# Is the transition kernel of a Metropolis-Hastings chain of the form $P(x,A)=\varrho(x)\tilde P(x,A)+(1-\varrho(x))1_A(x)$?

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