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Let $(X_t)_{t\in\Bbb R_{\ge0}}$ be a random process on a countable state space $I$.

For each state $i_{n-1}\in I$ let $E_1,E_2,...$ be independent random variables s.t. $E_j \sim Exp(\lambda_{i_{n-1},j})$ models the time to pass from state $i_{n-1}$ to $j$.

At each step the shortest $E_{i_{n-1},j}$ "wins the race" and the process goes from state $i_{n-1}$ to the state $j$ that corresponds to the winner.

My question is: why do we have $\Bbb P(X(T_n)=j\mid X(T_{n-1})=i)=\frac{\lambda_{i,j}}{\sum\limits_{k\ne i}\lambda_{k,j}}$, where $T_n$ are the jumping times?

YuiTo Cheng
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John Cataldo
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1 Answers1

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Please note, that there must be some conditions on $\lambda_{ij}$, which were not stated. For example, $\sum_{j} \lambda_{ij} < \infty$ for each $i$.

According to me, the properties of the exponential distribution suggests the answer to our question. Assume that you are in the state $i_{n-1}$. In this state you wait time = minimum time until the next transition. Since all transitions are exponentially dsitributed, minimum has also exponential distribution with the parameter $\Lambda_{i_{n-1}}=\sum_{j, j\neq i_{n-1}} \lambda_{i_{n-1},j}$. Then the probability that this minimal time is equal to the the time to jump to $i_n$ is equal to $\lambda_{i_{n-1},i_n}/\Lambda_{i_{n-1}}$. There are several ways to show this. You can find strict proofs in many books. For example, see Lemma 1.2.3 in this book (P. P. Bocharov, C. D'Apice, A. V. Pechinkin. Queueing Theory. Modern Probability and Statistics. Walter de Gruyter, 2011.)

rrv
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