Let $f_{X}$ be a density we would like to sample from. For some reasons, $f_{X}$ may be analytically intractable or expensive to evaluate. A solution consists in considering a density $(x,y) \in X \times Y \, \mapsto \, f(x,y)$ on $X \times Y$ such that $f_{X}$ is a marginal density of $f$. Then, we define a Markov chain $(X_{n})_{n \geq 0}$ as follows :
Given the current state of the chain $X_{n} = x$ :
- $ Y_n \sim f_{Y \mid X}(\cdot \mid x)$
- $X_{n+1} \sim f_{X \mid Y}(\cdot \mid Y_n)$
where $f_{X \mid Y}(\cdot \mid y)$ (resp. $f_{Y \mid X}(\cdot \mid x)$) denotes the conditional density of $X$ given $Y=y$ (resp. of $Y$ given $X=x$). I would like to determine the transition kernel of the Markov chain $(X_{n})_{n \geq 0}$. For that, let $\varphi$ be a bounded measurable function on $X$. I would like to compute $\mathbb{E}\big[ \varphi(X_{n+1}) \mid \mathcal{F}_{n} \big]$ where $\mathcal{F}_{n} = \sigma(X_{0},\ldots,X_{n})$ is the natural filtration.
Let $g(Y_n) = \mathbb{E}\big[ \varphi(X_{n+1}) \mid Y_n \big]$. Then :
$$ g(y) = \int_{X} \varphi(u) f_{X \mid Y}(u \mid y) \, du. $$
Therefore :
$$ \mathbb{E}\big[ \varphi(X_{n+1}) \mid \mathcal{F}_{n} \big] = \mathbb{E}\big[ g(Y_n) \mid \mathcal{F}_{n} \big] = \int_{Y} \Bigg( \int_{X} \varphi(u) f_{X \mid Y}(u \mid y) \, du \Bigg) f_{Y \mid X}(y \mid X_{n}) \, dy.$$
This can be rewritten :
$$ \int_{X} \varphi(u) \Bigg( \int_{Y} f_{X \mid Y}(u \mid y) f_{Y \mid X}(y \mid X_{n}) \, dy \Bigg) \, du \tag{$\star$}$$
We know that $\mathbb{E}\big[ \varphi(X_{n+1}) \mid \mathcal{F}_{n} \big] = P\varphi(X_{n})$ where the transition kernel $P$ satisfies $\displaystyle P\varphi(x) = \int_{X} \varphi(u)P(x,du)$. So we can use $(\star)$ to identify the transition kernel.
I am not sure this is correct and I am having trouble deriving the transition kernel from the last identity.