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Soujourn time of a CTMC at time $t$ is defined as :

$$T(t)= \inf\{ s > 0 : X(t+s) \neq X(t)\}$$

My question is why "inf", not min ? Here $T(t)$ belongs to the set $\{ s > 0 : X(t+s) \neq X(t)\}$. Then we can write minimum. Is this correct ?

Martin Sleziak
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1 Answers1

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If your chain is known to be right continuous, then indeed you can replace the inf with min. Otherwise you cannot. Let $S \sim \mathrm{Exp}(\lambda)$ be an exponential random variable, and consider the chain on the state space $\{a,b\}$ defined by $$X(t) = \begin{cases} a, & t \le S \\ b, & t > S. \end{cases}$$ (This chain starts at $a$, jumps from $a$ to $b$ at rate $\lambda$, and from $b$ to $a$ at rate 0.) We have $T(0) = S$, but the infimum is not achieved since $X(S) = X(0)=a$.

Nate Eldredge
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  • understood, nice explanation –  Mar 30 '13 at 07:02
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    Actually, one must use an infimum for càdlàg processes as well, to take into account the event that $X(t)$ stays forever at a same site. In most cases this has probability zero but, to define $T(t)$ on the whole probability space one needs to use an infimum. (Or, one could define $T(t)$ only on a subset of probability $1$ of the probability space but this choice becomes awkward as soon as one considers the process $(T(t))_t$, indexed by an uncountable set, hence nobody does this.) – Did Mar 30 '13 at 10:32