Consider a Continuous-Time Markov Chain (CTMC) with $P_{i,i+1}=1$ and $v_i=i^2$. Here, $P_{i,i+1}$ is the probability of jumping from state $i$ to state $i+1$, and $v_i$ is the rate of holding times. We know a regular CTMC is defined to be a CTMC such that with probability $1$, the number of jumps in any finite time interval is finite. Prove that the above CTMC is not regular.
I appreciate any comment/hint.
my attempt: I want to use the fact that the holding time in state $i$, denoted by $\tau_i$ has an exponential distribution with rate $v_i$. Now, consider the interval $[0,t]$. The probability that the number of jumps is finite (say less than a finite number $n$) is equivalent to the probability that that $\sum_{i=1}^n \tau_i > t$. Now, $\sum_{i=1}^n \tau_i$ is the sum of exponential random variables with rates $v_i$. I need to find the pdf of $\sum_{i=1}^n \tau_i$. I do not know how to proceed.
Best Answer
As suggested by John Dawkins, we can show that $\sum_{i=1}^\infty \tau_i < \infty$. Note that \begin{align} \mathbb{E}[\sum_{i=1}^\infty \tau_i] = \sum_{i=1}^\infty \mathbb{E}[\tau_i] = \sum_{i=1}^\infty \frac{1}{i^2} = \frac{\pi^2}{6} < \infty. \end{align} Since the expectation $\mathbb{E}[\sum_{i=1}^\infty \tau_i] $ is finite, we have $\mathbb{P}(\sum_{i=1}^\infty \tau_i < \infty) = 1$, and this completes the proof.