im looking at the holding times for a markov process where c(.,.) are the transition rates. My question comes at the end of this working, can you explain when the sentance at the end means? I get that this quality (product to sum) holds for the exponential function but is it unique to the exponential function?
Let
$$\tau_{x}:=\inf\{t\geq 0 \hspace{1.5mm} : \hspace{1.5mm} \omega_{t} \neq x \}$$
then if $\sum_{y\in \Omega}c(x,y)>0$ it turns out that $\tau_{x}\sim Exp(\sum_{y\in \Omega}c(x,y)) $. Proof:\
Since our process is Markov it only cares about its current state and not how much time it has spent in its current state. i.e the holding time $\tau$ has the 'loss of memory property $$P^{x}(\tau_{x}>s+t \hspace{1.5mm} | \hspace{1.5mm} \tau_{x}>s)=P^{x}(\tau_{x}>s+t \hspace{1.5mm} | \hspace{1.5mm} \omega_{s}=x)=P^{x}(\tau_{x}>t).$$
Hence $$P^{x}(\tau_{x}>s+t)=P^{x}(\tau_{x}>s+t|\tau_{x}>s)P^{x}(\tau_{x}>s)=P^{x}(\tau_{x}>t)\cdot P^{x}(\tau_{x}>s)$$
???? Then this is the functional equation for the exponential and implies that $P^{x}(\tau_{x}>t)=e^{\lambda t}$ ????