1 Erratum to: Stat Comput (2016) 26:1187–1211 DOI 10.1007/s11222-015-9604-3

The statement of Proposition 4.1 is incorrect. We present the corrected result with its proof.

Proposition 4.1

Assume (P1), (P2), (W4), (W5). Alternatively, assume (P1*), (P2) and (W5). Then, there exists \(D_{k}>0\) and \(N_{0}\in \mathbb {N}^{+}\) such that for all \(N\ge N_{0}\),

$$\begin{aligned} \Vert \tilde{\pi }_{N}(\cdot )-\pi (\cdot )\Vert _{ TV} \le D_{k}\frac{\log \left( N\right) }{N^{\frac{\tau }{2+k}}}, \end{aligned}$$

where \(\tau =k\) if \(k\in (0,1)\) and \(\tau =\frac{1+k}{2}\) if \(k\ge 1\). If in addition (W5) holds for all \(k>0\), then for any \(\varepsilon \in (0,1/6)\) there will exist \(D_{\varepsilon }>0\) and \(N_{0}\in \mathbb {N}^{+}\) such that for all \(N\ge N_{0}\),

$$\begin{aligned} \Vert \tilde{\pi }_{N}(\cdot )-\pi (\cdot )\Vert _{TV} \le D_{\varepsilon }\frac{\log \left( N\right) }{N^{\frac{1}{2}-\varepsilon }}. \end{aligned}$$

Proof

The proof is identical to the original version up to the inequality

$$\begin{aligned}&\sup _{x\in \mathcal {X}}\Vert \tilde{P}_{N} (x,\cdot )-P(x,\cdot )\Vert _{ TV}\\&\quad \le 3\delta +\frac{2^{3+k}}{\delta ^{1+k}}\sup _{x\in \mathcal {X}}\mathbb {E}\left[ \Big \vert W_{x,N}-1\Big \vert ^{1+k}\right] . \end{aligned}$$

By the Marcinkiewicz–Zygmund inequality for i.i.d random variables (see, e.g., Gut 2012, Chapter 3, Corollary 8.2), there exists \(B_k<\infty \) such that

$$\begin{aligned} \mathbb {E}\left[ \Big \vert W_{x,N}-1\Big \vert ^{1+k}\right]&\le B_k \mathbb {E}\left[ \Big \vert W_{x}-1\Big \vert ^{1+k}\right] N^{-\tau }, \end{aligned}$$

where

$$\begin{aligned} \tau ={\left\{ \begin{array}{ll} k &{} \quad \text {if}\,k\in \left( 0,1\right) \\ \frac{1+k}{2} &{} \quad \text {if}\,k\ge 1. \end{array}\right. } \end{aligned}$$

Therefore,

$$\begin{aligned}&\sup _{x\in \mathcal {X}}\Vert \tilde{P}_{N} (x,\cdot )-P(x,\cdot )\Vert _{TV} \\&\quad \le 3\delta +\frac{2^{3+k}B_k}{\delta ^{1+k}N^{\tau }}\sup _{x\in \mathcal {X}}\mathbb {E}\left[ \Big \vert W_{x}-1\Big \vert ^{1+k}\right] . \end{aligned}$$

The first part of the result follows from the original proof by taking

$$\begin{aligned} C_{k} = B_k \sup _{x\in \mathcal {X}}\mathbb {E}\left[ \Big \vert W_{x}-1\Big \vert ^{1+k}\right] \end{aligned}$$

and considering \(N^\tau \) instead of \(N^k\).

For the second claim, take \(k_{\varepsilon }\ge (2\varepsilon )^{-1}-2\ge 1\) and apply the first part. \(\square \)