Markov function theorem
markov_function_theorem intertwining_relation
The Markov function theorem can be found in [{rogers-pitman81}].
Definitions and proof
Theorem(discrete time discrete space version of Markov function theorem). Let \(X\) be a Markov process with transition kernel \(P\) on state space \(S\), \(f : S \to T\) for some space \(T\), and \(Y\) a process on space \(T\) defined by \(Y_t = f(X_t)\) for some \(f\). If there is a kernel \(K: T \times S \to [0, 1]\) such that
- \(K(y, x) = 0 \forall x \notin f^{-1}(y)\)
- \(\sum_{x \in S} K(y, x) = 1 \forall y \in T\).
Then if stochastic matrix \(Q: T \times T \to [0,1]\) satisfies the intertwining relation
\[ Q K = K P \qquad (1) \]then if \(X_0 \sim K(Y_0, \cdot)\) then
- \(Y\) evolves as a Markov chain with kernel \(Q\).
- \(\prob(X_n = x|Y_n = y, Y_{n - 1}, ..., Y_0) = K(y, x)\).
Remark. Condition 1 means \((K(y, \cdot))_y\) has disjoint supports \(f^{-1}(y)\). Therefore the LHS of of (1) has only one term and (1) becomes
\[ Q(y, f(x')) K(f(x'), x') = \sum_{x \in f^{-1}(y)} K(y, x) P(x, x'). \]The intertwining relation (1) is the most important part of the theorem. Once it is verified, the results of the theorem should follow.
Proof. Let \(y_{0: n} \subset T\). Denote \(A_k := f^{- 1} (y_k)\).
\begin{align} \mathbb P_{Y_0 = y_0} (Y_{0: n} = y_{0: n}) &= \mathbb P_{X_0 \sim K(y_0, \cdot)} (X_0 \in A_0, X_1 \in A_1, ..., X_n \in A_n) \\ &= \sum_{x_{0: n} \in A_{0: n}} K(y_0, x_0) P(x_0, x_1) P(x_1, x_2) ... P(x_{n - 1}, x_n)\\ &= \sum_{y_1' \in T, x_{1: n} \in A_{1: n}} Q(y_0, y_1') K(y_1', x_1) P(x_1, x_2) ... P(x_{n - 1}, x_n) \\ &\overset{\dagger}{=} \sum_{x_{1: n} \in A_{1: n}} Q(y_0, y_1) K(y_1, x_1) P(x_1, x_2) ... P(x_{n - 1}, x_n)\\ &= ... = \sum_{x_n \in A_n} Q(y_0, y_1) Q(y_1, y_2) ... Q(y_{n - 1}, y_n) K(y_n, x_n) \\ &\overset{\dagger\dagger}{=} Q(y_0, y_1) Q(y_1, y_2) ... Q(y_{n - 1}, y_n). \end{align}where \(\dagger\) and \(\dagger\dagger\) we use the first and second conditions respectively. Therefore \(Y\) is a Markov process with transition kernel \(Q\). The proof for the continuous case should be the same. \(\square\)
Both assumptions that \(X\) and \(Y\) are probabilistic, as well as the positivity of \(K\) can be relaxed, if we only focus on the algebraic structure.
For continuous-time Markov processes / chains besides the transition kernel there are also generators, and we have a corresponding version of the Markov function theorem as well.
Theorem(continuous-time version of the Markov function theorem). Let \(X\) be a Markov process with generator \(L^X\) on space \(S\), and \(Y\) a process on \(T\) where \(Y_t = f(X_t)\) for some \(f\). If there is a kernel \(K: T \times S \to [0, 1]\) such that
- \(K(y, x) = 0 \forall x \notin f^{-1}(y)\)
- \(\int K(y, x) dx = 1\) for all \(y \in T\)
Then if Markov generator \(L^Y\) satisfies the intertwining relation
\[ K L^X = L^Y K \]then \(Y\) is Markov with generator \(L^Y\).
Relation between the continuous and discrete Markov chain versions
This bit is not verified by many examples. But just the RSK.
Suppose we have continuous-time Markov chains \(\hat X\) and \(\hat Y\) with generator \(\hat P\) and \(\hat Q\), and their embedded discrete-time chains \(X\) and \(Y\) with transition matrices \(P\) and \(Q\), such that
\[ P(x, x') = \begin{cases} - {\hat P(x, x') \over \hat P(x, x)}, & x \neq x' \\ 0, &x = x' \end{cases};\qquad Q(x, x') = \begin{cases} - {\hat Q(x, x') \over \hat Q(x, x)}, & x \neq x' \\ 0, &x = x' \end{cases} \]In one case we can say that \(P\) and \(Q\) intertwines with \(K\) iff \(\hat P\) and \(\hat Q\) with the same kernel.
Claim. Assuming the following is satisfied:
- \(\hat P(x, x) = \hat Q(y, y) = c \forall x, y\)
- \(\hat P(x, x') = 0\) whenever \(f(x) = f(x')\) and \(x \neq x'\),
then
\[ \hat Q K = K \hat P \Leftrightarrow Q K = K P. \]Proof. Quite straightforward. Consider the cases \(y = y'\) and \(y \neq y'\) separately. \(\square\)
This Claim allows us to verify the intertwining relations of the embedded chains. One example are the RS models, where the continuous-time version has an input of \(\ell\) Poisson processes, and \(X, Y\) are two levels of shapes \((\lambda^{\ell - 1}, \lambda^\ell), \lambda^\ell\). The two conditions correspond to
- The total rate of change of \(X\) and \(Y\) are the sum of rates of the input Poisson processes
- Whenever \(\lambda^{\ell - 1}\) changes the next level changes as well
References
- [rogers-pitman81] Markov functions, , Annals of Probability, Vol. 9, No. 4, p.573–582 1981.