TITLE: How to understand the rate in a Continuous Time Markov Chain.
QUESTION [3 upvotes]: Suppose there is a CTMC with three status $\{0,1,2\}$, the rate of transition $1\to2$ is $p$, and the rate of $1\to0$ is $q$.  I know that it means the time of the status stay in $1$ before transfer to $2$ is exponentially distributed with parameter $p$, and if transfer to $0$, the parameter is $q$. 
My problem is, why $p$ and $q$ also implies the probability of the status transfer from $1\to 0$ or $1\to2$ ?  For example, the probability of $1$ transfer to $0$ is $\frac{q}{p+q}$.
If it isn't, what is the relation between the time of a status last and the probability of the transition?
Thanks so much!

REPLY [0 votes]: Let $X_1 \sim \exp(p)$ and let $Y_1 \sim \exp(q)$ be independent random variables,
then
$$P(X_1 \leq Y_1) =  \iint_{x<y} pe^{-xp}q e^{-yq} \, dx \, dy= \int_0^\infty \int_0^y pe^{-xp} \, dx q e^{-yq}\, dy = \int_0^\infty  (1 - e^{-yp}) q e^{-yq}\, dy = 1 - \frac{q}{p+q} = \frac{p}{q+p}$$
So you are right when you say  that the probability of 1 transfer to 0 is $\frac{q}{p+q}$. and the probability of 1 transfer to 2 is $\frac{p}{p+q}$
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