# Why is it legitimate to assume that the Chapman-Kolmogorov equations hold everywhere?

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space
• $$(\mathcal F_t)_{t\ge0}$$ be a filtration on $$(\Omega,\mathcal A)$$
• $$E$$ be a Polish space and $$\mathcal E:=\mathcal B(E)$$
• $$(X_t)_{t\ge0}$$ be an $$E$$-valued $$\mathcal F$$-Markov process on $$(\Omega,\mathcal A,\operatorname P)$$
• $$\kappa_{s,\:t}$$ be a regular version of the conditional probability of $$X_t$$ given $$X_s$$, i.e. $$\kappa_{s,\:t}$$ is a Markov kernel on $$(E,\mathcal E)$$ with $$\operatorname P\left[X_t\in B\mid X_s\right]=\kappa_{s,\:t}(X_s,B)\;\;\;\text{almost surely for all }B\in\mathcal E\tag1$$ for $$s,t\ge0$$

By the Markov property and $$(1)$$, $$\operatorname P\left[X_t\in B\mid\mathcal F_s\right]=\kappa_{s,\:t}(X_s,B)\;\;\;\text{almost surely for all }B\in\mathcal E\text{ and }0\le s\le t.\tag2$$

Usually, we want $$(\kappa_{s,\:t}:0\le s\le t)$$ to satisfy the Chapman-Kolmogorov equation $$\kappa_{r,\:t}=\kappa_{r,\:s}\kappa_{s,\:t},\tag3$$ where the right-hand side denotes the composition of transiton kernels, for all $$0\le r\le s\le t$$. However, with the definition of $$\kappa_{s,\:t}$$ as the conditional probability of $$X_t$$ given $$X_s$$, we've only got$$^1$$ $$\kappa_{r,\:t}(x,B)=(\kappa_{r,\:s}\kappa_{s,\:t})(x,B)\;\;\;\text{for all }B\in\mathcal E\text{ and }\operatorname P\circ\:X_r^{-1}\text{-almost all }x\in E\tag4$$ for all $$0\le s\le t$$. However, in the literature, one is usually assuming that $$(2)$$ and $$(3)$$ hold together. Why is that possible?

I could imagine that the reason is the following: Since $$E$$ is Polish, given $$(\kappa_{s,\:t}:0\le s\le t)$$ with $$(3)$$ there is always a Markov process $$\tilde X$$ on an other probability space with transition semigroup $$(\kappa_{s,\:t}:0\le s\le t)$$ and initial distribution $$\operatorname P\circ\:X_0^{-1}$$. Is that the correct argument?

$$^1$$ Note that there is a crucial selection of a common null set for all $$B$$ happening in $$(4)$$. I guess this is legitimate as long as $$\mathcal E$$ is countably generated. Maybe someone could comment on this.

• Something seems amiss in equation (3), as it would have a stray X_s variable on the right that is not on the left. If we stick to countable state space then (3) should be augmented by summing over the intermediate states x_s – Michael Jan 21 at 1:09
• @Michael What should be missing in equation $(3)$? The right-hand side is the composition of $\kappa_{r,\:s}$ and $\kappa_{s,\:t}$, which is again a Markov kernel on $(E,\mathcal E)$. – 0xbadf00d Jan 21 at 9:12
• As I mentioned about summing over intermediate states for countable state spaces, $$P[X_{r+s+t} = j| X_r=i] = \sum_{w \in S}P[X_{r+s+t}=j|X_{r+s}=w]P[X_{r+s}=w|X_r=i] \neq \underbrace{P[X_{r+s+t}=j|X_{r+s}=w]P[X_{r+s}=w|X_r=i]}_{\mbox{stray w index}}$$ Now it seems to me that writing a sum/integral/inner-product ("composition?") as if it is a multiplication leaves a lot of room for confusion. – Michael Jan 21 at 10:47
• @Michael Actually, I have no idea what you mean. Are you saying that $\kappa_{r,\:s}\kappa_{s,\:t}$ is a confusing notation for the composition or are you saying that the whole concept of composition of transition kernels is confusing? Just to be sure, the definition is: $$(\kappa_{r,\:s}\kappa_{s,\:t})(x,B):=\int\kappa_{r,\:s}(x,{\rm d}y)\kappa_{s,\:t}(y,B)\;\;\;\text{for }(x,B)\in E\times\mathcal E.$$ – 0xbadf00d Jan 21 at 10:47
• All of the above and more: $k_{r,s}k_{s,t}$ seems a multiplication of two scalar-valued functions...I would assume a "composition" is when you plug the output of a function into the input of another, like $f(g(x))$...I would not think of an integration as such...the integral definition you give above seems different from the linked definition which seems to require the second function have a differently-structured domain...plugging "dy" into a function argument is just way-too-scary for me. – Michael Jan 21 at 10:53

My (perhaps naive) interpretation of your question is that it deals with "surely" versus "almost surely" issues. For fixed times $$r < s < t$$, the following equation only holds in an “almost sure” sense: $$k_{r,t} = k_{r,s} \cdot k_{s,t} \quad (Eq. 1)$$ To see this, suppose it holds surely (for all corresponding points $$x$$ and sets $$B$$ for which we can evaluate $$k_{r,t}(x,B)$$). Let $$\tilde{k}_{r,t}$$ be another “version” of $$k_{r,t}$$ that is different from $$k_{r,t}$$ on a set of measure 0. Then we want to also have $$\tilde{k}_{r,t} = k_{r,s} \cdot k_{s,t} \quad (Eq. 2)$$ surely, but this is impossible since we have changed the left-hand-side of equation (1) without changing the right-hand-side. The only way it makes sense is that both equations (1) and (2) are only guaranteed to hold in an almost-sure sense.
An example way to change $$k_{r,t}$$ to a different version $$\tilde{k}_{r,t}$$ is this: Suppose we can identify an $$x^*$$ such that $$P[X_r=x^*]=0$$. Then define for all corresponding $$x, B$$: $$\tilde{k}_{r,t}(x,B) = \left\{ \begin{array}{ll} k_{r,t}(x,B) &\mbox{ if x \neq x^*} \\ \mu(B) & \mbox{ if x =x^*} \end{array} \right.$$ where $$\mu(B)$$ is some measure over events $$B$$ that is different from $$k_{r,t}(x^*,B)$$.
• Alternatively, perhaps your question deals generally with the facts (i) $P[f(X)\neq g(X)]=0$ if and only if $P[X^{-1}(\{x : f(x) = g(x)\})]=1$; (ii) If $h(x)$ and $r(x)$ differ only on a set of measure zero then $\int h(x) d\mu = \int r(x) d\mu$. – Michael Jan 21 at 21:16
• Another interpretation of your question is that, instead of fixing times $r,s,t$ and working with those in the distribution equations, you are packaging the uncountably infinite set of times in the real number line as part of the "almost surely" statements (which seems to be overkill). If none of these, then I do not know what you are concerned with and/or how you feel (2), (3), (4) conflict. – Michael Jan 22 at 14:28
• Sorry for my late response. Look, the problem is the following: Assume for simplicity that $X$ is time-homogeneous. Then, $$\operatorname P\left[X_{s+t}\in B\mid\mathcal F_s\right]=\kappa_t(X_s,B)\;\;\;\text{almost surely}$$ for all $B\in\mathcal E$ and $s,t\ge0$ (where $\kappa_t:=\kappa_{0,\:t}$). Since $\mathcal E$ is countably generated, we are able to conclude $$\kappa_{s+t}(x,\;\cdot\;)=(\kappa_s\kappa_t)(x,\;\cdot\;)\;\;\;\text{for }\operatorname P\circ X_0^{-1}\text{-all }x\in E$$ for all $s,t\ge0$. – 0xbadf00d Feb 18 at 15:16
• However, in every book I know, it's always assumed that $\kappa_{s+t}=\kappa_s\kappa_t$ (everywhere). Kallenberg, for example, is explicitly mentioning that he's assuming this on page 143 of his book (Foundations of mondern probability, 2nd edition). It's obvious that this won't hold. So, the question is why it's legitimate to assume it. – 0xbadf00d Feb 18 at 15:19