I am having difficulty with the Chapman-Kolmogorov equation on pg 16 Chapter 4 of the course notes concerning Markov jump processes. The student is advised to use (6.5) which is straightforward but the second step using equation (6.7) in not clear to me. Clearly the summation is split for k=j and k not equal to j but i am not sure how the final equation on the RHS is obtained.
I've only just spotted your question but I was also bothered by this point. I think the second part of the expression makes sense for the k not equal to j case, but the first part isn't clear to me. Would any of the tutors be able to respond please?
If we take pij(t+s) and rewrite it using the suggested substitutions, we get: pij(h+t)=pi1(t)p1j(h) + pi2(t)p2j(h) + ... For all these terms except where k=j, we can write p1j(h) = h sigma 1j, p2j(h) = h sigma 2j, and so on. This gives us the summation term on the right hand side. When k=j, we write pjj(h) = 1+h sigma jj. This tends to 1 as h tends to zero. So this gives us the pij(t) term.
