Let $(X_n)_{n\in\mathbb N_0}$ be a stationary time-homogeneous Markov chain with $X_0\sim f\mu$ for some measure space $(E,\mathcal E,\mu)$ and $\mathcal E$-measurable $f:E\to[0,\infty)$ with $$\int f\:{\rm d}\mu=1.$$ We know that the evolution of $(X_n)_{n\in\mathbb N_0}$ is uniquely determined by its transition kernel, which is a Markov kernel $\kappa$ on $(E,\mathcal E)$ with $$\operatorname P\left[X_1\in B\mid X0\right]=\kappa(X_0,B)\;\;\;\text{almost surely for all }B\in\mathcal E\tag1.$$ Since we know that $X_0,X_1\sim f\mu$, are we able to determine $\kappa$ explicitly?


No, you cannot determine a kernel just from knowledge of the distributions of $X_0$ and $X_1$. You must know the joint distribution of $(X_0,X_1)$ to determine the kernel.

For example, consider $X_0$ and $X_1$ uniformly distributed on $\{1,\ldots,n\}$ and let $\sigma$ be any deterministic permutation. Then $\sigma$ induces a kernel sending $i$ to the dirac measure at $\sigma(i)$, and this kernel sends $X_0$ to $X_1$.

If you want a continuous version of this example, simply consider any permuton (i.e. a continuous analog of a random permutation).

  • $\begingroup$ If the chain is constructed by the Metropolis-Hastings algorithm (and we assume it already reached stationarity), then we should be able to determine $\kappa$, right? $\endgroup$
    – 0xbadf00d
    Aug 12 '19 at 6:44
  • $\begingroup$ I don't believe even then it is determined, since there is a great amount of freedom in choosing the parameters in the Metropolis-Hastings algorithm. $\endgroup$
    – pre-kidney
    Aug 12 '19 at 6:46
  • $\begingroup$ The transition kernel of the Metropolis-Hastings algorithm is determined explicitly. See, for example, equation 2.2.3 on page 5 here: math.wustl.edu/~sawyer/hmhandouts/MetropHastingsEtc.pdf. $\endgroup$
    – 0xbadf00d
    Aug 12 '19 at 6:54
  • 1
    $\begingroup$ The acceptance criteria $a(x,y)$ is an arbitrary function in the notes you linked, and plays a role in determining the transition kernel. Note that "explicit" does not mean "uniquely determined". In any case, the discussion in these comments is tangential to the question you asked... $\endgroup$
    – pre-kidney
    Aug 12 '19 at 6:58

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