Furthermore, let

  • $(E,\mathcal E,\lambda)$ be a $\sigma$-finite measure space;
  • $Q$ be a Markov kernel on $(E,\mathcal E)$ with density $q$ with respect to $\lambda$;
  • $\mu$ be a probability measure on $(E,\mathcal E)$ with density $p$ with respect to $\lambda$
  • $$\alpha(x,y):=\left.\begin{cases}\min\left(1,\displaystyle\frac{p(y)q(y,x)}{p(x)q(x,y)}\right)&\text{, if }p(x)q(x,y)>0\\1&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x,y\in E$$
  • $$r(x):=1-\int Q(x,{\rm d}y)\alpha(x,y)\;\;\;\text{for }x\in E;$$
  • $\kappa$ denote the transition kernel of the Metropolis-Hastings algorithm with proposal kernel $Q$ and target distribution $\mu$; i.e. $$\kappa(x,B)=\underbrace{\int_BQ(x,{\rm d}y)\alpha(x,y)}_{=:\;\pi(x,\;B)}+r(x)1_B(x)\;\;\;\text{for all }(x,B)\in E\times\mathcal E.$$

In Proposition 5.6.3 of this paper, it is claimed that, if $x\in E$ with $p(x)>0$, then $$\left\|\kappa(x,\;\cdot\;)-Q(x,\;\cdot\;)\right\|=r(x),\tag1$$ where $\|\;\cdot\|$ denotes the total variation norm. How do we proof this result?

A proof is given, but I don't understand it. The paper is considering MALA, instead of the general setup here, but I don't think this is crucial. I'm willing to impose further assumptions on $Q$, if necessary.

If $\nu_i$ is a probability measure on $(E,\mathcal E)$, the total variation distance between $\nu_1$ and $\nu_2$ is defined as $$\|\nu_1-\nu_2\|:=\sup_{B\in\mathcal E}(\nu_1-\nu_2)(B).$$

Now, what I get is that $$\kappa(x,B)-Q(x,B)=\begin{cases}Q(x,B^c)-\pi(x,B^c)&\text{, if }x\in B;\\\pi(x,B)-Q(x,B)&\text{, otherwise}\end{cases}\tag2$$ for all $B\in\mathcal E$.

For $B=\{x\}$, we obtain $$\kappa(x,B)-Q(x,B)=r(x)\tag3.$$ So, the question reduces to why we maximize $(2)$ by choosing $B=\{x\}$?

For this $B$, we clearly got $x\in B$ and hence $\kappa(x,B)-Q(x,B)=Q(x,B^c)-\pi(x,B^c)$. Clearly, the first summand $Q(x,B^c)$ will increase when we reduce $B$. But it's not clear to me, why this increase is larger than the decrease from $\pi(x,B^c)$. Seems like we need some kind of monotonicity between $Q$ and $\pi$ here ...

  • $\begingroup$ The (coupling) inequality$$\left\|\kappa(x,\;\cdot\;)-Q(x,\;\cdot\;)\right\|\le r(x),\tag1$$is straightforward as the proposal from $Q(x,\cdot)$ is accepted on average with probability $1-r(x)$. $\endgroup$
    – Xi'an
    Apr 15 at 7:39
  • $\begingroup$ @Xi'an Wouldn't your $(1)$ and my $(3)$ together already imply $(1)$? But I don't get your argument. I know what "the" coupling inequality is and I also know that the proposals from $Q(x,\;\cdot\;)$ are accepted on average with probability $1-r(x)$. $\endgroup$
    – 0xbadf00d
    Apr 15 at 8:37

1 Answer 1


Conditional on $x$, the "coupling" interpretation is to start with the move attached to $Q(x,\cdot)$ and to propose to couple the output with $\kappa(x,\cdot)$, a coupling that is accepted with probability $1-r(x)$. Hence the total variation distance between $Q(x,\cdot)$ and $\kappa(x,\cdot)$ is zero with probability $1-r(x)$. And $1$ (i.e., one) with probability $r(x)$ since $$\mathbb I_{\{x\}^c}(x) - Q(x,\{x\}^c)=1$$ if one assumes that $Q$ is free of atoms.

  • $\begingroup$ I see. The coupling inequality makes it really straightforward, while I was struggling to show it directly from the definition of the total variation norm. Do you think a similar result can be obtained for the difference $$\langle f,f-Qf\rangle_{L^2(\mu)}-\langle f,f-\kappa f\rangle_{L^2(\mu)}$$ for $f\in L^2(\mu)$? $\endgroup$
    – 0xbadf00d
    Apr 16 at 15:11
  • $\begingroup$ From the variational characterization of the total variation I only get $$|(\kappa(x,\;\cdot\;)-Q(x,\;\cdot\;))f|\le2\|f\|_\infty\|\kappa(x,\;\cdot\;)-Q(x,\;\cdot\;)\|\tag1,$$ which requires $f$ to be bounded. What I get from this is $$\left|\langle f,(\kappa-Q)f\rangle_{L^2(\mu)}\right|\le2\|f\|^2_\infty\mu r.$$ Is there a nice expression for $\mu r$? In any case, maybe this bound is not really sharp. Would it be better to consider $\chi^2$-distance? $\endgroup$
    – 0xbadf00d
    Apr 16 at 19:23
  • $\begingroup$ I would still be interested in your thoughts on this. $\endgroup$
    – 0xbadf00d
    Apr 24 at 11:55

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