Inhomogeneous Markov chains and the product-integral as a solution to the Kolmogorov forward equation

Question

We have a inhomogeneous continous $K$-State Markov chain $X(t)$ with transition intensity matrix $Q(t)$. Therefore its entries are: $$q_{ij}(t)= \lim_{\delta \to 0} \frac{1}{\delta} \mathbb{P}(X(t+\delta) = j | X(t) = i).$$ The transistion probabilities are in a matrix $P(s,t)$ with entries $$P_{ij}(s,t) = \mathbb{P}(X(t) = j | X(s) = i).$$

If we use the Chapman-Kolmogorov equation we can get the Kolmogorov forward equation as $$\frac{d}{dt}P(s,t) = P(s,t)Q(t)$$ This can also be written as \begin{align} P(s,t) &= I+\int_s^tP(s,u-)Q(u)du \\ &= I+\int_s^tP(s,u-)dA(u), \end{align} where $I$ is the identity matrix and $A(u)$ the cumulative transition intensities.

Now every literature states that one can find a solution to this Integral as the product-limit $$P(s,t) = \prod_{u \in (s,t]}\bigl(I+dA(u)\bigr).$$ I don´t see why this is true and how someone can derive this product-limit of the integral representation from the Kolmogorov forward equation.

Answer 1

I see it as follows: $$\frac{\partial}{\partial t}P(s,t)=P(s,t)Q(t) \rightarrow P(x,y+\Delta y)=P(x,y)+P(x,y)Q(y)\Delta y+o(\Delta y)\ ,$$ so we also have $$P(x,y+\Delta y)=P(x,y)[I+Q(y)\Delta y]+o(\Delta y) \rightarrow P(y,y+\Delta y)=I+Q(y)\Delta y+o(\Delta y)\ .$$Now, as $P(a,c)=P(a,b)P(b,c)$ whenever $a\le b \le c$, if we divide the interval between $s$ and $t$ in $N$ equal intervals with amplitude $\Delta t$ we can write $$P(s,t)=P(s,s+\Delta t)P(s+\Delta t,s+2\Delta t)\ldots P(t-\Delta t,t)=\\ [I+Q(s)\Delta t +o(\Delta t)][I+Q(s+\Delta t)\Delta t +o(\Delta t)]\ldots [I+Q(t-\Delta t)\Delta t +o(\Delta t)]= \\ = [I+Q(s)\Delta t][I+Q(s+\Delta t)\Delta t]\ldots [I+Q(t-\Delta t)\Delta t)]+N\; o(\Delta t) .$$Since $\Delta t=(s-t)/N$ the quantity $N\; o(\Delta t)=o(s-t)$ tends to zero as $\Delta t$ tends to zero, and so we can write: $$ P(s,t)=\lim_{N\to\infty}\prod_{k=0}^{N-1}[I+Q(s+k\Delta t)\Delta t]$$where $\Delta t=(s-t)/N$ as usual, which is, I think, equivalent to $$P(s,t) = \prod_{u \in (s,t]}\bigl(I+dA(u)\bigr).$$ It is not difficult to prove, starting from the productorial formula, that, if $Q(x)Q(y)=Q(y)Q(x)$ for every $x$ and $y$ in $(s,t)$ then we also have $$P(s,t)=\exp\left[\int_s^t Q(u)du\right]\ .$$