Analysis of an interacting particle method for rare event estimation

Abstract

We present a large deviations analysis for the performance of an interacting particle method for rare event estimation. The analysis is restricted to a one-dimensional setting, though even in this restricted setting a number of new techniques must be developed. In contrast to the large deviations analyses of related algorithms, for interacting particle schemes it is an occupation measure analysis that is relevant, and within this framework many standard assumptions (stationarity, Feller property) can no longer be assumed. The methods developed are not limited to the question of performance analysis, and in fact give the full large deviations principle for such systems.

Appendix

Proof of Lemma 5.2

Let $t$ $\in [0,1]$ be given, and let $\theta _{i}=\theta _{i}(t)$ be iid with distribution $\mu (\cdot |t)$ (for simplicity we omit the $t$ -dependence of $\theta _{i}$). For $v\in [0,1)$, define

$$\begin{aligned} R_{v}\doteq \sum _{i=1}^{\infty }\left[ v+\sum _{k=1}^{i}\theta _{k}-1\right] 1_{\left\{ v+\sum _{k=1}^{\ell }\theta _{k}<1 \text{ for} \ell =1,\ldots ,i-1\right\} }1_{\left\{ v+\sum _{k=1}^{i}\theta _{k}\ge 1\right\} }. \end{aligned}$$

Then with the understanding that $v+\sum _{k=1}^{\ell }\theta _{k}<1$ for all $\ell <\infty$ means $R_{v}=\Delta$, the distribution of $R_{v}$ is $\gamma (\cdot |v,t).$ Thus for a bounded and continuous function $f:\mathbb{R }\rightarrow \mathbb{R }$

$$\begin{aligned}&\int \limits _{\mathbb{R }}f(x)\gamma (\text{ d}x|v,t)\\&\quad =\mathbb{E }\left[ \sum _{i=1}^{\infty }f\left( v+\sum _{k=1}^{i}\theta _{k}-1\right) 1_{\left\{ v+\sum _{k=1}^{\ell }\theta _{k}<1 \text{ for} \ell =1,\ldots ,i-1\right\} }1_{\left\{ v+\sum _{k=1}^{i}\theta _{k}\ge 1\right\} }\right] . \end{aligned}$$

Suppose that $v$ is not an element of $(0,1]\cap \left[ \mathbb{Z }/q\right] .$ Then for each $i$ the indicator functions in the last display are continuous at $v$ w.p.1, and so by the Lebesgue Dominated Convergence Theorem, if $v_{j}\rightarrow v$ as $j\rightarrow \infty$ then

$$\begin{aligned} \int \limits _{\mathbb{R }}f(x)\gamma (\text{ d}x|v_{j},t)\rightarrow \int \limits _{\mathbb{R }}f(x)\gamma (\text{ d}x|v,t). \end{aligned}$$

The analogous argument applies when $v\in (0,1]\cap \left[ \mathbb{Z }/q \right]$ if we restrict to $v_{j}\downarrow v$, which proves the right continuity. $\square$

Proof of Lemma 5.3

To simplify notation we omit the $t$ dependence. We need to show for bounded and continuous $g$ and $h$ that

$$\begin{aligned} \int g(v)h(y)(\mu _{n}\otimes \gamma )(\text{ d}v\times \text{ d}y)\rightarrow \int g(v)h(y)(\mu \otimes \gamma )(\text{ d}v\times \text{ d}y). \end{aligned}$$

Now $f(v)=\int h(y)\gamma (\text{ d}y|v)$ is continuous from the right for all $v$ and from the left save (possibly) those $v$ of the form $k/q.$ Let $f^{\delta }(v)$ be continuous, have the same uniform bound as $f$, and equal to $f$ outside $\cup _{k=1}^{q}([k/q]-\delta ,[k/q]).$ Then

$$\begin{aligned} \left| \int g(v)\mu _{n}(\text{ d}v)\int h(y)\gamma (\text{ d}y|v)-\int g(v)\mu _{n}(\text{ d}v)f^{\delta }(v)\right| \le 2\varepsilon \left\| {h}\right\| _{\infty }\left\| {g}\right\| _{\infty }. \end{aligned}$$

Since $\mu _{n}\Rightarrow \mu$ implies $\mu (\cup _{k=1}^{q}([k/q]-\delta ,[k/q]))\le \varepsilon$,

$$\begin{aligned} \left| \int g(v)\mu (\text{ d}v)\int h(y)\gamma (\text{ d}y|v)-\int g(v)\mu (\text{ d}v)f^{\delta }(v)\right| \le 2\varepsilon \left\| {h}\right\| _{\infty }\left\| {g}\right\| _{\infty }, \end{aligned}$$

and since $f^{\delta }(x)$ is continuous, $\mu _{n}\Rightarrow \mu$ implies

$$\begin{aligned} \limsup _{n\rightarrow \infty }\left| \int g(v)h(y)\mu _{n}(\text{ d}v)\gamma (\text{ d}y|v)-\int g(v)h(y)\mu (\text{ d}v)\gamma (\text{ d}y|v)\right| \le 4\varepsilon \left\| {h}\right\| _{\infty }\left\| {g}\right\| _{\infty }. \end{aligned}$$

The result now follows since $\varepsilon >0$ is arbitrary. $\square$

Proof of Lemma 7.7

Part 1. For fixed $T<\infty$, consider a modification of $\nu _{j,i}^{m}$, such that if $\bar{\sigma }_{j}^{m}<T,$ then we redefine $\nu _{j,i}^{m}=\mu$ for $\bar{\sigma }_{j}^{m}<i\le T-1.$ With this modified definition of the control, which does not change the distribution of the hitting time on the set $\{\bar{\sigma }_{j}^{m}<T\}$, we have

$$\begin{aligned} {\bar{\mathbb{E }}}\left[ \sum _{i=0}^{\bar{\sigma }_{j}^{m}-1}R(\nu _{j,i}^{m}\left\| \mu \right. )\right]&\ge {\bar{\mathbb{E }}}\left[ \sum _{i=0}^{(\bar{\sigma }_{j}^{m}\wedge T)-1}R(\nu _{j,i}^{m}\left\| \mu \right. )\right] \\&= R\left( \left. \mathbb{Q }^{T}\right| _{\left\{ 0,\ldots ,T-1\right\} }\left\| \left. \mathbb{P }\right| _{\left\{ 0,\ldots ,T-1\right\} }\right. \right) ,\nonumber \end{aligned}$$

(9.1)

where $\mathbb{Q }^{T}$ denotes the modified joint measure on the space of increments, $\mathbb{Q }^{T}|_{\left\{ 0,\ldots ,T-1\right\} }$ denotes the restriction to the first $T$ coordinates in the underlying product space, and $\left. \mathbb{P }\right| _{\left\{ 0,\ldots ,T-1\right\} }$ denotes product measure with marginal $\mu .$

We now consider the disjoint, finite partition of the space $\mathbb{R }^{T}=A_{j}^{T}\cup B_{j}^{T}\cup D_{j}^{T}$, where

$$\begin{aligned} A_{j}^{T}&= \left\{ \omega :\bar{Z}_{j,\sigma _{j}^{m}}^{m}\ge j+1,\bar{\sigma }_{j}^{m}<T\right\} , \\ B_{j}^{T}&= \left\{ \omega :\bar{Z}_{j,\sigma _{j}^{m}}^{m}\le -1,\bar{\sigma }_{j}^{m}<T\right\} , \\ D_{j}^{T}&= \left\{ \omega :\bar{\sigma }_{j}^{m}\ge T\right\} . \end{aligned}$$

Using the approximation property of relative entropy via sums over finite measurable partitions [8, Lemma 1.4.3(g)], we obtain

$$\begin{aligned} R\left( \left. \mathbb{Q }^{T}\right| _{\left\{ 0,\ldots ,T-1\right\} }\left\| \left. \mathbb{P }\right| _{\left\{ 0,\ldots ,T-1\right\} }\right. \right) \ge R\left( \tau _{j}^{m,T}\left\| \alpha _{j}^{m,T}\right. \right) , \end{aligned}$$

(9.2)

where $\tau _{j}^{m,T},\alpha _{j}^{m,T}\in \mathcal P \left( \left\{ (-\infty ,-1]\cup \left[ j+1,\infty \right) ,\Sigma \right\} \right)$ are the measures induced by $\mathbb{Q }^{T}|_{\left\{ 0,\ldots ,T-1\right\} }$ and $\mathbb{P }|_{\left\{ 0,\ldots ,T-1\right\} }$, and $\Sigma$ corresponds to the event $\sigma _{j}^{m}\ge T.$

We know that $\alpha _{j}^{m,T}(\Sigma )\rightarrow 0$ as $T\rightarrow \infty .$ Let $\bar{\alpha }_{j}^{m}$ denote the extension of $\alpha _{j}^{m}$ to $\left\{ (-\infty ,-1]\cup \left[ j+1,\infty \right) ,\Sigma \right\}$ with $\bar{\alpha }_{j}^{m}(\Sigma )=0.$ Let $\bar{\tau }_{j}^{m}$ denote the limit of $\tau _{j}^{m,T}$, which must exist by monotonicity. By the lower semi-continuity of relative entropy,

$$\begin{aligned} \liminf _{T\rightarrow \infty }R\left( \tau _{j}^{m,T}\left\| \alpha _{j}^{m,T}\right. \right) \ge R\left( \bar{\tau }_{j}^{m}\left\| \bar{ \alpha }_{j}^{m}\right. \right) . \end{aligned}$$

(9.3)

If $\bar{\tau }_{j}^{m}(\Sigma )>0$ then $R(\bar{\tau }_{j}^{m}\mathbf \parallel \bar{\alpha }_{j}^{m})=\infty$, and if $\bar{\tau }_{j}^{m}(\Sigma )=0$ then $R(\bar{\tau }_{j}^{m}{\parallel }\bar{\alpha } _{j}^{m})=R(\tau _{j}^{m}{\parallel }\alpha _{j}^{m}).$ The last sentence and (9.1), (9.2) and (9.2) imply

$$\begin{aligned} {\bar{\mathbb{E }}}\left[ \sum _{i=0}^{\bar{\sigma }_{j}^{m}-1}R(\nu _{j,i}^{m}\left\| \mu \right. )\right] \ge R\left( \tau _{j}^{m}\left\| \alpha _{j}^{m}\right. \right) . \end{aligned}$$

Part 2. Define

$$\begin{aligned} U(x;y)=-\log \left( \frac{\text{ d}\tau \left( \cdot \left| x\right. \right) }{\text{ d}\alpha _{j}^{m}\left( \cdot \left| x\right. \right) }(y)\right) . \end{aligned}$$

Since by assumption $U(x;y)$ is bounded from above, according to [8, Proposition 4.5.1]

$$\begin{aligned} -\log \int e^{-U(x;y)}\alpha _{j}^{m}\left( \text{ d}y|x\right) =\inf _{\beta :R(\beta {\parallel }\alpha _{j}^{m})<\infty }\left\{ R\left( \beta \left\| \alpha _{j}^{m}\right. \right) +\int U\left( x;y\right) \beta \left( \text{ d}y\right) \right\} . \end{aligned}$$

Since $R(\tau {\parallel }\alpha _{j}^{m})<\infty$, by the definition of $U$

$$\begin{aligned} \tau \left( \text{ d}y|x\right) =e^{-U\left( x;y\right) }\alpha _{j}^{m}\left( \text{ d}y|x\right) \end{aligned}$$

(9.4)

achieves the infimum in the variational formula, i.e.,

$$\begin{aligned} -\log \int e^{-U\left( x;y\right) }\alpha _{j}^{m}\left( \text{ d}y|x\right) =R\left( \tau \left\| \alpha _{j}^{m}\right. \right) +\int U\left( x;y\right) \tau \left( \text{ d}y|x\right) . \end{aligned}$$

We next consider the analogous relations on the space of increments of the process. We have

$$\begin{aligned} -\log \int e^{-U\left( x;\bar{Z}_{j,\bar{\sigma }_{j}^{m}}^{m}\left( \omega \right) \right) }\mathbb{P }\left( \text{ d}\omega \right) =\inf _{\mathbb{Q }}\left\{ R\left( \mathbb{Q }\left\| \mathbb{P }\right. \right) +\int U\left( x;\bar{Z }_{j,\bar{\sigma }_{j}^{m}}^{m}\left( \omega \right) \right) \mathbb{Q }\left( \text{ d}\omega \right) \right\} , \end{aligned}$$

where $\bar{\sigma }_{j}^{m}$ denotes the first hitting time of the process and $\mathbb{P }$ is the underlying product probability with marginal $\mu .$ The definition of $\alpha _{j}^{m}$ implies that for any integrable function $K$

$$\begin{aligned} \int K\left( \bar{Z}_{j,\bar{\sigma }_{j}^{m}}^{m}\left( \omega \right) \right) e^{-U\left( x;\bar{Z}_{j,\bar{\sigma }_{j}^{m}}^{m}\left( \omega \right) \right) }\mathbb{P }\left( \text{ d}\omega \right) =\int K(y)e^{-U\left( x;y\right) }\alpha _{j}^{m}\left( \text{ d}y|x\right) . \end{aligned}$$

Also, the minimum is achieved at

$$\begin{aligned} \mathbb{Q }\left( \text{ d}\omega \right) =e^{-U\left( x;\bar{Z}_{j,\bar{\sigma } _{j}^{m}}^{m}\left( \omega \right) \right) }\mathbb{P }\left( \text{ d}\omega \right) . \end{aligned}$$

(9.5)

Since the first hitting probability $\alpha _{j}^{m}$ is induced by $\mathbb{P }$, (9.4) and (9.5) imply

$$\begin{aligned} \int U\left( x;\bar{Z}_{j,\bar{\sigma }_{j}^{m}}^{m}\left( \omega \right) \right) \mathbb{Q }\left( \text{ d}\omega \right)&= \int U\left( x;\bar{Z}_{j,\bar{ \sigma }_{j}^{m}}^{m}\left( \omega \right) \right) e^{-U\left( x;\bar{Z}_{j, \bar{\sigma }_{j}^{m}}^{m}\left( \omega \right) \right) }\mathbb{P }\left( \text{ d}\omega \right) \\&= \int U\left( x;y\right) e^{-U\left( x;y\right) }\alpha _{j}^{m}\left( \text{ d}y|x\right) \\&= \int U\left( x;y\right) \tau \left( \text{ d}y|x\right) . \end{aligned}$$

Hence, by the chain rule

$$\begin{aligned} R\left( \tau \left\| \alpha _{j}^{m}\right. \right) =R\left( \mathbb{Q } \left\| \mathbb{P }\right. \right) =\mathbb{E }^\mathbb{Q }\left[ \sum _{i=0}^{\infty }R(\nu _{j,i}^{m}\left\| \mu \right. )\right] \ge \mathbb{E }^\mathbb{Q }\left[ \sum _{i=0}^{\bar{\sigma }_{j}^{m}-1}R(\nu _{j,i}^{m}\left\| \mu \right. )\right] , \end{aligned}$$

where $\{\nu _{j,i}^{m}\}$ are the controls defined by factoring $\mathbb{Q }.$

Combining this with the bound proved in part 1, we have constructed controls and a controlled process $\{\bar{Z}_{j,i}^{m}\}$ satisfying

$$\begin{aligned} \mathbb{E }^{\mathbb{Q }}\left[ \sum _{i=0}^{\bar{\sigma }_{j}^{m}-1}R(\nu _{j,i}^{m}\left\| \mu \right. )\right] =R\left( \tau \left\| \alpha _{j}^{m}\right. \right) , \end{aligned}$$

where $\tau$ is the distribution induced by the stopped, controlled process. Although here we have constructed the control on the canonical space $\mathbb{R }^{\infty }$, it can be applied in the general setting by making the obvious identifications. $\square$

Proof of Proposition 7.10

Since $\{{\hat{\mathbf{X }}}_{j}^{m}\}$ is tight, given $\eta >0$, there is compact $S_{1}\subset \mathcal T _{1}$ such that $\mathbb{P }\{{\hat{\mathbf{X }}}_{j}^{m}\notin S_{1}\}\le \eta$ for all $j\le m$, including the index that first exceeds $i/\kappa .$ Using standard results from stochastic stability, there is a compact set $S_{2}\subset \mathcal T _{1}$ such that, after starting in $S_{1}$, the probability of escaping from $S_{2}$ is also less than $\eta .$

To simplify the presentation, we denote the initial value of $j\in [i/\kappa ,(i+1)/\kappa )$ by zero and the final value by $m$, and write $\hat{J}^{m}$ for $\hat{J}_{i}^{m}.$ By the geometric ergodicity of $\bar{p}_{i}^{\kappa }$ and the compactness of $S_{2}$

$$\begin{aligned} \left\| \bar{p}_{i}^{\kappa ,(k)}\left( \mathbf x ,\cdot \right) -\bar{\lambda }_{i}^{\kappa }\left( \cdot \right) \right\| _{v}\rightarrow 0 \end{aligned}$$

uniformly in $\mathbf x \in S_{2}$, where $\bar{p}_{i}^{\kappa ,(k)}$ is the $k$-fold convolution of $\bar{p}_{i}^{\kappa }.$ Let $\delta >0$, and assume that $k$ is large enough that for such $\mathbf x$

$$\begin{aligned} \left\| \bar{p}_{i}^{\kappa ,(k)}\left( \mathbf x ,\cdot \right) -\bar{\lambda }_{i}^{\kappa }\left( \cdot \right) \right\| _{v}<\delta . \end{aligned}$$

(9.6)

Let $\hat{p}_{j,\ell }^{m}$ denote convolution of the transition kernels $\hat{p}_{jk+\ell }^{m},\hat{p}_{jk+\ell +1}^{m},\ldots \hat{p}_{(j+1)k+\ell -1}^{m}$, so that if ${\hat{\mathbf{X }}}_{kj+\ell }^{m}=\mathbf x$ then the distribution of ${\hat{\mathbf{X }}}_{k(j+1)+\ell }^{m}$ is given by $\hat{p} _{j,\ell }^{m}(\mathbf x ,\cdot ).$

Let $\zeta >0$ be given. For $m$ sufficiently large we know that

$$\begin{aligned} \left\| \hat{p}_{j}^{m}\left( \mathbf x ,\cdot \right) -\bar{p} _{i}^{\kappa }\left( \mathbf x ,\cdot \right) \right\| _{v}<\zeta . \end{aligned}$$

(9.7)

Iterating the (9.7) gives

$$\begin{aligned} \left\| \hat{p}_{j,\ell }^{m}\left( \mathbf x ,\cdot \right) -\bar{p} _{i}^{\kappa ,(k)}\left( \mathbf x ,\cdot \right) \right\| _{v}\le k\zeta \end{aligned}$$

for all $\ell$ and all $\mathbf x \in S_{2}$, and thus

$$\begin{aligned} \left\| \hat{p}_{j,\ell }^{m}\left( \mathbf x ,\cdot \right) -\bar{\lambda }_{i}^{\kappa }\left( \cdot \right) \right\| _{v}<k\zeta +\delta \end{aligned}$$

(9.8)

for all $\ell$ and all $\mathbf x \in S_{2}.$

Let $k\in \mathbb N$ be given. We will break the empirical measure $\hat{J} ^{m}$ into $k$ sums, and it follows from the fact that $k$ will be fixed as $m\rightarrow \infty$ that we can consider just those $m$ of the form $rk,r\in \mathbb N .$ We then write

$$\begin{aligned} \hat{J}^{m}=\frac{1}{m}\sum _{j=0}^{m-1}\delta _{{\hat{\mathbf{X }}}_{j}^{m}}= \frac{1}{m}\left[ \sum _{\ell =0}^{k-1}\sum _{j=0}^{[m/k]-1}\delta _{{\hat{\mathbf{X }}}_{kj+\ell }^{m}}\right] =\frac{1}{k}\sum _{\ell =0}^{k-1}\frac{k}{m} \sum _{j=0}^{[m/k]-1}\delta _{{\hat{\mathbf{X }}}_{kj+\ell }^{m}}\doteq \frac{1}{ k}\sum _{\ell =0}^{k-1}\hat{J}_{\ell }^{m}, \end{aligned}$$

where the last equality defines the $\hat{J}_{\ell }^{m}.$

Now fix $\ell \in \left\{ 0,1,\ldots ,k-1\right\}$ and consider any bound measurable function $f:\mathcal T _{1}\rightarrow \mathbb R$ with $\left\| {f}\right\| _{\infty }\le 1.$ As usual, we have that

$$\begin{aligned} f\left( {\hat{\mathbf{X }}}_{k(j+1)+\ell }^{m}\right) -\int f\left( \mathbf y \right) \hat{p}_{j,\ell }^{m}\left( {\hat{\mathbf{X }}}_{jk+\ell }^{m},\text{ d}\mathbf y \right) \end{aligned}$$

is a martingale difference sequence, and thus by (9.8)

$$\begin{aligned} \left[ f\left( {\hat{\mathbf{X }}}_{k(j+1)+\ell }^{m}\right) -\int f\left( \mathbf y \right) \bar{\lambda }_{i}^{\kappa }\left( \text{ d}\mathbf y \right) \right] 1_{\left\{ {\hat{\mathbf{X }}}_{jk+\ell }^{m}\in S_{2}\right\} }=\varepsilon _{1,j}^{m}+\varepsilon _{2,j}^{m}, \end{aligned}$$

where $|\varepsilon _{1,j}^{m}|\le k\zeta +\delta$ and $\varepsilon _{2,j}^{m}$ has conditional mean zero and uniformly (in $m$ and $j$) bounded second moment. It follows using a standard calculation that

$$\begin{aligned} \mathbb{P }\left\{ \bigcup _{\ell =0}^{k-1}\left\{ \left| \int f\left( \mathbf x \right) \hat{J}_{\ell }^{m}\left( \text{ d}\mathbf x \right) \!-\!\int f\left( \mathbf y \right) \bar{\lambda }_{i}^{\kappa }\left( \text{ d}\mathbf y \right) \right| \ge 2\left[ k\zeta \!+\!\delta \right] \right\} \right\} \le k \frac{k}{4\left[ k\zeta \!+\!\delta \right] ^{2}m}+2\eta . \end{aligned}$$

Letting first $m\rightarrow \infty$ shows that weak limits of the $\hat{J} _{\ell }^{m}$ are all within $2\left[ k\zeta +\delta \right]$ of $\bar{ \lambda }_{i}^{\kappa }$ in total variation norm save on a set of probability no more than $2\eta .$ We now send $\zeta \downarrow 0$, then $\delta \downarrow 0$, and finally $\eta \downarrow 0$ to complete the proof. $\square$

Proof of Part 3 of Lemma 7.6

Part 3. Let $h(\mathbf x,y )=(\text{ d}\bar{\Lambda }/\text{ d}\Lambda )(\mathbf x,y )$, where the Radon-Nikodym derivative is in $\mathbf y .$ For $k\in \mathbb N$ define

$$\begin{aligned} S^{k}\left( \mathbf x \right) =1-\int \limits _\mathcal{T _{2}}h^{k}\left( \mathbf x,y \right) \Lambda \left( \text{ d}\mathbf y \left| \mathbf x \right. \right) , \end{aligned}$$

where $h^{k}(\mathbf x,y )=k\wedge h(\mathbf x,y ).$ Also, let

$$\begin{aligned} \bar{\Lambda }^{k}\left( A\left| \mathbf x \right. \right) =\int \limits _{A}h^{k}\left( \mathbf x,y \right) \Lambda \left( \text{ d}\mathbf y \left| \mathbf x \right. \right) +S^{k}\left( \mathbf x \right) \tilde{ \Lambda }\left( A\left| \mathbf x \right. \right) \end{aligned}$$

for any Borel set $A$ of $\mathcal T _{2}.$ Obviously, $(\bar{\xi }^{k},\bar{\eta })$ is also admissible and the associated transition kernel is equivalent to $\bar{p}(\mathbf x,\cdot ).$ Hence this transition kernel is also geometrically ergodic. Furthermore, the relative entropy

$$\begin{aligned} R\left( \left. \bar{\Lambda }^{k}\left( \cdot \left| \mathbf x \right. \right) \right\| \Lambda \left( \cdot \left| \mathbf x \right. \right) \right) =\int \limits _\mathcal{T _{2}}\log h^{k}\left( \mathbf x,y \right) \bar{\Lambda }^{k}\left( \text{ d}\mathbf y \left| \mathbf x \right. \right) +S^{k}\left( \mathbf x \right) R\left( \left. \tilde{\Lambda }\left( \cdot \left| \mathbf x \right. \right) \right\| \Lambda \left( \cdot \left| \mathbf x \right. \right) \right) \end{aligned}$$

is uniformly bounded as a function of $\mathbf x .$ Since $S^{k}\left( \mathbf x \right) \rightarrow 0$ as $k\rightarrow \infty$, the Dominated Convergence Theorem implies (7.7).

It is easy to check that the construction implies $\Vert \bar{\Lambda } ^{k}(\cdot |\mathbf x )-\bar{\Lambda }(\cdot |\mathbf x )\Vert _{v}\rightarrow 0.$ It then follows from Proposition 7.10, which is stated and proved later in this section, that the first marginals of $\bar{\xi }^{k}$ converges to $\bar{\xi }_{1}$ in total variation, i.e., $\Vert \bar{\xi }_{1}^{k}-\bar{\xi }_{1}\Vert _{v}\rightarrow 0.$ $\square$