldsymbol y = \{y_{n} \}_{n\geq 0}$ in ${\cal Y}$.
Since $\|\xi'-\xi''\|\leq 1$ and $\|\zeta'-\zeta''\|\leq\|\zeta'\|+\|\zeta''\|$,
we have
\begin{align}\label{l3.2.503}
	\left\| \tilde{G}_{\theta,\boldsymbol y}^{m:n}(\xi',\zeta')
	-
	\tilde{G}_{\theta,\boldsymbol y}^{m:n}(\xi'',\zeta'') \right\|
	\leq &
	2\tilde{C}_{2}\rho_{2}^{n-m}(1 + \|\zeta'\| + \|\zeta''\| ).
\end{align}
Throughout the rest of the proof, the following notation is used.
$\theta$, $x$, $x'$, $\xi$, $\xi'$, $\xi''$,
$\boldsymbol y = \{y_{n} \}_{n\geq 0}$ have the same meaning as in (i).
$\zeta$, $\zeta'$, $\zeta''$ are any elements of
${\cal M}_{s}^{d}({\cal X} )$.
$\tilde{C}_{3}$, $\tilde{C}_{4}$, $\tilde{C}_{5}$, $C_{2}$ are the real numbers defined by
$\tilde{C}_{3}=2K\varepsilon^{-2}$,
$\tilde{C}_{4}=4d\tilde{C}_{2}\tilde{C}_{3}\rho_{2}^{-1}$,
$\tilde{C}_{5}=C_{1}\tilde{C}_{3}\tilde{C}_{4}$,
$C_{2}=d\tilde{C}_{3}\tilde{C}_{5}$
($\varepsilon$, $K$ are specified in Assumptions 2.1, 2.2).

When $m=n$, the left-hand sides of (4.9), (4.10)
reduce to $\|\zeta'-\zeta''\|$ and $\|\zeta\|$
(respectively).
Hence, it is sufficient to show (4.9), (4.10) for $n>m\geq 0$.
In the rest of the proof, $n$, $m$ are any integers satisfying
$n>m\geq 0$.

Owing to Assumptions 2.1, 2.2, we have
\begin{align}\label{l3.2.301}
	\int \|\nabla_{\theta} p_{\theta}(x'|x) \| \mu(dx')
	=
	\int \left\|\frac{\nabla_{\theta} p_{\theta}(x'|x) }{p_{\theta}(x'|x) } \right\|
	p_{\theta}(x'|x) \mu(dx')
	\leq
	\frac{K}{\varepsilon}
	\leq
	\tilde{C}_{3}.
\end{align}
Due to the same assumptions and (\ref{3.29}), we have
\begin{align*}
	\left\|
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(B|\xi,\zeta)
	\right\|
	\leq &
	\frac{\int_{B} q_{\theta}(y_{m}|x) |\zeta|(dx)
	+ \int_{B} \|\nabla_{\theta} q_{\theta}(y_{m}|x)\| \xi(dx) }
	{\int q_{\theta}(y_{m}|x) \xi(dx) }
	\\
	\leq &
	\frac{K}{\varepsilon}
	+
	\frac{\|\zeta\| }{\varepsilon^{2} }
	\leq
	\tilde{C}_{3} (1 + \|\zeta\| ).
\end{align*}
Hence, we get
\begin{align*}
	\left\| \tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi,\zeta) \right\|
	\leq
	d\tilde{C}_{3}(1 + \|\zeta\| ).
\end{align*}
Consequently, (\ref{l3.2.503}) implies
\begin{align}\label{l3.2.23}
	&
	\left\|
	\tilde{G}_{\theta,\boldsymbol y}^{m:n-1}
	\big(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi'),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi',\zeta') \big)
	-
	\tilde{G}_{\theta,\boldsymbol y}^{m:n-1}
	\big(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi''),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi'',\zeta'') \big)
	\right\|
	\\ \nonumber
	&
	\begin{aligned}[b]
	\leq &
	\tilde{C}_{2}\rho_{2}^{n-m-1}
	\left(
	1
	+
	\big\|\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi',\zeta') \big\|
	+
	\big\|\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi'',\zeta'') \big\|
	\right)
	\end{aligned}
	\\ \nonumber
	&\leq
	4d\tilde{C}_{2}\tilde{C}_{3}\rho_{2}^{n-m-1}
	(1 + \|\zeta'\| + \|\zeta''\| )
	\leq
	\tilde{C}_{4} \rho_{2}^{n-m}
	(1 + \|\zeta'\| + \|\zeta''\| ).
\end{align}
Then, Lemma \ref{lemma3.1} and (\ref{l3.2.3}), (\ref{l3.2.301}) yield
\begin{align*}
	&
	\left\|
	G_{\theta,\boldsymbol y}^{m:n}(B|\xi',\zeta')
	-
	G_{\theta,\boldsymbol y}^{m:n}(B|\xi'',\zeta'')
	\right\|
	\\ \nonumber
	&
	\begin{aligned}[b]
	\leq &
	\int\int I_{B}(x') \|\nabla_{\theta} p_{\theta}(x'|x) \|
	\mu(dx')
	\left|
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}
	(dx| \tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi') )
	-
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}(dx| \tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi'') )
	\right|
	\\ \nonumber
	&\begin{aligned}[t]
	+
	\int\int I_{B}(x') p_{\theta}(x'|x) \mu(dx')
	\Big|
	&
	\tilde{G}_{\theta,\boldsymbol y}^{m:n-1}
	(dx| \tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi',\zeta') )
	\\
	&-
	\tilde{G}_{\theta,\boldsymbol y}^{m:n-1}
	(dx| \tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi'',\zeta'') )
	\Big|
	\end{aligned}
	\end{aligned}
	\\ \nonumber
	&\begin{aligned}[t]
	\leq &
	\tilde{C}_{3}
	\left\|
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}
	(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi') )
	-
	\tilde{F}_{\theta,\boldsymbol y}^{m:n-1}(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi'') )
	\right\|
	\\
	&+
	\left\|
	\tilde{G}_{\theta,\boldsymbol y}^{m:n-1}
	(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi',\zeta') )
	-
	\tilde{G}_{\theta,\boldsymbol y}^{m:n-1}
	(\tilde{\alpha}_{\theta,\boldsymbol y}^{m}(\xi),
	\tilde{\beta}_{\theta,\boldsymbol y}^{m}(\xi'',\zeta'') )
	\right\|
	\end{aligned}
	\\
	&\leq
	C_{1}\tilde{C}_{3}\rho_{1}^{n-m}
	+
	\tilde{C}_{4}\rho_{2}^{n-m}(1 + \|\zeta'\| + \|\zeta'' \| )
	\leq
	\tilde{C}_{5}\rho_{2}^{n-m}(1 + \|\zeta'\| + \|\zeta'' \| ).
\end{align*}
Thus, we get
\begin{align*}
	\left\|
	G_{\theta,\boldsymbol y}^{m:n}(\xi',\zeta')
	-
	G_{\theta,\boldsymbol y}^{m:n}(\xi'',\zeta'')
	\right\|
	\leq &
	d\tilde{C}_{5}(1 + \|\zeta'\| + \|\zeta'' \| )
	\leq
	C_{2}\rho_{2}^{n-m}(1 + \|\zeta'\| + \|\zeta'' \| ).
\end{align*}

Owing to (2.2), (4.1), we have
\begin{align*}
	s_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	=
	r_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\left(
	\sum_{k=m+1}^{n}
	\frac{\nabla_{\theta} r_{\theta,\boldsymbol y}^{k}(x_{k}|x_{k-1}) }
	{r_{\theta,\boldsymbol y}^{k}(x_{k}|x_{k-1}) }
	\right)
\end{align*}
for $x_{m},\dots,x_{n}\in{\cal X}$.
Moreover, due to Assumptions 2.1, 2.2, we have
\begin{align}\label{l3.2.25}
	\left\|
	\frac{\nabla_{\theta} r_{\theta,\boldsymbol y}^{n}(x'|x) }
	{r_{\theta,\boldsymbol y}^{n}(x'|x) }
	\right\|
	\leq
	\left\|
	\frac{\nabla_{\theta} p_{\theta}(x'|x) }{p_{\theta}(x'|x) }
	\right\|
	+
	\left\|
	\frac{\nabla_{\theta} q_{\theta}(y_{n-1}|x) }{q_{\theta}(y_{n-1}|x) }
	\right\|
	\leq
	\frac{2K}{\varepsilon}
	\leq
	\tilde{C}_{3}.
\end{align}
Therefore, we get
\begin{align*}
	\left\| s_{\theta,\boldsymbol y}^{m:n}(x_{m:n} ) \right\|
	\leq &
	r_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\left(
	\sum_{k=m+1}^{n}
	\left\|
	\frac{\nabla_{\theta} r_{\theta,\boldsymbol y}^{k}(x_{k}|x_{k-1}) }
	{r_{\theta,\boldsymbol y}^{k}(x_{k}|x_{k-1}) }
	\right\|
	\right)
	\\
	\leq &
	\tilde{C}_{3}(n-m)r_{\theta,\boldsymbol y}^{m:n}(x_{m:n} ).
\end{align*}
Consequently, (4.3), (\ref{l3.2.303}) imply
\begin{align}\label{l3.2.27}
	\left\|
	\int S_{\theta,\boldsymbol y}^{m:n}(B|x) \xi(dx)
	\right\|
	\leq &
	\begin{aligned}[t]
	&\int\!\cdots\!\int\!\!\int
	I_{B}(x_{n} ) \left\| s_{\theta,\boldsymbol y}^{m:n}(x_{m:n} ) \right\|
	\mu(dx_{n} )\cdots\mu(dx_{m+1} )\xi(dx_{m} )
	\end{aligned}
	\\ \nonumber
	\leq &
	\frac{\tilde{C}_{3}(n\!-\!m) }{\varepsilon^{2} }\!
	\begin{aligned}[t]
	&\int\!\cdots\!\int\!\!\int\!
	r_{\theta,\boldsymbol y}^{m+1:n}(x_{m:n} )
	\mu(dx_{n} )\cdots\mu(dx_{m+1} )
	\end{aligned}
	\\ \nonumber
	= &
	\frac{\tilde{C}_{3}(n\!-\!m) }{\varepsilon^{2} }
	\int R_{\theta,\boldsymbol y}^{m:n}({\cal X}|x) \mu(dx).
\end{align}
Similarly, (4.2), (\ref{l3.2.303}) imply
\begin{align}\label{l3.2.29}
	\left\|
	\int R_{\theta,\boldsymbol y}^{m:n}(B|x) \zeta(dx)
	\right\|
	\leq &
	\begin{aligned}[t]
	&\int\!\cdots\!\int\!\!\int
	I_{B}(x_{n} ) r_{\theta,\boldsymbol y}^{m:n}(x_{m:n} )
	\mu(dx_{n} )\cdots\mu(dx_{m+1} )|\zeta|(dx_{m} )
	\end{aligned}
	\\ \nonumber
	\leq &
	\frac{\|\zeta\| }{\varepsilon^{2} }\!
	\begin{aligned}[t]
	&\int\!\cdots\!\int\!\!\int\!
	r_{\theta,\boldsymbol y}^{m+1:n}(x_{m:n} )
	\mu(dx_{n} )\cdots\mu(dx_{m+1} )
	\end{aligned}
	\\ \nonumber
	= &
	\frac{\|\zeta\|}{\varepsilon^{2} }
	\int R_{\theta,\boldsymbol y}^{m:n}({\cal X}|x) \mu(dx).
\end{align}
Combining (4.5), (\ref{l3.2.307}), (\ref{l3.2.27}), (\ref{l3.2.29}), we get
\begin{align*}
	\left\| H_{\theta,\boldsymbol y}^{m:n}(B|\xi,\zeta)\right\|
	\leq
	\frac{\left\|
	\int R_{\theta,\boldsymbol y}^{m:n}(B|x) \zeta(dx)
	\right\|
	+
	\left\|
	\int S_{\theta,\boldsymbol y}^{m:n}(B|x) \xi(dx)
	\right\|}
	{\int R_{\theta,\boldsymbol y}^{m:n}({\cal X}|x) \xi(dx) }
	\leq
	\tilde{C}_{3} (n-m+\|\zeta\| ).
\end{align*}
Hence, we have
\begin{align*}
	\left\| H_{\theta,\boldsymbol y}^{m:n}(\xi,\zeta)\right\|
	\leq
	d\tilde{C}_{3}(n-m+\|\zeta\| )
	\leq
	C_{2} (n-m+\|\zeta\| ).
\end{align*}
\end{proof}

\section{\nopunct}

In this section, we prove Lemma 6.1.
The proof is mainly based on the same arguments as Lemma \ref{lemma3.1}.

\begin{proof}[Proof of Lemma 6.1]
(i) Throughout this part of the proof, the following notation is used.
$\theta$ is any element of $\Theta$,
while $B$ is any element of ${\cal B}({\cal X} )$.
When $m=0$, (6.13) is trivially satisfied.
Hence, it is sufficient to show (6.13) for $n\geq m>0$.
In this part of the proof, $n$, $m$ are any integers satisfying $n\geq m>0$.

Owing to (2.2), we have
\begin{align}\label{l5.3.303}
	r_{\theta,\boldsymbol Y}^{m-1:n}(x_{m-1:n} )
	=
	r_{\theta,\boldsymbol Y}^{m:n}(x_{m:n} ) r_{\theta,\boldsymbol Y}^{m-1:m}(x_{m-1:m} )
\end{align}
for $x_{m-1},\dots,x_{n}\in{\cal X}$.
Consequently, (4.2) implies
\begin{align*}
	\int R_{\theta,\boldsymbol Y}^{m-1:n}(B|x)\hat{\xi}_{m-1}^{\theta}(dx)
	=&
	\begin{aligned}[t]
	&\int\cdots\int\int
	I_{B}(x_{n}) r_{\theta,\boldsymbol Y}^{m:n}(x_{m:n} ) r_{\theta,\boldsymbol Y}^{m-1:m}(x_{m-1:m} )
	\\
	&\cdot\mu(dx_{n})\cdots\mu(dx_{m})\hat{\xi}_{m-1}^{\theta}(dx_{m-1})
	\end{aligned}
	\\
	=&
	\int\int R_{\theta,\boldsymbol Y}^{m:n}(B|x') R_{\theta,\boldsymbol Y}^{m-1:m}(dx'|x)
	\hat{\xi}_{m-1}^{\theta}(dx).
\end{align*}
Combining this with (4.4), (6.5), we get
\begin{align}\label{l5.3.1}
	\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx)
	=&
	\frac{\int\int R_{\theta,\boldsymbol Y}^{m:n}(B|x')
	R_{\theta,\boldsymbol Y}^{m-1:m}(dx'|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	\\ \nonumber
	=&
	\frac{\int R_{\theta,\boldsymbol Y}^{m-1:n}(B|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }.
\end{align}
Thus, we have
\begin{align}\label{l5.3.1'}
	\int R_{\theta,\boldsymbol Y}^{m:n}({\cal X}|x) \hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx)
	=
	\frac{\int R_{\theta,\boldsymbol Y}^{m-1:n}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }.
\end{align}
Then, (4.4), (6.5) yield
\begin{align*}
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:n}(B)
	=
	\frac{\int R_{\theta,\boldsymbol Y}^{m-1:n}(B|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:n}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	=
	\frac{\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m:n}({\cal X}|x) \hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx)}.
\end{align*}
Hence, (6.13) holds.

(ii) Let $\theta$, $B$ have the same meaning as in (i).
Moreover, let $\xi$ be any element of ${\cal P}({\cal X} )$,
while $n$, $m$ are any integers satisfying $n\geq m\geq 0$.

Owing to (6.4), (6.8), we have
\begin{align}\label{l5.3.301}
	\int \hat{\Psi}_{\theta,\boldsymbol Y}^{m:n}(B|x) \xi(dx)
	=&
	\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{v}_{\theta,\boldsymbol Y}^{m}(x) \xi(dx)
	+
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x) \xi(dx)
	\\ \nonumber
	=&
	\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\alpha}_{\theta,\boldsymbol Y}^{m}(dx|\xi)
	+
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x) \xi(dx).
\end{align}
Combining this with (4.5), (6.10), we get
\begin{align}\label{l5.3.501'}
	\hat{C}_{\theta, \boldsymbol Y}^{m:n}(B|\xi)
	=
	\frac{\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\alpha}_{\theta,\boldsymbol Y}^{m}(dx|\xi)
	+
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x) \xi(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \xi(dx) }
	=
	H_{\theta,\boldsymbol Y}^{m:n}(B|\xi,\hat{\alpha}_{\theta,\boldsymbol Y}^{m}(\xi) ).
\end{align}
Consequently, (4.6), (6.12) imply
\begin{align*}
	\hat{A}_{\theta,\boldsymbol Y}^{m:n}(B|\xi)
	=&
	H_{\theta,\boldsymbol Y}^{m:n}(B|\xi,\hat{\alpha}_{\theta,\boldsymbol Y}^{m}(\xi) )
	-
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:n}(B)
	H_{\theta, \boldsymbol Y}^{m:n}({\cal X}|\xi,\hat{\alpha}_{\theta,\boldsymbol Y}^{m}(\xi) )
	\\
	=&
	\big(
	F_{\theta,\boldsymbol Y}^{m:n}(B|\xi)
	-
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:n}(B)
	\big)
	H_{\theta, \boldsymbol Y}^{m:n}({\cal X}|\xi,\hat{\alpha}_{\theta,\boldsymbol Y}^{m}(\xi) )
	\\
	&+
	G_{\theta,\boldsymbol Y}^{m:n}(B|\xi,\hat{\alpha}_{\theta,\boldsymbol Y}^{m}(\xi) ).
\end{align*}
Hence, (6.14), (6.16) hold.

Due to (2.1), (2.6), (2.7), (5.3), (6.3), we have
\begin{align*}
	W_{k,i}^{\theta}
	=&
	\frac{
	\sum_{j=1}^{N}
	r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |\hat{X}_{k-1,j}^{\theta} \big)
	W_{k-1,j}^{\theta}
	+
	\sum_{j=1}^{N}
	\nabla_{\theta}
	r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |\hat{X}_{k-1,j}^{\theta} \big) }
	{\sum_{j=1}^{N}
	r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |\hat{X}_{k-1,j}^{\theta} \big) }
	\\
	=&
	\frac{
	\sum_{j=1}^{N}
	r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |\hat{X}_{k-1,j}^{\theta} \big)
	V_{k-1,j}^{\theta}
	+
	\sum_{j=1}^{N}
	\nabla_{\theta}
	r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |\hat{X}_{k-1,j}^{\theta} \big) }
	{\sum_{j=1}^{N}
	r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |\hat{X}_{k-1,j}^{\theta} \big) }
	+
	\frac{1}{N} \!\sum_{j=1}^{N} W_{k-1,j}^{\theta}
	\\
	=&
	\frac{
	\int r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |x \big)
	\hat{\zeta}_{k-1}^{\theta}(dx)
	+
	\int
	\nabla_{\theta}
	r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |x \big)
	\hat{\xi}_{k-1}^{\theta}(dx) }
	{\int r_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} |x \big)
	\hat{\xi}_{k-1}^{\theta}(dx) }
	+
	\frac{1}{N} \!\sum_{j=1}^{N} W_{k-1,j}^{\theta}
	\\
	=&
	\hat{v}_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} \big)
	+
	\frac{1}{N} \!\sum_{j=1}^{N} W_{k-1,j}^{\theta}
\end{align*}
for $1\leq i\leq N$, $k\geq 1$.
Hence, (2.6), (5.3), (6.4) imply
\begin{align}\label{l5.3.7'}
	V_{k,i}^{\theta}
	=
	\hat{v}_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} \big)
	-
	\frac{1}{N} \sum_{j=1}^{N}
	\hat{v}_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,j}^{\theta} \big)
	=&
	\hat{v}_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} \big)
	-
	\int \hat{v}_{\theta,\boldsymbol Y}^{k}(x)
	\hat{\xi}_{k}^{\theta}(dx)
	\\ \nonumber
	=&
	\hat{v}_{\theta,\boldsymbol Y}^{k}\big(\hat{X}_{k,i}^{\theta} \big)
	-
	\hat{\beta}_{\theta,\boldsymbol Y}^{k}({\cal X}).
\end{align}
Since $W_{0,i}^{\theta}=w_{\theta}(\hat{X}_{0,i}^{\theta} )$,
(5.3), (6.1), (6.3), (6.4) yield
\begin{align}\label{l5.3.9'}
	V_{0,i}^{\theta}
	=
	u_{\theta}\big( \hat{X}_{0,i}^{\theta} \big)
	-
	\frac{1}{N} \sum_{j=1}^{N}
	u_{\theta}\big( \hat{X}_{0,j}^{\theta} \big)
	=&
	\hat{v}_{\theta,\boldsymbol Y}^{0}\big(\hat{X}_{0,i}^{\theta} \big)
	-
	\frac{1}{N} \sum_{j=1}^{N}
	\hat{v}_{\theta,\boldsymbol Y}^{0}\big(\hat{X}_{0,j}^{\theta} \big)
	\\ \nonumber
	=&
	\hat{v}_{\theta,\boldsymbol Y}^{0}\big(\hat{X}_{0,i}^{\theta} \big)
	-
	\int \hat{v}_{\theta,\boldsymbol Y}^{0}(x)
	\hat{\xi}_{0}^{\theta}(dx)
	\\ \nonumber
	=&
	\hat{v}_{\theta,\boldsymbol Y}^{0}\big(\hat{X}_{0,i}^{\theta} \big)
	-
	\hat{\beta}_{\theta,\boldsymbol Y}^{0}({\cal X}).
\end{align}
Using (2.6), (5.3), (6.4), (\ref{l5.3.7'}), (\ref{l5.3.9'}), we conclude
\begin{align}\label{l5.3.11'}
	\hat{\zeta}_{n}^{\theta}(B)
	=
	\frac{1}{N} \sum_{i=1}^{N}
	\big(
	\hat{v}_{\theta,\boldsymbol Y}^{n}\big(\hat{X}_{n,i}^{\theta} \big)
	-
	\hat{\beta}_{\theta,\boldsymbol Y}^{n}({\cal X} )
	\big)
	\delta_{\hat{X}_{n,i}^{\theta} }(B)
	=&
	\!\int_{B} \hat{v}_{\theta,\boldsymbol Y}^{n}(x)
	\hat{\xi}_{n}^{\theta}(dx)
	-
	\hat{\xi}_{n}^{\theta}(B)
	\hat{\beta}_{\theta,\boldsymbol Y}^{n}({\cal X} )
	\\ \nonumber
	=&
	\hat{\beta}_{\theta,\boldsymbol Y}^{n}(B)
	-
	\hat{\xi}_{n}^{\theta}(B)
	\hat{\beta}_{\theta,\boldsymbol Y}^{n}({\cal X} ).
\end{align}
Thus, (6.4), (\ref{l5.3.301}) imply
\begin{align*}
	\int \hat{\Psi}_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\xi}_{m}^{\theta}(dx)
	=&
	\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\alpha}_{\theta,\boldsymbol Y}^{m}(dx|\hat{\xi}_{m})
	+
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\xi}_{m}^{\theta}(dx)
	\\
	=&
	\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\beta}_{\theta,\boldsymbol Y}^{m}(dx)
	+
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\xi}_{m}^{\theta}(dx)
	\\
	=&
	\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\zeta}_{m}^{\theta}(dx)
	+
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\xi}_{m}^{\theta}(dx)
	\\
	&+
	\hat{\beta}_{\theta,\boldsymbol Y}^{m}({\cal X})
	\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\xi}_{m}^{\theta}(dx).
\end{align*}
Combining this with (4.4), (4.5), (6.5), (6.7), we get
\begin{align}\label{l5.3.501}
	\hat{C}_{\theta, \boldsymbol Y}^{m:n}(B|\hat{\xi}_{m}^{\theta} )
	=&
	\frac{\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\zeta}_{m}^{\theta}
	+
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\xi}_{m}^{\theta} }
	{\int R_{\theta,\boldsymbol Y}^{m:n}({\cal X}|x) \hat{\xi}_{m}^{\theta} }
	+
	\hat{\beta}_{\theta,\boldsymbol Y}^{m}({\cal X} )
	\frac{\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{\xi}_{m}^{\theta} }
	{\int R_{\theta,\boldsymbol Y}^{m:n}({\cal X}|x) \hat{\xi}_{m}^{\theta} }
	\\ \nonumber
	=&
	\hat{H}_{\theta,\boldsymbol Y}^{m:n}(B)
	+
	\hat{F}_{\theta,\boldsymbol Y}^{m:n}(B) \hat{\beta}_{\theta,\boldsymbol Y}^{m}({\cal X} ).
\end{align}
As $\hat{F}_{\theta,\boldsymbol Y}^{m:n}({\cal X} )=1$,
(4.6), (6.5), (6.6), (6.11), (6.12), (\ref{l5.3.501})
yield
\begin{align*}
	\hat{A}_{\theta,\boldsymbol Y}^{m:n}(B|\hat{\xi}_{m}^{\theta} )
	+
	\hat{B}_{\theta,\boldsymbol Y}^{m:n}(B|\hat{\xi}_{m}^{\theta} )
	=&
	\hat{C}_{\theta,\boldsymbol Y}^{m:n}(B|\hat{\xi}_{m}^{\theta} )
	-
	\hat{F}_{\theta,\boldsymbol Y}^{m:n}(B)
	\hat{C}_{\theta,\boldsymbol Y}^{m:n}({\cal X}|\hat{\xi}_{m}^{\theta} )
	\\
	=&
	\hat{H}_{\theta,\boldsymbol Y}^{m:n}(B)
	-
	\hat{F}_{\theta,\boldsymbol Y}^{m:n}(B)
	\hat{H}_{\theta,\boldsymbol Y}^{m:n}({\cal X})
	\\
	=&
	\hat{G}_{\theta,\boldsymbol Y}^{m:n}(B).
\end{align*}
Hence, (6.15) holds.

Owing to (6.1) -- (6.3), we have
$\hat{\zeta}_{-1}^{\theta}(dx)=\hat{v}_{\theta,\boldsymbol Y}^{0}(x) \hat{\xi}_{-1}^{\theta}(dx)$.
Since $\hat{F}_{\theta,\boldsymbol Y}^{-1:0}(dx)=\hat{\xi}_{-1}^{\theta}(dx)$,
(4.5), (6.7), (6.8) imply
\begin{align*}
	\hat{H}_{\theta,\boldsymbol Y}^{-1:n}(B)
	=
	H_{\theta,\boldsymbol Y}^{0:n}
	\big(B|\hat{\xi}_{-1}^{\theta}, \hat{\zeta}_{-1}^{\theta} \big)
	=&
	\frac{\int R_{\theta,\boldsymbol Y}^{0:n}(B|x) \hat{\zeta}_{-1}^{\theta}(dx)
	+
	\int S_{\theta,\boldsymbol Y}^{0:n}(B|x) \hat{\xi}_{-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{0:n}({\cal X}|x) \hat{\xi}_{-1}^{\theta}(dx) }
	\\ \nonumber
	=&
	\frac{\int \big(
	R_{\theta,\boldsymbol Y}^{0:n}(B|x) \hat{v}_{\theta,\boldsymbol Y}^{0}(x) + S_{\theta,\boldsymbol Y}^{0:n}(B|x)
	\big)
	\hat{\xi}_{-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{0:n}({\cal X}|x) \hat{\xi}_{-1}^{\theta}(dx) }
	\\ \nonumber
	=&
	\frac{\int
	\hat{\Psi}_{\theta,\boldsymbol Y}^{0:n}(B|x)
	\hat{F}_{\theta,\boldsymbol Y}^{-1:0}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{0:n}({\cal X}|x) \hat{F}_{\theta,\boldsymbol Y}^{-1:0}(dx) }.
\end{align*}
Consequently, (4.6), (6.6), (6.9) imply
\begin{align*}
	\hat{G}_{\theta,\boldsymbol Y}^{-1:n}(B)
	=&
	H_{\theta,\boldsymbol Y}^{0:n}(B|\hat{\xi}_{-1}^{\theta}, \hat{\zeta}_{-1}^{\theta} )
	-
	F_{\theta,\boldsymbol Y}^{0:n}(B|\hat{\xi}_{-1}^{\theta} )
	H_{\theta,\boldsymbol Y}^{0:n}({\cal X}|\hat{\xi}_{-1}^{\theta}, \hat{\zeta}_{-1}^{\theta} )
	\\ \nonumber
	=&
	\frac{\int
	\big(
	\hat{\Psi}_{\theta,\boldsymbol Y}^{0:n}(B)
	-
	\hat{F}_{\theta,\boldsymbol Y}^{-1:n}(B)
	\hat{\Psi}_{\theta,\boldsymbol Y}^{0:n}({\cal X}|x)
	\big)
	\hat{F}_{\theta,\boldsymbol Y}^{-1:0}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{0:n}({\cal X}|x) \hat{F}_{\theta,\boldsymbol Y}^{-1:0}(dx) }
	\\ \nonumber
	=&
	\frac{\int \hat{\Phi}_{\theta,\boldsymbol Y}^{0:n}(B|x) \hat{F}_{\theta,\boldsymbol Y}^{-1:0}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{0:n}({\cal X}|x) \hat{F}_{\theta,\boldsymbol Y}^{-1:0}(dx) }.
\end{align*}
Hence, (6.17) holds for $m=0$.
Therefore, to complete the proof of (ii),
it is sufficient to show (6.17) for $n\geq m>0$.
In what follows in this part of the proof,
we assume $n\geq m>0$.

Since $r_{\theta,\boldsymbol Y}^{m-1:m}(x_{m-1:n} ) = r_{\theta,\boldsymbol Y}^{m}(x_{m}|x_{m-1} )$,
(\ref{l5.3.303}) implies
\begin{align}\label{l5.3.305}
	r_{\theta,\boldsymbol Y}^{m-1:n}(x_{m-1:n} )
	=
	r_{\theta,\boldsymbol Y}^{m:n}(x_{m:n} ) r_{\theta,\boldsymbol Y}^{m}(x_{m}|x_{m-1})
\end{align}
for $x_{m-1},\dots,x_{n}\in{\cal X}$.
Differentiating (\ref{l5.3.305}) in $\theta$ and using (4.1), we get
\begin{align*}
	s_{\theta,\boldsymbol Y}^{m-1:n}(x_{m-1:n} )
	=
	r_{\theta,\boldsymbol Y}^{m:n}(x_{m:n} ) \nabla_{\theta} r_{\theta,\boldsymbol Y}^{m}(x_{m}|x_{m-1})
	+
	s_{\theta,\boldsymbol Y}^{m:n}(x_{m:n} ) r_{\theta,\boldsymbol Y}^{m}(x_{m}|x_{m-1}).
\end{align*}
Consequently, (4.3) yields
\begin{align}\label{l5.3.23}
	\int S_{\theta,\boldsymbol Y}^{m-1:n}(B|x) \hat{\xi}_{m-1}^{\theta}(dx)
	=&
	\begin{aligned}[t]
	&\int\cdots\int\int
	I_{B}(x_{n}) r_{\theta,\boldsymbol Y}^{m:n}(x_{m:n} ) \nabla_{\theta} r_{\theta,\boldsymbol Y}^{m}(x_{m}|x_{m-1})\\
	&\cdot\mu(dx_{n})\cdots\mu(dx_{m})\hat{\xi}_{m-1}^{\theta}(dx_{m-1})
	\end{aligned}
	\\ \nonumber
	&+
	\begin{aligned}[t]
	&\int\cdots\int\int
	I_{B}(x_{n}) s_{\theta,\boldsymbol Y}^{m:n}(x_{m:n} ) r_{\theta,\boldsymbol Y}^{m}(x_{m}|x_{m-1})\\
	&\cdot\mu(dx_{n})\cdots\mu(dx_{m})\hat{\xi}_{m-1}^{\theta}(dx_{m-1})
	\end{aligned}
	\\ \nonumber
	=&
	\int\int R_{\theta,\boldsymbol Y}^{m:n}(B|x')
	\nabla_{\theta} r_{\theta,\boldsymbol Y}^{m}(x'|x) \mu(dx') \hat{\xi}_{m-1}^{\theta}(dx)
	\\ \nonumber
	&+
	\int\int S_{\theta,\boldsymbol Y}^{m:n}(B|x')
	r_{\theta,\boldsymbol Y}^{m}(x'|x) \mu(dx') \hat{\xi}_{m-1}^{\theta}(dx).
\end{align}
Similarly, (4.2) implies
\begin{align}\label{l5.3.25}
	\int R_{\theta,\boldsymbol Y}^{m-1:n}(B|x) \hat{\zeta}_{m-1}^{\theta}(dx)
	=&
	\begin{aligned}[t]
	&\int\cdots\int\int
	I_{B}(x_{n}) r_{\theta,\boldsymbol Y}^{m:n}(x_{m:n} ) r_{\theta,\boldsymbol Y}^{m}(x_{m}|x_{m-1})\\
	&\cdot\mu(dx_{n})\cdots\mu(dx_{m})\hat{\zeta}_{m-1}^{\theta}(dx_{m-1})
	\end{aligned}
	\\ \nonumber
	=&
	\int\int R_{\theta,\boldsymbol Y}^{m:n}(B|x')
	r_{\theta,\boldsymbol Y}^{m}(x'|x) \mu(dx') \hat{\zeta}_{m-1}^{\theta}(dx).
\end{align}

As $r_{\theta,\boldsymbol Y}^{m-1:m}(x_{m-1:n} ) \!=\! r_{\theta,\boldsymbol Y}^{m}(x_{m}|x_{m-1} )$,
we have $R_{\theta,\boldsymbol Y}^{m-1:m}(dx'|x) \!=\! r_{\theta,\boldsymbol Y}^{m}(x'|x)\mu(dx')$.
Then, using (4.4), (6.5), we conclude
\begin{align}\label{l5.3.29}
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx)
	=&
	\frac{\int\int S_{\theta,\boldsymbol Y}^{m:n}(B|x')
	R_{\theta,\boldsymbol Y}^{m-1:m}(dx'|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	\\ \nonumber
	=&
	\frac{\int\int S_{\theta,\boldsymbol Y}^{m:n}(B|x') r_{\theta,\boldsymbol Y}^{m}(x'|x)
	\mu(dx') \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }.
\end{align}
Relying on the same arguments, we deduce
\begin{align}\label{l5.3.27'}
	\int \!\!R_{\theta,\boldsymbol Y}^{m:n}\!(B|x) \hat{v}_{\theta,\boldsymbol Y}^{m}\!(x)
	\hat{F}_{\theta,\boldsymbol Y}^{m\!-\!1:m}\!(dx)
	\!=&
	\frac{\int\!\int R_{\theta,\boldsymbol Y}^{m:n}(B|x') \hat{v}_{\theta,\boldsymbol Y}^{m}(x')
	R_{\theta,\boldsymbol Y}^{m\!-\!1:m}(dx'|x) \hat{\xi}_{m\!-\!1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m\!-\!1:m}({\cal X}|x) \hat{\xi}_{m\!-\!1}^{\theta}(dx) }
	\\ \nonumber
	\!=&
	\frac{\int \!R_{\theta,\boldsymbol Y}^{m:n}(B|x') \hat{v}_{\theta,\boldsymbol Y}^{m}(x')
	\!\left( \!\int r_{\theta,\boldsymbol Y}^{m}(x'|x) \hat{\xi}_{m\!-\!1}^{\theta}(dx) \!\right) \!\mu(dx') }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m\!-\!1}^{\theta}(dx) }.
\end{align}
Moreover, due to (6.3), we have
\begin{align*}
	\hat{v}_{\theta,\boldsymbol Y}^{m}(x) \left( \int r_{\theta,\boldsymbol Y}^{m}(x|x') \hat{\xi}_{m-1}^{\theta}(dx') \right)
	= &
	\int r_{\theta,\boldsymbol Y}^{m}(x|x') \hat{\zeta}_{m-1}^{\theta}(dx')
	\\
	&+
	\int \nabla_{\theta} r_{\theta,\boldsymbol Y}^{m}(x|x') \hat{\xi}_{m-1}^{\theta}(dx').
\end{align*}
Combining this with (\ref{l5.3.27'}), we get
\begin{align}\label{l5.3.27}
	\int \!\!R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{v}_{\theta,\boldsymbol Y}^{m}(x)
	\hat{F}_{\theta,\boldsymbol Y}^{m\!-\!1:m}(dx)
	\!=&
	\frac{\int\int \!R_{\theta,\boldsymbol Y}^{m:n}(B|x') r_{\theta,\boldsymbol Y}^{m}(x'|x)
	\mu(dx') \hat{\zeta}_{m\!-\!1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m\!-\!1:m}({\cal X}|x) \hat{\xi}_{m\!-\!1}^{\theta}(dx) }
	\\ \nonumber
	&+
	\frac{\int\int R_{\theta,\boldsymbol Y}^{m:n}(B|x')
	\nabla_{\theta} r_{\theta,\boldsymbol Y}^{m}(x'|x)
	\mu(dx') \hat{\xi}_{m\!-\!1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m\!-\!1:m}({\cal X}|x) \hat{\xi}_{m\!-\!1}^{\theta}(dx) }.
\end{align}

Owing to (6.8), (\ref{l5.3.29}), (\ref{l5.3.27}), we have
\begin{align*}
	\int \hat{\Psi}_{\theta,\boldsymbol Y}^{m:n}(B|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx)
	=&
	\int R_{\theta,\boldsymbol Y}^{m:n}(B|x) \hat{v}_{\theta,\boldsymbol Y}^{m}(x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx)
	\\ \nonumber
	&+
	\int S_{\theta,\boldsymbol Y}^{m:n}(B|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx)
	\\ \nonumber
	=&
	\frac{\int\int R_{\theta,\boldsymbol Y}^{m:n}(B|x') r_{\theta,\boldsymbol Y}^{m}(x'|x)
	\mu(dx') \hat{\zeta}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	\\ \nonumber
	&+
	\frac{\int\int
	R_{\theta,\boldsymbol Y}^{m:n}(B|x')
	\nabla_{\theta} r_{\theta,\boldsymbol Y}^{m}(x'|x)
	\mu(dx') \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	\\ \nonumber
	&+
	\frac{\int\int
	S_{\theta,\boldsymbol Y}^{m:n}(B|x') r_{\theta,\boldsymbol Y}^{m}(x'|x)
	\mu(dx') \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }.
\end{align*}
Then, due to (\ref{l5.3.23}), (\ref{l5.3.25}), we also have
\begin{align*}
	\int \hat{\Psi}_{\theta,\boldsymbol Y}^{m:n}(B|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx)
	=&
	\frac{\int R_{\theta,\boldsymbol Y}^{m-1:n}(B|x) \hat{\zeta}_{m-1}^{\theta}(dx)
	+ \int S_{\theta,\boldsymbol Y}^{m-1:n}(B|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:m}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }.
\end{align*}
Consequently, (4.5), (6.7), (\ref{l5.3.1'}) imply
\begin{align*}
	\hat{H}_{\theta,\boldsymbol Y}^{m-1:n}(B)
	=&
	\frac{\int R_{\theta,\boldsymbol Y}^{m-1:n}(B|x) \hat{\zeta}_{m-1}^{\theta}(dx)
	+ \int S_{\theta,\boldsymbol Y}^{m-1:n}(B|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m-1:n}({\cal X}|x) \hat{\xi}_{m-1}^{\theta}(dx) }
	\\ \nonumber
	=&
	\frac{\int \hat{\Psi}_{\theta,\boldsymbol Y}^{m:n}(B|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m:n}({\cal X}|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx) }.
\end{align*}
Then, (4.5), (4.6), (6.6), (6.7), (6.9) yield
\begin{align*}
	\hat{G}_{\theta,\boldsymbol Y}^{m-1:n}(B)
	\!=&
	\hat{H}_{\theta,\boldsymbol Y}^{m-1:n}(B)
	-
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:n}(B)
	\hat{H}_{\theta,\boldsymbol Y}^{m-1:n}({\cal X})
	\\ \nonumber
	\!=&
	\frac{\int
	\big(
	\hat{\Psi}_{\theta,\boldsymbol Y}^{m:n}(B|x)
	\!-
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:n}(B)
	\hat{\Psi}_{\theta,\boldsymbol Y}^{m:n}({\cal X}|x)
	\big)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m:n}({\cal X}|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx) }
	\\ \nonumber
	\!=&
	\frac{\int \hat{\Phi}_{\theta,\boldsymbol Y}^{m:n}(B|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx) }
	{\int R_{\theta,\boldsymbol Y}^{m:n}({\cal X}|x)
	\hat{F}_{\theta,\boldsymbol Y}^{m-1:m}(dx) }.
\end{align*}
Hence, (6.17) holds.

\end{proof}

config: arxiv


text:
\begin{document}
In this section, we prove that dominant constraint graphs on $H(\alpha_{Z_m},2q)$ are well-behaved.
\subsection{The Set of Graphs $R(H(\alpha_{Z_m},2q))$}
In order to analyze constraint graphs on $H(\alpha_{Z_m},2q)$, we need to analyze a more general class of $H$. In particular, we need to analyze all $H$ which can be obtained by taking isolated vertices which are not incident to any spokes in $H(\alpha_{Z_m},2q)$ and merging their neighbors together.
\begin{defn}
Define $R(H(\alpha_{Z_m},2q))$ to be the set of graphs $H$ which can be obtained by starting from $H(\alpha_{Z_m},2q)$ and repeatedly applying the following operation:
\begin{enumerate}
\item Choose a vertex $v \in V(H)$ which is in a wheel with at least $4$ vertices and is not incident to any spokes. Merge the two neighbors of $v$, delete $v$ from the graph, and delete any pairs of edges in $H$ which coincide.
\end{enumerate}
\end{defn}
\begin{lemma}
For any $H \in R(H(\alpha_{Z_m},2q))$, we can decompose $H$ as $H = \alpha_{1} \circ \ldots \circ \alpha_{2q'}$ where the following statements are true:
\begin{enumerate}
\item For all odd $i$, $\alpha_i$ consists of trivial top layers, a multi-$Z$ shape in the middle layers, and trivial bottom layers.
\item For all even $i$, $\alpha_i$ consists of trivial top layers, a multi-$Z^T$ shape in the middle layers, and trivial bottom layers.
\item For any two neighboring wheels, the spokes connect with each other and alternate between going up and to the right and down and to the right.
\item For any layer, the intervals where this layer is trivial have even length.
\end{enumerate}
\end{lemma}
\begin{proof}
To prove this, we show that this structure is preserved when we make a new contraction. Without loss of generality, we can assume that the isolated vertex $v$ is the bottom right vertex in some multi-$Z$ shape $\alpha_i$ where $i$ is odd, as cases when the isolated vertex is the top left vertex in a multi-$Z$ shape, the bottom left vertex in a multi-$Z^T$ shape, or the top right vertex in a multi-$Z^T$ shape can be handled in a similar way. Let $j$ be the next index where this wheel is non-trivial. Note that $j$ must be even. Moreover, all layers below $v$ must be trivial in $\alpha_i$ and $\alpha_j$ as otherwise $v$ would not be isolated.

Let $u$ be the vertex preceding $v$ and let $w$ be the vertex after $v$. We merge $u$ and $w$ together and delete $v$. This deletes the edges $(u,v),(v,w)$ and may delete spokes incident to $u$ and $w$. We have the following cases.
\begin{enumerate}
\item There is a spoke $(u,t)$ in $\alpha_i$ and a spoke $(t,w)$ in $\alpha_j$. In this case, merging $u$ and $w$ together also deletes the spokes $(u,t),(t,w)$. We account for this by making $v$'s layer trivial in $\alpha_i$ and $\alpha_j$, replacing it with the single vertex $u = w$.
\item There is a spoke $(u,t)$ in $\alpha_i$ but no spoke incident to $w$ in $\alpha_j$. In this case, the edge $(v,w)$ is the only non-trivial part of $\alpha_j$. Let $k$ be the next index such that this layer is non-trivial in $\alpha_k$. Observe that $\alpha_k$ is a multi-$Z$ shape where all layers above this layer are trivial. We account for merging $u$ and $w$ together and deleting $v$ as follows:
\begin{enumerate}
\item Glue $\alpha_i$ and $\alpha_k$ together at the vertex $u = w$.
\item Create a copy of $\alpha_{i} \circ \ldots \circ \alpha_{k-1}$ which only contains the part below the current layer and is trivial at this layer and above. Put these copies to the left of the glued shape.
\item Create a copy of $\alpha_{i+1} \circ \ldots \circ \alpha_{k}$ which only contains the part above the current layer and is trivial at this layer and above. Put these copies to the right of the glued shape.
\end{enumerate}
\item There is no spoke incident to $u$ in $\alpha_i$ but there is a spoke $(t,w)$ in $\alpha_j$. This can be handled in a similar way to the previous case.
\item There are no spokes incident to $u$ or $w$. In this case, the edges $(u,v)$ and $(v,w)$ are the only non-trivial parts of $\alpha_i$ and $\alpha_j$. We again account for this by making $v$'s layer trivial in $\alpha_i$ and $\alpha_j$, replacing it with the single vertex $u = w$.
\end{enumerate}
\end{proof}
\subsection{Proof that dominant constraint graphs are well-behaved}
With this structural result on $R(H(\alpha_{Z_m},2q))$ , we can now prove that dominant constraint graphs for $H(\alpha_{Z_m},2q)$ are well behaved. To do this, we use ideas from Appendix B of \cite{AMP20}. First, we modify our constraint graph as follows:
\begin{defn}
Let $H = \alpha_{1} \circ \ldots \circ \alpha_{2q}$ where we set $V_{\alpha_{2q}} = U_{\alpha_1}$. Given a constraint graph $C$ on $H$, we define the constraint graph $C_{aug}$ as follows:
\begin{enumerate}
\item Draw the constraint edges so that all paths in $C$ go from left to right.
\item For each vertex $u \in U_{\alpha_1}$, letting $v$ be the rightmost vertex such that there is a path of constraint edges from $u$ to $v$, we add an auxiliary constraint edge from $v$ to $u$. We treat this edge as going from $v$ on the left to $u$ on the right (i.e. we think of $u$ as both on the left side of $H$ and on the right side of $H$ as $u \in U_{\alpha_1} = V_{\alpha_{2q}}$). If $u$ is isolated, this means that we add an auxiliary loop from $u$ to itself.
\end{enumerate}
\end{defn}
\begin{defn}
Given $H$ and $C_{aug}$ as described above, for each $\alpha_i$ we define $S_{\alpha_i}$ to be union of $U_{\alpha_i} \cap V_{\alpha_i}$ and the set of vertices $v$ such that there exists a path $P$ in $\alpha_i$ from $U_{\alpha_i}$ to $V_{\alpha_i}$ such that $v$ is the first vertex on $P$ where either
\begin{enumerate}
\item There is a constraint edge from $v$ to the right (i.e. to a vertex in some $\alpha_j$ where $j > i$).
\item $v \in V_{\alpha_i}$.
\end{enumerate}
\end{defn}
\begin{prop}
$S_{\alpha_i}$ is a vertex separator of $\alpha_i$.
\end{prop}
\begin{rmk}
Alternatively, we could have started from $V_{\alpha_i}$ and taken the first vertex on each path in $\alpha_i$ from $V_{\alpha_i}$ to $U_{\alpha_i}$ which has a constraint edge to the left or is in $U_{\alpha_i}$.
\end{rmk}
\begin{lemma}
Let $P$ be a path in $\alpha_i$ from $U_{\alpha_i}$ to $V_{\alpha_i}$ and let $v$ be the first vertex on $P$ where either
\begin{enumerate}
\item There is a constraint edge from $v$ to the right (i.e. to a vertex in some $\alpha_j$ where $j > i$).
\item $v \in V_{\alpha_i}$.
\end{enumerate}
For any vertex $u \in V(P) \setminus U_{\alpha_i}$ which is equal to $v$ or comes before $v$, $u$ has an edge to the left.
\end{lemma}
\begin{proof}
Let $l$ be the vertex which comes before $u$ in $P$. Observe that $l$ does not have any constraint edges to the right. Thus, in order for the edge $(l,u)$ to be duplicated, $u$ must have a constraint edge to the left.
\end{proof}
\begin{cor}
$C$ is a dominant constraint graph for $H$ if and only if the following statements are true for each $\alpha_i$ and $S_{\alpha_i}$:
\begin{enumerate}
\item $S_{\alpha_i}$ is a minimum vertex separator of $\alpha_i$.
\item Each vertex in $V(\alpha_i) \setminus (U_{\alpha_i} \cup V_{\alpha_i} \cup S_{\alpha_i})$ is incident with exactly one constraint edge.
\item Each vertex in $U_{\alpha_i} \setminus S_{\alpha_i}$ is not incident with any constraint edges to the right.
\item Each vertex in $V_{\alpha_i} \setminus S_{\alpha_i}$ is not incident with any constraint edges to the left.
\end{enumerate}
\end{cor}
\begin{thm}\label{dominantiswellbehavedtheorem}
For any $H \in R(H(\alpha_{Z_m},2q))$, all dominant constraint graphs $C$ on $H$ are well-behaved (i.e. wheel-respecting and parity preserving).
\end{thm}
\begin{proof}[Proof sketch]
This theorem can be proved by induction using the following lemma.
\begin{lemma}\label{findingisolatedvertexlemma}
For any $H \in R(H(\alpha_{Z_m},2q))$ and any dominant constraint graph $C$ on $H$, there is a vertex $v$ which is isolated and is not incident to any spokes in $H$.
\end{lemma}
\begin{proof}
We prove this lemma with a series of observations.
\begin{prop}\label{norightleftcombo}
In any dominant constraint graph $C$, if $u$ precedes $v$ on some wheel then either $u$ does not have a constraint edge to the right or $v$ does not have a constraint edge to the left.
\end{prop}
\begin{proof}
Assume that $C$ is dominant, $u$ has an edge to the right, and $v$ has an edge to the left. Consider the separator $S$ for the segment containing $u$ and $v$. Since $u$ has a constraint edge to the right, $u \in S$. Now either $v \in S$ or $v \notin S$. If $v \in S$ then $S$ is not a minimum vertex separator so $C$ is not dominant. If $v \notin S$ then the constraint edge to the left from $v$ is not accounted for by $S$ so $C$ is not dominant. Thus, in either case $C$ is not dominant, which is a contradiction.
\end{proof}
\begin{cor}\label{isolatedmiddlevertex}
In any dominant constraint graph, if $l$ and $r$ are two vertices on the same wheel such that $l < r$, $l$ has a constraint edge to the right, and $r$ has a constraint edge to the left then there is a vertex $m$ such that $l < m < r$ and $m$ is isolated.
\end{cor}
\begin{proof}
Assume that there exist two vertices $l$ and $r$ on the same wheel such that $l < r$, $l$ has a constraint edge to the right, $r$ has a constraint edge to the left, and there is no vertex $m$ such that $l < m < r$ and $m$ is isolated. Choose $l$ and $r$ such that $d(l,r)$ is minimized. Let $v$ be the vertex after $l$ on this wheel.

By Proposition \ref{norightleftcombo}, since $l$ precedes $v$ and $l$ has a constraint edge to the right, $v$ does not have a constraint edge to the left. This implies that either $v$ is isolated or $v$ only has a constraint edge to the right. However, we cannot have that $v$ is isolated as otherwise we could take $m = v$ and we would have that $l < m < r$ and $m$ is isolated. Thus, $v$ must only have a constraint edge to the right. But then if we take $l' = v$, $l'$ and $r$ are on the same wheel, $l'$ has a constraint edge to the right, $r$ has a constraint edge to the left, there is no vertex $m$ such that $l < m < r$ and $m$ is isolated, and $d(l',r) < d(l,r)$. This contradicts the fact that we chose $l$ and $r$ to minimize $d(l,r)$.
\end{proof}
With these observations, we can now prove Lemma \ref{findingisolatedvertexlemma}. Consider the highest wheel such that there exist vertices $l < r$ on this wheel satisfying the following properties:
\begin{enumerate}
\item $l$ has a constraint edge to the right and $r$ has a constraint edge to the left.
\item For any vertex $m$ between $l$ and $r$, $m$ is not incident to any spokes in $H$ going from $m$ to the wheel below $m$.
\end{enumerate}
Observe that the bottom wheel has these properties, so this wheel always exists. By Corollary \ref{isolatedmiddlevertex}, there is a vertex $m$ such that $l < m < r$ and $m$ is isolated. There are now two cases to consider. Either $m$ is not incident to any spokes in $H$ going from $m$ to the wheel above $m$, or $m$ is incident to two such spokes. Note that $m$ is not incident to any spokes in $H$ going from $m$ to the wheel below $m$, so if $m$ is not incident to any spokes in $H$ going from $m$ to the wheel above $m$, then we have found an isolated vertex which is not incident to any spokes in $H$. If $m$ is incident to two spokes in $H$ from $m$ to the wheel above $m$, let $l' < r'$ be the other endpoints of these spokes. Since $m$ is isolated, because of the structure of $H$, $l'$ must have an edge to the right, $r'$ must have an edge to the left, and there are no vertices $m'$ such that $l' < m' < r'$ and $m'$ is incident to spokes in $H$ going from $m'$ to the wheel below $m'$. However, this is a contradiction, as this implies that we did not start with the highest wheel such that there exist vertices $l < r$ on this wheel, where $l$ has an edge to the right, $r$ has an edge to this left, and for any vertex $m$ between $l$ and $r$, $m$ is not incident to any spokes in $H$ going from $m$ to the wheel below $m$.
\end{proof}
\end{proof}
\end{document}


TITLE: How would one show the evaluation map to $K(\alpha)$ is surjective directly?
QUESTION [0 upvotes]: Let $K$ be a field and $\alpha$ be an algebraic element. Consider the following ring homomorphism by which $\phi:K[X]\to K(\alpha)$ by which $\phi(X)=\alpha.$ I was wondering how would one show such $\phi$ must be a surjection directly? ($K(\alpha)$ denotes the smallest field to contain both $K$ and $\alpha$.)
I wanted to show this is a surjection thus the minimal polynomial of $\alpha$ must be irreducible. (I know other ways of proving this statement but I thought it was interesting trying out this way)
I know that we can use FIT and thus $\frac{K[X]}{\langle m_{\alpha}(X)\rangle}\cong K(\alpha)$; however, via this route we need to use $m_\alpha$, the minimal polynomial, is irreducible and thus I do not want to end up in a circular argument. Hence I was wondering if there exists other clever way of showing such map must be surjective.
Many thanks in advance!

REPLY [0 votes]: The image of $\phi$ is by definition $K[\alpha]$, a integral domain that is an integral extension of the field $K$. Hence $K[\alpha]$ is a field, so $K(\alpha)\subseteq K[\alpha]\subseteq K(\alpha)$.