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Recall the classic Chapman-Kolmogorov equation for a Markov process $(X_t)_t$: if $p(s,x;t,I)=\mathbb{P}(X_t \in I | X_s = x)$ is the transition probability (where $I$ is a Borel set), then $$ p(s,x;t,I)=\int_{\mathbb{R}} p(s,x;r,dz)p(r,z;t,I) \qquad (*) $$

I am asked to rewrite the equation $(*)$ in the case that the probability measure $I \rightarrow \mathbb{P}(X_t \in I | X_s=x)$ admits density, that is $\mathbb{P}(X_t \in I | X_s=x) = \int_{I} p(s,x;t,y) \, dy$. I already know that the result should be: $$ p(s,x;t,y)=\int_{\mathbb{R}} p(s,x;z,u)p(z,u;t,y) \, du \qquad (**) $$

Back to the proof, in these hypothesis, the LHS of $(*)$ is just $\int_{I} p(s,x;t,y) \, dy$, while the RHS is:

$$ \int_{\mathbb{R}} p(s,x;r,dz)p(r,z;t,I)=\int_{\mathbb{R}} p(s,x;r,dz) \int_{I} p(r,z;t,y) \, dy $$

I'm pretty sure that the conclusion relies on the fact that we should have two integrals that are equal for every Borel set $I$, hence the argument maps coincide. By the way, I cannot see the right way to "get rid" of that $dz$ as an argument of $p$ (the notation of a transition probability with a differential variable is already unclear to me) and/or switch integration order (probably Fubini will suffice for that). Thanks in advance.

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  • $\begingroup$ Are you sure of eqn. $(**)$? it seems to me that you should be integrating over space ($u$ in this case) not time.. $\endgroup$ – Kore-N Feb 11 '17 at 13:50
  • $\begingroup$ Soon checking, thanks for pointing out the doubt! $\endgroup$ – GaC Feb 11 '17 at 13:55
  • $\begingroup$ You are right, we must integrate over space, fixed! Can you provide a proof of $(**)$? $\endgroup$ – GaC Feb 11 '17 at 14:06
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We can easily show that $(**)$ follows from $(*)$ under the assumption that the transition probabilities admit a density w.r.t the Lebesgue measure. All we need to do is rewrite $(*)$ in the following form (here we slightly change notation: let $A$ be any borel subset of $\mathbb{R}$ - just because $I$ reminds me so much of an interval. Furthermore, just for completeness, let $s \le r \le t$):

$$ \int_A p(s,x; \ t, y)dy = p(s,x; \ t,A) = \int_{\mathbb{R}}\left(p(s,x; \ r,z) \int_{A}p(r,z; \ t,y)dy\right)dz$$

Now we exchange the integral signs and we find that for any Borel $A \subset \mathbb{R}$:

$$\int_A p(s,x; \ t, y)dy = \int_A \int_{\mathbb{R}}\left(p(s,x; \ r,z) p(r,z; \ t,y)dz\right)dy$$

Now it is assumed to be of common knowledge that if $$\int_A f = \int_A g$$ holds for all subsets $A$, then the two functions coincide almost surely. Hence you get $(**)$ - if you want to get rid of the a.s. you can argue that densities are actually defined almost surely.

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