Hi everyone I'm trying to prove the following theorem about continous time, finite Markov's chains:
Let $J=\min\left\{t:X_{t}\neq i\right\}$. Under the law $\mathbb{P}_{i}$ of the Markov chain started in $X_{0}=i$ the random variables $J$ and $X(J)$ are independent. The distribution of $J$ is exponential with rate $q_{i}:=\sum_{j\neq i}q_{ij}$, i.e. $$\forall t>0;\,\,\,\,\,\,\,\,\mathbb{P}_{i}\left\{J>t\right\}=e^{-q_{i}t}.$$ Moreover, $$\mathbb{P}_{i}\left\{X(J)=j\right\}=\frac{q_{ij}}{q_{i}}$$.
The probability law $\mathbb{P}_{i}$ used in the theorem is simply defined in the following way: $$\forall\,A\in\mathcal{F};\,\,\,\,\,\,\mathbb{P}_{i}\left\{A\right\}:=\mathbb{P}\left\{A|X(0)=i\right\}.$$
What one could observe is that every Markov's process of this form can be considered a Poisson process by simply considering a one to one function between the state set $I$ and the natural numbers, so every state is the number of occurences. But how can I prove that the sojourn time in the first state is exponentially distributed and how can the rate be exactly $q_{i}$?