This is a rather long question, I will make some notation clear before I post the question. Some of the assumptions is from a book and is not mine.
We assume that we have a continuous time Markov chain. That is, we have a finite number of states we can be in.
Assumption 1: We assume that for any states i and j and any times t and t+s where $s\ge 0$, the conditional probability $P(Y(t+s)=j|Y(t)=i)$ is well-defined in the sense that its value does not depend on any information about the process before time t.
Assumption 2: We assume that for any positive time interval h:
P(Two or more transitions within a time period of length h)=o(h).
That a function $f$ is $o(h)$ means that $\lim\limits_{h\rightarrow 0}\frac{f(h)}{h}=0$.
Definition 1:
$_tp_x^{ij}=P(Y(x+t=j)|Y(x)=i)$
Definition 2:
$\mu_x^{ij}=\lim\limits_{h \rightarrow 0^+}\frac{_hp_x^{ij}}{h}$ for $i\ne j$.
Assumption 3: For all states i and j and all $x \ge 0$ , we assume that $_tp_x^{ij}$ is a differentiable function in t.
Lemma 1: $_hp_x^{ij}=\mu_x^{i,j}h+o(h)$. Proof: Omitted
Now I am ready for my question. Assume that you are aged x right now. Assume also that you are in state $i$ now and that up to time $T$ you get 1 unit of money if you transfer from state $j$ to $k$. I want to show that the expted value of this payout is:
$${\int\limits_0^{T}} {}_tp_x^{ij}{ \cdot}\mu_{x+t}^{jk}dt$$
What I am having trouble with is showing that some funcitons go to zero. My book has a rather non-stringent way of showing this quantity by saying that the probability of 2 or more transitions in a timeinterval dt is negligible, but I am not able to show that it is also negligible when we integrate.
I will now show my attempt at the sollution.
My attempt: We will create a Riemann sum and let the norm of the partition go to zero.
Assume that you have a sequence of partitions $\{\Pi_n\}$ of $[0,T]$ where the norm of the partition goes to zero. For a partition $\Pi_f$ the expected value of the quantitiy is:
$E\bigg[\sum\limits_{l=0}^{N(f)-2}I(\text{transition from j to k in the interval }[t_l^{(f)},t_{l+1}^{(f)}])\Bigg]=\sum\limits_{l=0}^{N(f)-2}E\bigg[I(\text{transition from j to k in the interval }[t_l^{(f)},t_{l+1}^{(f)}])\Bigg]=$
$\sum\limits_{l=0}^{N(f)-2}{}_{t_l^{(f)}}p_x^{ij}\cdot{}_{\Delta t_l^{(f)}}p_{x+t_{l+1}^{(f)}}^{jk}=$
$\sum\limits_{l=0}^{N(f)-2}{}_{t_l^{(f)}}p_x^{ij}\cdot{}\left(\mu_{x+t_l^{(f)}}^{jk}\Delta t_l^{(f)}+o(\Delta t_l^{(f)})\right)$.
What I am having problems with is showing that the quantity $\sum\limits_{l=0}^{N(f)-2}{}_{t_l^{(f)}}p_x^{ij}\cdot{}o(\Delta t_l^{(f)})$
goes to zero.
Here are some issues I have.
When $f$ goes to infinity we get new $o$ functions for each partition. These also depend on what time the time-points in the partition is. So when we get new partitions the new functions might not be that small?
If we define ${}_tp_x^{ij,2}=P(Y(x+t=j)|Y(x)=i \text{ and 2 or more transitions take place in the interval [x,x+t]})$ we then get
$\sum\limits_{l=0}^{N(f)-2}{}_{t_l^{(f)}}p_x^{ij}\cdot{}o(\Delta t_l^{(f)})=\sum\limits_{l=0}^{N(f)-2}{}_{t_l^{(f)}}p_x^{ij}\cdot{}_{\Delta t_l^{(f)}}p_{x+t_l^{(f)}}^{j,k,2}$. Is it easier showing that this quanity goes to zero?
So what I have trouble with is showing the formula propery and showing that indeed the quantity goes to zero. Do we need any more assumptions than we had in the start? Do we need some uniformly boundness on the o-functions? Can you please help me solve this problem?