Metropolized backward sampling for conditional sequential Monte Carlo: Lean formalization

2 The forced move and the coupling kernel

Definition 10 Forced move, eq. (4)
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For a probability vector \(\mathbf{p}\) on \([N]\),

\begin{align*} F_{\mathbf{p}}(i, j) & = \frac{p_j}{\max \{ 1 - p_i, 1 - p_j\} }, \quad j \neq i, \\ F_{\mathbf{p}}(i, i) & = 1 - \sum _{j \neq i} F_{\mathbf{p}}(i, j), \end{align*}

and \(F_{\mathbf{p}}(i, \cdot ) \mathrel {\vcenter {:}}=\delta _i\) if \(p_i = 1\). For \(\mathbf{u} \in [0, \infty )^N\), \(F_{\mathbf{u}} \mathrel {\vcenter {:}}=F_{\mathbf{u} / (u_1 + \dotsb + u_N)}\) if \(\mathbf{u} \neq \mathbf{0}\) and \(F_{\mathbf{0}}(i, \cdot ) \mathrel {\vcenter {:}}=\delta _i\).

For a probability vector \(\mathbf{p}\), \(F_{\mathbf{p}}\) is a Markov kernel with \(F_{\mathbf{p}}(i, i) \leq p_i\). For \(\mathbf{u} \in [0, \infty )^N\), \(F_{\mathbf{u}}\) is a Markov kernel and \(u_i F_{\mathbf{u}}(i, j) = u_j F_{\mathbf{u}}(j, i)\) for all \(i, j \in [N]\).

Proof ▶

For \(j \neq i\), \(F_{\mathbf{p}}(i, j) \leq p_j / (1 - p_i)\), so that the off-diagonal entries of a row sum to at most one. The bound \(F_{\mathbf{p}}(i, i) \leq p_i\) follows from the second part of Lemma 12. The statements on \(F_{\mathbf{u}}\) follow from those on \(F_{\mathbf{p}}\).

The kernel \(F_{\mathbf{p}}\) is reversible with respect to \(\mathbf{p}\). For \(i \neq j\), \(F_{\mathbf{p}}(i, j) \geq p_j\). For every \(\varphi \colon [N] \to \mathbb {R}\), \(\langle \varphi , F_{\mathbf{p}} \varphi \rangle _{\mathbf{p}} \leq \mathbf{p}(\varphi )^2\).

Proof ▶

The first two parts are due to Liu (1996, Section 2) when \(p_i {\lt} 1\) for all \(i \in [N]\). If \(p_i = 1\) for some \(i\), then they follow from \(p_j = 0\) and \(F_{\mathbf{p}}(j, i) = 1\) for \(j \neq i\). The third part follows from the first two by Lemma 6 with \(q = p\), or by Lemma 7.

Lemma 13 The forced move for two particles
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For \(N = 2\) and \(p_1, p_2 {\gt} 0\), \(F_{\mathbf{p}}(1, 2) = \min \{ 1, p_2 / p_1\} \) is the Metropolis acceptance probability.

Proof ▶

By (4), \(F_{\mathbf{p}}(1, 2) = p_2 / \max \{ p_1, p_2\} \).

Definition 14 Coupling kernel, eq. (5)
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With \(\mathbf{p} \odot \mathbf{q} \mathrel {\vcenter {:}}=(p_1 q_1, \dotsc , p_N q_N)\), \(C_{\mathbf{p},\mathbf{q}}(i, \cdot ) \mathrel {\vcenter {:}}=\mathbf{q} + q_i \{ F_{\mathbf{p} \odot \mathbf{q}}(i, \cdot ) - \delta _i\} \), \(i \in [N]\).

Lemma 15 The coupling kernel is a Markov kernel
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For all \(i \in [N]\), \(C_{\mathbf{p},\mathbf{q}}(i, i) = q_i F_{\mathbf{p} \odot \mathbf{q}}(i, i)\), so that, for probability vectors \(\mathbf{p}\) and \(\mathbf{q}\), the entries of \(C_{\mathbf{p},\mathbf{q}}\) are nonnegative and \(C_{\mathbf{p},\mathbf{q}}\) is a Markov kernel.

Proof ▶

By (5) and Lemma 11.

Let \(\mathbf{p}\) and \(\mathbf{q}\) be probability vectors on \([N]\). For all \(i, i' \in [N]\), \(p_i C_{\mathbf{p},\mathbf{q}}(i, i') = q_{i'} C_{\mathbf{q},\mathbf{p}}(i', i)\). For \(i \neq i'\), \(C_{\mathbf{p},\mathbf{q}}(i, i') \geq q_{i'}\). For every \(\varphi \colon [N] \to \mathbb {R}\), \(\langle \varphi , C_{\mathbf{p},\mathbf{q}} \varphi \rangle _{\mathbf{p}} \leq \mathbf{p}(\varphi ) \mathbf{q}(\varphi )\).

Proof ▶

As in the proof of Lemma 2 in Section S1, the first part follows from the reversibility of \(F_{\mathbf{p} \odot \mathbf{q}}\) with respect to \(\mathbf{p} \odot \mathbf{q}\). The second part follows from \(C_{\mathbf{p},\mathbf{q}}(i, i') = q_{i'} + q_i F_{\mathbf{p} \odot \mathbf{q}}(i, i')\) for \(i \neq i'\). The third part follows from the first two by Lemma 6.