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Let $(E,\mathcal E)$ be a measurable space and $\kappa$ be a Markov kernel on $(E,\mathcal E)$. Assume that $$\kappa(x,B)=\int_Bp(x,y)\:\lambda({\rm d}y)\;\;\;\text{for all }(x,B)\in E\times\mathcal E$$ for some $\mathcal E^{\otimes2}$-measurable $p:E^2\to[0,\infty)$ with $$\int p(x,y)\:\lambda({\rm d}y)=1\;\;\;\text{for all }x\in E\tag1$$ and a measure $\lambda$ on $(E,\mathcal E)$.

Assume $\kappa$ is the transition kernel of a time-homogeneous Markov chain $(X_n)_{n\in\mathbb N_0}$ on $(E,\mathcal E)$. I would like to assume that $$p(x,z)=\int p(x,y)p(y,z)\:\lambda({\rm d}y)\tag2\;\;\;\text{for all }x,z\in E,$$ but I wonder how strong this assumption really is.

As usual, $\kappa$ can be thought as a bounded linear operator on the space $\mathcal E_b$ of bounded $\mathcal E$-measurable functions equipped with the supremum norm. $(2)$ obviously implies that $$\kappa^2f=\kappa f\;\;\;\text{for all }f\in\mathcal E_b\tag3;$$ i.e. $\kappa$ is a projection. What does this imply for $(X_n)_{n\in\mathbb N_0}$?

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    $\begingroup$ If $E = [0,1]$ and $\lambda$ is Lebesgue, do you know of an example with $(2)$? $\endgroup$
    – Gerald Edgar
    Jul 10, 2022 at 0:10
  • $\begingroup$ This readily implies that $X_n$ is distributed like $X_1$ for every $n$, so this is not very interesting from the perspective of random walks. $\endgroup$
    – M. Dus
    Jan 4, 2023 at 23:49
  • $\begingroup$ @M.Dus How does that imply $X_n\sim X_1$? $\endgroup$
    – 0xbadf00d
    Jan 5, 2023 at 12:18
  • $\begingroup$ Well the right hand side in (2) is the convolution of p with itself, denoted by $p^{(2)}$. It is the distribution of $X_2$. If $p=p^{(2)}$, then by induction, $p=p^{(n)}$ for all $n$ and so $X_1$ is distributed like $X_n$. $\endgroup$
    – M. Dus
    Jan 17, 2023 at 10:58

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