After equation (1) at page 3 of this paper it is claimed that the transition kernel of a Markov chain generated by the Metropolis-Hastings algorithm is of the form $$P(x,A)=\varrho(x)\tilde P(x,A)+(1-\varrho(x))1_A(x)\;,\;\;\;(x,A)\in E\times\mathcal E\tag1,$$ where $\tilde P$ is a Markov kernel on a measurable space $(E,\mathcal E)$ and $\varrho:E\to(0,1]$ is $\mathcal E$-measurable.

How do we see this? The transition kernel $\kappa$ of the Metropolis-Hastings algorithm with proposal kernl $Q$ and acceptance function $\alpha$ is given by $$\kappa(x,B)=\int_BQ(x,{\rm d}y)\alpha(x,y)+r(x)1_B(x)\;,\;\;\;(x,B)\in E\times\mathcal E,$$ where $$r(x):=1-\int Q(x,{\rm d}y)\alpha(x,y)\;\;\;\text{for }x\in E.$$

I don't see how $\tilde P$ should be defined to match the form.


The claim is equivalent to saying that $P$ generates a chain where,

  • at $x$, with probability $1 - \rho ( x )$, you stay at $x$, and
  • otherwise, you do move, and you do so according to $\tilde{P}$.

Thus, in the Metropolis-Hastings case, you take

\begin{align} \rho ( x ) &= 1 - r ( x), \\ \tilde{P} ( x, dy ) &= Q ( x, dy ) \cdot \frac{ \alpha ( x, y ) }{ 1 - r ( x ) } \end{align}

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